I am trying to achieve this.
I have a string of 9char (always the same). But i also know that the first and last char is always a aplhabetic, it must be. the rest in the middle are numbers. How to check for that.
I got this logic so far, syntax is my problem
string samplestring;
samplestring = a1234567B
If(samplestring.length() == 9 && samplestring.substring(0,1).uppercase && samplestring.substring(8,9) && samplestring.THE REST OF THE CHAR IN THE MIDDLE ARE DIGITS)
{
println("yes this is correct");
}
else
{
println("retype");
}
Please dont mind about the simple english just want to know the syntax but the logic is there i hope..
Also can please show me those lowercase ones how to convert to uppercase?
A regular expression would be suitable:
String s = new String("A2345678Z");
if (s.matches("[A-Z][0-9]{7}[A-Z]")))
{
}
Regular expression explained:
[A-Z] means any uppercase letter
[0-9]{7} means 7 digits
Pattern p = Pattern.compile("^[A-Za-z]\\d+[A-Za-z]$");
Matcher m = p.match("A1234567B");
if (m.matches()) {
//
}
Edit:
If there are always seven digits, you can replace the \\d+ with \\d{7}
String str="A12345678B";
char first = str.charAt(0);
char second = str.charAt(str.length()-1);
if(Character.isUpperCase(first)&& Character.isUpperCase(second)){
//do something
}
Related
I'm unsure of the code for this, but if one were to input "oooooooooo" after a prompt (like in an if-statement or something where the program registers "o" as "one" or something), how could you make "oooooooooo" translate into "o"?
Would one have to write down manually various iterations of "o" (like, "oo" and "ooo" and "oooo"...etc.). Would it be similar to something like the ignore case method where O and o become the same? So "ooo..." and "o" end up as the same string.
Although probably overkill for this one use-case, it would be helpful to learn how to use regexes in the future. Java provides a regex library to use called Pattern. For example, the regex /o+ne/ would match any string "o...ne" with at least one "o".
using regex:
public static String getSingleCharacter(String input){
if(input == null || input.length() == 0) return null;
if(input.length() == 1) return input;
if(!input.toLowerCase().matches("^\\w*?(\\w)(?!\\1|$)\\w*$")){
return Character.toString(input.toLowerCase().charAt(0));
}
return null;
}
if the method returns null then the characters are not all the same, else it will return that single char represented as a string.
Use the regular expression /(.)\1+/ and String#replaceAll() to match runs of two or more of the same character and then replace the match with the value of the first match group identified with $1 as follows:
public static String squeeze(String input) {
return input.replaceAll("(.)\\1+", "$1");
}
String result = squeeze("aaaaa bbbbbbb cc d");
assert(result.equals("a b c d"));
public string condense(String input) {
if(input.length >= 3) {
for(int i=0; i< input.length-2; i++){
if(input.substring(i,i+1) != input.substring(i+1,i+2)){
return input;
}
}
}
return input.substring(0,1);
}
This checks if the string is 3 characters or longer, and if so it loops through the entire string. If every character in the string is the same, then it returns a condensed version of the string.
String always consists of two distinct alternating characters. For example, if string 's two distinct characters are x and y, then t could be xyxyx or yxyxy but not xxyy or xyyx.
But a.matches() always returns false and output becomes 0. Help me understand what's wrong here.
public static int check(String a) {
char on = a.charAt(0);
char to = a.charAt(1);
if(on != to) {
if(a.matches("["+on+"("+to+""+on+")*]|["+to+"("+on+""+to+")*]")) {
return a.length();
}
}
return 0;
}
Use regex (.)(.)(?:\1\2)*\1?.
(.) Match any character, and capture it as group 1
(.) Match any character, and capture it as group 2
\1 Match the same characters as was captured in group 1
\2 Match the same characters as was captured in group 2
(?:\1\2)* Match 0 or more pairs of group 1+2
\1? Optionally match a dangling group 1
Input must be at least two characters long. Empty string and one-character string will not match.
As java code, that would be:
if (a.matches("(.)(.)(?:\\1\\2)*\\1?")) {
See regex101.com for working examples1.
1) Note that regex101 requires use of ^ and $, which are implied by the matches() method. It also requires use of flags g and m to showcase multiple examples at the same time.
UPDATE
As pointed out by Austin Anderson:
fails on yyyyyyyyy or xxxxxx
To prevent that, we can add a zero-width negative lookahead, to ensure input doesn't start with two of the same character:
(?!(.)\1)(.)(.)(?:\2\3)*\2?
See regex101.com.
Or you can use Austin Anderson's simpler version:
(.)(?!\1)(.)(?:\1\2)*\1?
Actually your regex is almost correct but problem is that you have enclosed your regex in 2 character classes and you need to match an optional 2nd character in the end.
You just need to use this regex:
public static int check(String a) {
if (a.length() < 2)
return 0;
char on = a.charAt(0);
char to = a.charAt(1);
if(on != to) {
String re = on+"("+to+on+")*"+to+"?|"+to+"("+on+to+")*"+on+"?";
System.out.println("re: " + re);
if(a.matches(re)) {
return a.length();
}
}
return 0;
}
Code Demo
I want to get the number of substrings out of a string.
The inputs are excel formulas like IF(....IF(...))+IF(...)+SUM(..) as a string. I want to count all IF( substrings. It's important that SUMIF(...) and COUNTIF(...) will not be counted.
I thought to check that there is no capital letter before the "IF", but this is giving (certainly) index out of bound. Can someone give me a suggestion?
My code:
for(int i = input.indexOf("IF(",input.length());
i != -1;
i= input.indexOf("IF(,i- 1)){
if(!isCapitalLetter(tmpFormulaString, i-1)){
ifStatementCounter++;
}
}
Although you can do the parsing by yourself as you were doