How can I write a piece of Java code that checks if a string is convertible to an integer, that is if any character of it is convertible to an integer between 0 and 9 ? I thought of something like this:
String s = "...";
for (int i = 0; i < s.length(); i++)
{
int h = Integer.parseInt(s.substring(i, i + 1));
if (h < 0 || h > 9)
throw new IntegerFormatException();
}
catch (IntegerFormatException e)
{
System.out.println("This is not an integer");
}
where IntegerFormat Exception is something like
public class IntegerFormatException extends Exception
{
public IntegerFormatException()
{
super ("This string isn't convertible to an integer");
}
public IntegerFormatException(String message)
{
super (message);
}
}
But if I then try the code with, say, "8&35" as the string I don't get my message "This is not an integer", but I get and IDE automatic red ink message stating
Exception in thread "main" java.lang.NumberFormatException: For input string: "&"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:569)
at java.lang.Integer.parseInt(Integer.java:615)
at project.pkg9.pkg2.Project92.main(Project92.java:60)
Java Result: 1
Is my code wrong ? If yes, how can I fix it ? Many thanks in advance for all your answers.
There is already a run time exception for that, use it!
String s = "8&35";
try {
int result = Integer.parseInt(s);
System.out.println("This is parsable: " + result);
} catch (NumberFormatException e) {
System.err.println("s is NOT parsable....");
}
That's because Integer.parseInt will throw a NumberFormatException exception if you try to parse something that's not an int, and you don't catch that in your try/catch block. You could catch that in your catch block or use a regex to check what the character is at that location before you try to parse (or, as pointed out in the comments, Character.isDigit()).
Also, there's no need to see if a single digit is less than 0 or greater than 9 - how could a single digit possibly be anything but an integer in the range of 0 - 9? This won't really tell you if the character is a valid int or not because, unless you're comparing the ASCII values, it doesn't make sense to talk about whether & is greater than 9, for example.
I would do it with a Regex :
if(s.matches("[0-9]"))
{
//it's an Integer,
//execute logic
}
Instead of looping you can use regex. This will work and would throw the IntegerFormatException for an empty string as well.
String s = "...";
if (!s.matches("[0-9]+")) {
throw new IntegerFormatException();
}
It won't cover Integer type capacity overflow - I mean a huge number above Integer.MAX_VALUE. Not sure do you need that though.
Integer.parseInt(s.substring(i, i + 1)); throw an exception by itself.
In particular, it throws the java.lang.NumberFormatException: which is not caught by your catch.
If you really need to use your exception (there is already one so why not to use that one?) use the following instead:
public static void main(String[] args) {
String s = "1234/&&56";
try{
for (int i = 0; i < s.length(); i++)
{
char c = s.charAt(i);
int h = c-'0';
if (h < 0 || h > 9)
throw new IntegerFormatException();
}
}catch(IntegerFormatException e)
{
System.out.println("This is not an integer");
}
}
Note that you should also include the check for the very first character (it could be the - sign) in the case you are trying to parse signed numbers.
try this
String regex = "^-?[0-9]+$";
if (s.matches(regex) {
// it's a number
} else {
// not a number
}
You should refer to the documentation for the Integer.parseInt(String s) method. The full header for this method is public static int valueOf(String s) throws NumberFormatException and the documentation states that the NumberFormatException is thrown when the string does not contain a parsable integer.
To obtain the functionality you described, you should consider that passing a single char into the function will always result in an integer from 0 to 9 (inclusive) if the char is parsable to an int, so, your if statement is probably unecessary. To make your implementation work, you should catch the NumberFormatException and use that to determine if the char parsed correctly.
Also, to simplify your code, you can also use the s.charAt(i) method rather than s.substring(i, i + 1). You might also want to look into regex to obtain the behavior you explained.
You can use Character class isdigit api or other api to test if it is digit or not
String s = "1234/&&56";
for(int i=0;i<s.length();i++){
char ch=s.charAt(i);
if((Character.isDigit(ch))){
System.out.println(ch);
}
}
The function chr($i) returns a one-character string that contains the character equivalent of $i, where $i is an integer between 0 and 255.
CS student here. I want to write a program that will decompress a string that has been encoded according to a modified form of run-length encoding (which I've already written code for). For instance, if a string contains 'bba10' it would decompress to 'bbaaaaaaaaaa'. How do I get the program to recognize that part of the string ('10') is an integer?
Thanks for reading!
A simple regex will do.
final Matcher m = Pattern.compile("(\\D)(\\d+)").matcher(input);
final StringBuffer b = new StringBuffer();
while (m.find())
m.appendReplacement(b, replicate(m.group(1), Integer.parseInt(m.group(2))));
m.appendTail(b);
where replicate is
String replicate(String s, int count) {
final StringBuilder b = new StringBuilder(count);
for (int i = 0; i < count; i++) b.append(s);
return b.toString();
}
Not sure whether this is one efficient way, but just for reference
for (int i=0;i<your_string.length();i++)
if (your_string.charAt(i)<='9' && your_string.charAt(i)>='0')
integer_begin_location = i;
I think you can divide chars in numeric and not numeric symbols.
When you find a numeric one (>0 and <9) you look to the next and choose to enlarge you number (current *10 + new) or to expand your string
Assuming that the uncompressed data does never contain digits: Iterate over the string, character by character until you get a digit. Then continue until you have a non-digit (or end of string). The digits inbetween can be parsed to an integer as others already stated:
int count = Integer.parseInt(str.substring(start, end));
Here is a working implementation in python. This also works fine for 2 or 3 or multiple digit numbers
inputString="a1b3s22d4a2b22"
inputString=inputString+"\0" //just appending a null char
charcount=""
previouschar=""
outputString=""
for char in inputString:
if char.isnumeric():
charcount=charcount+char
else:
outputString=outputString
if previouschar:
outputString=outputString+(previouschar*int(charcount))
charcount=""
previouschar=char
print(outputString) // outputString= abbbssssssssssssssssssssssddddaabbbbbbbbbbbbbbbbbbbbbb
Presuming that you're not asking about the parsing, you can convert a string like "10" into an integer like this:
int i = Integer.parseInt("10");
Im in highschool and this is an assignment i have, you guys are out of my league but im willing to learn and understand. I looked all over the place but all i could find was complicated syntax i dont know yet. This is what i have, it takes a String and reverses it. I managed to get it to ignore Capitals, but i cannot figure out how to make it ignore symbols. The numbers i have there are from the ANSI Characters, there is a list on textpad im using. Dont be afraid to be harsh, im not good at this and i only want to improve so have at it.
import java.util.Scanner;
public class PalindromeV2
{
public static void main(String[] args)
{
//declare
Scanner sc = new Scanner(System.in);
String fwd, rev;
String result;
//input
System.out.println("What word would you like to Palindrome test?");
fwd = sc.next();
rev = reverseString(fwd);
result = stripPunctuation(fwd);
if(stripPunctuation(rev).equals(stripPunctuation(fwd)))
{
System.out.println("That is a palindrome");
}
else
System.out.println("That is not a palindrome");
}
public static String reverseString(String fwd)
{
String rev = "";
for(int i = fwd.length()-1; i >= 0; i--)
{
rev += fwd.charAt(i);
}
return rev.toUpperCase();
}
public static String stripPunctuation(String fwd)
{
String result = "";
fwd = fwd.toUpperCase();
for(int i = fwd.length()-1; i >= 0; i--)
{
if((fwd.charAt(i)>=65 && fwd.charAt(i)<=90)||(fwd.charAt(i) >= 48 && fwd.charAt(i) <= 58));
result = result + fwd.charAt(i);
}
return result;
}
}
You can use this as a checking condition
if (Character.isLetter(fwd.charAt(i)) {
// do something
}
This will check to make sure the character is a letter, so you don't have to worry about case, numbers, or other symbols.
If you want to strip your string out of some set of characters than do something like that
clearString=targetStringForStripping.replaceAll([type_characters_for_stripping],"");
this will remove all characters you will provide inside square brackets.
There is even more. If you want to let say leave only letters (because in palindromes nothing matters except letters - spaces are not important to) than you simply can use predefine character set - letters.
To conclude all if you do
clearString=targetStringForStripping.replaceAll("[\w]","");
or
clearString=targetStringForStripping.replaceAll("[^a-zA-Z]","");
you will get clear string with white characters in first example, and only letters in second one. Perfect situation for isPalindrom resolution.
if((fwd.charAt(i)>=65 && fwd.charAt(i)<=90)||(fwd.charAt(i) >= 48 && fwd.charAt(i) <= 58));
you have semicolon at last. so i think if condition is no use here
Since this is a highschool assignment, I'll just give some pointers, you'll figure it out on your own.
Think about what you want to include / exclude, then write the code.
Keep in mind, that you can compare char variables using < or > operators as long as you do not want to handle complex character encodings.
A String is really just a sequence of chars which one by one you can compare or reorder, include or exclude.
A method should only do one thing, not a lot of things. Have a look at your reverseString method. This is doing an toUpperCase to your string at the same time. If your programs get more complex, this way of doing things is not to easy to follow.
Finally, if you e.g. just want to include capital letters in your palindrome check, then try some code like this:
char[] toCheck = fwd.toCharArray();
for (char c : toCheck) {
if (c >= 'A' && c <= 'Z') {
result = result + c;
}
}
Depending on your requirements this might do what you want. If you want something different, have a look at the hints I gave above.
Java golf?
public static String stripPunctuation(String stripThis) {
return stripThis.replaceAll("\\W", "");
}
I need to implement a very crude language identification algorithm. In my world, there are only two languages: English and not-English. I have ArrayList and I need to determine if each String is likely in English or the other language which has its Unicode chars in a certain range. So what I want to do is to check each String against this range using some type of "presence" test. If it passes the test, I say the String is not English, otherwise it's English. I want to try two type of tests:
TEST-ANY: If any char in the string falls within the range, the string passes the test
TEST-ALL: If all chars in the string fall within the range, the string passes the test
Since the array might be very long, I need to implement this very efficiently. What would be the fastest way of doing this in Java?
Thx
UPDATE: I am specifically checking for non-English by looking at a specific range of Unicodes rather then checking for whether the characters are ASCII, in part to take care of the "resume" problem mentioned below. What I am trying to figure out is whether Java provides any classes/methods that essentially implement TEST-ANY or TEST-ALL (or another similar test) as efficiently as possible. In other words, I am trying to avoid reinventing the wheel especially if the wheel invented before me is better anyway.
Here's how I ended up implementing TEST-ANY:
// TEST-ANY
String str = "wordToTest";
int UrangeLow = 1234; // can get range from e.g. http://www.utf8-chartable.de/unicode-utf8-table.pl
int UrangeHigh = 2345;
for(int iLetter = 0; iLetter < str.length() ; iLetter++) {
int cp = str.codePointAt(iLetter);
if (cp >= UrangeLow && cp <= UrangeHigh) {
// word is NOT English
return;
}
}
// word is English
return;
I really don't think that this solution is ideal for determining language, but if you want to check to see if a string is all ascii, you could do something like this:
public static boolean isASCII(String s){
boolean ret = true;
for(int i = 0; i < s.length() ; i++) {
if(s.charAt(i)>=128){
ret = false;
break;
}
}
return ret;
}
So then if you try this:
boolean r = isASCII("Hello");
r would equal true. But if you try:
boolean r = isASCII("Grüß dich");
then r would equal false. I haven't tested performance, but this would work reasonably fast, because all it does is compare a character to the number 128.
But as #AlexanderPogrebnyak mentioned in the comments above, this will return false if you give it "résumé". Be aware of that.
Update:
I am specifically checking for non-English by looking at a specific range of Unicodes rather then checking for whether the characters are ASCII
But ASCII is a range in Unicode (well at least in UTF-8). Unicode is just an extension of ASCII. What the code #mP. and I provided does is it checks to see whether each character is in a certain range. I chose that range to be ASCII, which is any Unicode character that has a decimal value of less than 128. You can just as well choose any other range. But the reason I chose ASCII is because it's the one with the Latin alphabet, the Arabic numbers, and some other common characters that would normally be in an 'English' string.
public static boolean isAscii( String s ){
int length = s.length;
for( int i = 0; i < length; i++){
final char c = s.charAt( i );
if( c > 'z' ){
return false;
}
}
return true;
}
#Hassan thanks for picking the typo replaced test against big Z with little z.
I know variants of this question have been asked frequently before (see here and here for instance), but this is not an exact duplicate of those.
I would like to check if a String is a number, and if so I would like to store it as a double. There are several ways to do this, but all of them seem inappropriate for my purposes.
One solution would be to use Double.parseDouble(s) or similarly new BigDecimal(s). However, those solutions don't work if there are commas present (so "1,234" would cause an exception). I could of course strip out all commas before using these techniques, but that would seem to pose loads of problems in other locales.
I looked at Apache Commons NumberUtils.isNumber(s), but that suffers from the same comma issue.
I considered NumberFormat or DecimalFormat, but those seemed far too lenient. For instance, "1A" is formatted to "1" instead of indicating that it's not a number. Furthermore, something like "127.0.0.1" will be counted as the number 127 instead of indicating that it's not a number.
I feel like my requirements aren't so exotic that I'm the first to do this, but none of the solutions does exactly what I need. I suppose even I don't know exactly what I need (otherwise I could write my own parser), but I know the above solutions do not work for the reasons indicated. Does any solution exist, or do I need to figure out precisely what I need and write my own code for it?
Sounds quite weird, but I would try to follow this answer and use java.util.Scanner.
Scanner scanner = new Scanner(input);
if (scanner.hasNextInt())
System.out.println(scanner.nextInt());
else if (scanner.hasNextDouble())
System.out.println(scanner.nextDouble());
else
System.out.println("Not a number");
For inputs such as 1A, 127.0.0.1, 1,234, 6.02e-23 I get the following output:
Not a number
Not a number
1234
6.02E-23
Scanner.useLocale can be used to change to the desired locale.
You can specify the Locale that you need:
NumberFormat nf = NumberFormat.getInstance(Locale.GERMAN);
double myNumber = nf.parse(myString).doubleValue();
This should work in your example since German Locale has commas as decimal separator.
You can use the ParsePosition as a check for complete consumption of the string in a NumberFormat.parse operation. If the string is consumed, then you don't have a "1A" situation. If not, you do and can behave accordingly. See here for a quick outline of the solution and here for the related JDK bug that is closed as wont fix because of the ParsePosition option.
Unfortunately Double.parseDouble(s) or new BigDecimal(s) seem to be your best options.
You cite localisation concerns, but unfortunately there is no way reliably support all locales w/o specification by the user anyway. It is just impossible.
Sometimes you can reason about the scheme used by looking at whether commas or periods are used first, if both are used, but this isn't always possible, so why even try? Better to have a system which you know works reliably in certain situations than try to rely on one which may work in more situations but can also give bad results...
What does the number 123,456 represent? 123456 or 123.456?
Just strip commas, or spaces, or periods, depending on locale specified by user. Default to stripping spaces and commas. If you want to make it stricter, only strip commas OR spaces, not both, and only before the period if there is one. Also should be pretty easy to check manually if they are spaced properly in threes. In fact a custom parser might be easiest here.
Here is a bit of a proof of concept. It's a bit (very) messy but I reckon it works, and you get the idea anyways :).
public class StrictNumberParser {
public double parse(String numberString) throws NumberFormatException {
numberString = numberString.trim();
char[] numberChars = numberString.toCharArray();
Character separator = null;
int separatorCount = 0;
boolean noMoreSeparators = false;
for (int index = 1; index < numberChars.length; index++) {
char character = numberChars[index];
if (noMoreSeparators || separatorCount < 3) {
if (character == '.') {
if (separator != null) {
throw new NumberFormatException();
} else {
noMoreSeparators = true;
}
} else if (separator == null && (character == ',' || character == ' ')) {
if (noMoreSeparators) {
throw new NumberFormatException();
}
separator = new Character(character);
separatorCount = -1;
} else if (!Character.isDigit(character)) {
throw new NumberFormatException();
}
separatorCount++;
} else {
if (character == '.') {
noMoreSeparators = true;
} else if (separator == null) {
if (Character.isDigit(character)) {
noMoreSeparators = true;
} else if (character == ',' || character == ' ') {
separator = new Character(character);
} else {
throw new NumberFormatException();
}
} else if (!separator.equals(character)) {
throw new NumberFormatException();
}
separatorCount = 0;
}
}
if (separator != null) {
if (!noMoreSeparators && separatorCount != 3) {
throw new NumberFormatException();
}
numberString = numberString.replaceAll(separator.toString(), "");
}
return Double.parseDouble(numberString);
}
public void testParse(String testString) {
try {
System.out.println("result: " + parse(testString));
} catch (NumberFormatException e) {
System.out.println("Couldn't parse number!");
}
}
public static void main(String[] args) {
StrictNumberParser p = new StrictNumberParser();
p.testParse("123 45.6");
p.testParse("123 4567.8");
p.testParse("123 4567");
p.testParse("12 45");
p.testParse("123 456 45");
p.testParse("345.562,346");
p.testParse("123 456,789");
p.testParse("123,456,789");
p.testParse("123 456 789.52");
p.testParse("23,456,789");
p.testParse("3,456,789");
p.testParse("123 456.12");
p.testParse("1234567.8");
}
}
EDIT: obviously this would need to be extended for recognising scientific notation, but this should be simple enough, especially as you don't have to actually validate anything after the e, you can just let parseDouble fail if it is badly formed.
Also might be a good idea to properly extend NumberFormat with this. have a getSeparator() for parsed numbers and a setSeparator for giving desired output format... This sort of takes care of localisation, but again more work would need to be done to support ',' for decimals...
Not sure if it meets all your requirements, but the code found here might point you in the right direction?
From the article:
To summarize, the steps for proper input processing are:
Get an appropriate NumberFormat and define a ParsePosition variable.
Set the ParsePosition index to zero.
Parse the input value with parse(String source, ParsePosition parsePosition).
Perform error operations if the input length and ParsePosition index value don't match or if the parsed Number is null.
Otherwise, the value passed validation.
This is an interesting problem. But perhaps it is a little open-ended? Are you looking specifically to identify base-10 numbers, or hex, or what? I'm assuming base-10. What about currency? Is that important? Or is it just numbers.
In any case, I think that you can use the deficiencies of Number format to your advantage. Since you no that something like "1A", will be interpreted as 1, why not check the result by formatting it and comparing against the original string?
public static boolean isNumber(String s){
try{
Locale l = Locale.getDefault();
DecimalFormat df = new DecimalFormat("###.##;-##.##");
Number n = df.parse(s);
String sb = df.format(n);
return sb.equals(s);
}
catch(Exception e){
return false;
}
}
What do you think?
This is really interesting, and I think people are trying to overcomplicate it. I would really just break this down by rules:
1) Check for scientific notation (does it match the pattern of being all numbers, commas, periods, -/+ and having an 'e' in it?) -- if so, parse however you want
2) Does it match the regexp for valid numeric characters (0-9 , . - +) (only 1 . - or + allowed)
if so, strip out everything that's not a digit and parse appropriately, otherwise fail.
I can't see a shortcut that's going to work here, just take the brute force approach, not everything in programming can be (or needs to be) completely elegant.
My understanding is that you want to cover Western/Latin languages while retaining as much strict interpretation as possible. So what I'm doing here is asking DecimalFormatSymbols to tell me what the grouping, decimal, negative, and zero separators are, and swapping them out for symbols Double will recognize.
How does it perform?
In the US, it rejects: "1A", "127.100.100.100"
and accepts "1.47E-9"
In Germany it still rejects "1A"
It ACCEPTS "1,024.00" but interprets it correctly as 1.024. Likewise, it accepts "127.100.100.100" as 127100100100.0
In fact, the German locale correctly identifies and parses "1,47E-9"
Let me know if you have any trouble in a different locale.
import java.util.Locale;
import java.text.DecimalFormatSymbols;
public class StrictNumberFormat {
public static boolean isDouble(String s, Locale l) {
String clean = convertLocaleCharacters(s,l);
try {
Double.valueOf(clean);
return true;
} catch (NumberFormatException nfe) {
return false;
}
}
public static double doubleValue(String s, Locale l) {
return Double.valueOf(convertLocaleCharacters(s,l));
}
public static boolean isDouble(String s) {
return isDouble(s,Locale.getDefault());
}
public static double doubleValue(String s) {
return doubleValue(s,Locale.getDefault());
}
private static String convertLocaleCharacters(String number, Locale l) {
DecimalFormatSymbols symbols = new DecimalFormatSymbols(l);
String grouping = getUnicodeRepresentation( symbols.getGroupingSeparator() );
String decimal = getUnicodeRepresentation( symbols.getDecimalSeparator() );
String negative = getUnicodeRepresentation( symbols.getMinusSign() );
String zero = getUnicodeRepresentation( symbols.getZeroDigit() );
String clean = number.replaceAll(grouping, "");
clean = clean.replaceAll(decimal, ".");
clean = clean.replaceAll(negative, "-");
clean = clean.replaceAll(zero, "0");
return clean;
}
private static String getUnicodeRepresentation(char ch) {
String unicodeString = Integer.toHexString(ch); //ch implicitly promoted to int
while(unicodeString.length()<4) unicodeString = "0"+unicodeString;
return "\\u"+unicodeString;
}
}
You're best off doing it manually. Figure out what you can accept as a number and disregard everything else:
import java.lang.NumberFormatException;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class ParseDouble {
public static void main(String[] argv) {
String line = "$$$|%|#|1A|127.0.0.1|1,344|95|99.64";
for (String s : line.split("\\|")) {
try {
System.out.println("parsed: " +
any2double(s)
);
}catch (NumberFormatException ne) {
System.out.println(ne.getMessage());
}
}
}
public static double any2double(String input) throws NumberFormatException {
double out =0d;
Pattern special = Pattern.compile("[^a-zA-Z0-9\\.,]+");
Pattern letters = Pattern.compile("[a-zA-Z]+");
Pattern comma = Pattern.compile(",");
Pattern allDigits = Pattern.compile("^[0-9]+$");
Pattern singleDouble = Pattern.compile("^[0-9]+\\.[0-9]+$");
Matcher[] goodCases = new Matcher[]{
allDigits.matcher(input),
singleDouble.matcher(input)
};
Matcher[] nanCases = new Matcher[]{
special.matcher(input),
letters.matcher(input)
};
// maybe cases
if (comma.matcher(input).find()){
out = Double.parseDouble(
comma.matcher(input).replaceFirst("."));
return out;
}
for (Matcher m : nanCases) {
if (m.find()) {
throw new NumberFormatException("Bad input "+input);
}
}
for (Matcher m : goodCases) {
if (m.find()) {
try {
out = Double.parseDouble(input);
return out;
} catch (NumberFormatException ne){
System.out.println(ne.getMessage());
}
}
}
throw new NumberFormatException("Could not parse "+input);
}
}
If you set your locale right, built in parseDouble will work with commas. Example is here.
I think you've got a multi step process to handle here with a custom solution, if you're not willing to accept the results of DecimalFormat or the answers already linked.
1) Identify the decimal and grouping separators. You might need to identify other format symbols (such as scientific notation indicators).
http://download.oracle.com/javase/1.4.2/docs/api/java/text/DecimalFormat.html#getDecimalFormatSymbols()
2) Strip out all grouping symbols (or craft a regex, be careful of other symbols you accept such as the decimal if you do). Then strip out the first decimal symbol. Other symbols as needed.
3) Call parse or isNumber.
One of the easy hacks would be to use replaceFirst for String you get and check the new String whether it is a double or not. In case it's a double - convert back (if needed)
If you want to convert some string number which is comma separated decimal to double, you could use DecimalSeparator + DecimalFormalSymbols:
final double strToDouble(String str, char separator){
DecimalFormatSymbols s = new DecimalFormatSymbols();
s.setDecimalSeparator(separator);
DecimalFormat df = new DecimalFormat();
double num = 0;
df.setDecimalFormatSymbols(s);
try{
num = ((Double) df.parse(str)).doubleValue();
}catch(ClassCastException | ParseException ex){
// if you want, you could add something here to
// indicate the string is not double
}
return num;
}
well, lets test it:
String a = "1.2";
String b = "2,3";
String c = "A1";
String d = "127.0.0.1";
System.out.println("\"" + a + "\" = " + strToDouble(a, ','));
System.out.println("\"" + a + "\" (with '.' as separator) = "
+ strToDouble(a, '.'));
System.out.println("\"" + b + "\" = " + strToDouble(b, ','));
System.out.println("\"" + c + "\" = " + strToDouble(c, ','));
System.out.println("\"" + d + "\" = " + strToDouble(d, ','));
if you run the above code, you'll see:
"1.2" = 0.0
"1.2" (with '.' as separator) = 1.2
"2,3" = 2.3
"A1" = 0.0
"127.0.0.1" = 0.0
This will take a string, count its decimals and commas, remove commas, conserve a valid decimal (note that this is based on US standardization - in order to handle 1.000.000,00 as 1 million this process would have to have the decimal and comma handling switched), determine if the structure is valid, and then return a double. Returns null if the string could not be converted. Edit: Added support for international or US. convertStoD(string,true) for US, convertStoD(string,false) for non US. Comments are now for US version.
public double convertStoD(string s,bool isUS){
//string s = "some string or number, something dynamic";
bool isNegative = false;
if(s.charAt(0)== '-')
{
s = s.subString(1);
isNegative = true;
}
string ValidNumberArguements = new string();
if(isUS)
{
ValidNumberArguements = ",.";
}else{
ValidNumberArguements = ".,";
}
int length = s.length;
int currentCommas = 0;
int currentDecimals = 0;
for(int i = 0; i < length; i++){
if(s.charAt(i) == ValidNumberArguements.charAt(0))//charAt(0) = ,
{
currentCommas++;
continue;
}
if(s.charAt(i) == ValidNumberArguements.charAt(1))//charAt(1) = .
{
currentDec++;
continue;
}
if(s.charAt(i).matches("\D"))return null;//remove 1 A
}
if(currentDecimals > 1)return null;//remove 1.00.00
string decimalValue = "";
if(currentDecimals > 0)
{
int index = s.indexOf(ValidNumberArguements.charAt(1));
decimalValue += s.substring(index);
s = s.substring(0,index);
if(decimalValue.indexOf(ValidNumberArguements.charAt(0)) != -1)return null;//remove 1.00,000
}
int allowedCommas = (s.length-1) / 3;
if(currentCommas > allowedCommas)return null;//remove 10,00,000
String[] NumberParser = s.split(ValidNumberArguements.charAt(0));
length = NumberParser.length;
StringBuilder returnString = new StringBuilder();
for(int i = 0; i < length; i++)
{
if(i == 0)
{
if(NumberParser[i].length > 3 && length > 1)return null;//remove 1234,0,000
returnString.append(NumberParser[i]);
continue;
}
if(NumberParser[i].length != 3)return null;//ensure proper 1,000,000
returnString.append(NumberParser[i]);
}
returnString.append(decimalValue);
double answer = Double.parseDouble(returnString);
if(isNegative)answer *= -1;
return answer;
}
This code should handle most inputs, except IP addresses where all groups of digits are in three's (ex: 255.255.255.255 is valid, but not 255.1.255.255). It also doesn't support scientific notation
It will work with most variants of separators (",", "." or space). If more than one separator is detected, the first is assumed to be the thousands separator, with additional checks (validity etc.)
Edit: prevDigit is used for checking that the number uses thousand separators correctly. If there are more than one group of thousands, all but the first one must be in groups of 3. I modified the code to make it clearer so that "3" is not a magic number but a constant.
Edit 2: I don't mind the down votes much, but can someone explain what the problem is?
/* A number using thousand separator must have
groups of 3 digits, except the first one.
Numbers following the decimal separator can
of course be unlimited. */
private final static int GROUP_SIZE=3;
public static boolean isNumber(String input) {
boolean inThousandSep = false;
boolean inDecimalSep = false;
boolean endsWithDigit = false;
char thousandSep = '\0';
int prevDigits = 0;
for(int i=0; i < input.length(); i++) {
char c = input.charAt(i);
switch(c) {
case ',':
case '.':
case ' ':
endsWithDigit = false;
if(inDecimalSep)
return false;
else if(inThousandSep) {
if(c != thousandSep)
inDecimalSep = true;
if(prevDigits != GROUP_SIZE)
return false; // Invalid use of separator
}
else {
if(prevDigits > GROUP_SIZE || prevDigits == 0)
return false;
thousandSep = c;
inThousandSep = true;
}
prevDigits = 0;
break;
default:
if(Character.isDigit(c)) {
prevDigits++;
endsWithDigit = true;
}
else {
return false;
}
}
}
return endsWithDigit;
}
Test code:
public static void main(String[] args) {
System.out.println(isNumber("100")); // true
System.out.println(isNumber("100.00")); // true
System.out.println(isNumber("1,5")); // true
System.out.println(isNumber("1,000,000.00.")); // false
System.out.println(isNumber("100,00,2")); // false
System.out.println(isNumber("123.123.23.123")); // false
System.out.println(isNumber("123.123.123.123")); // true
}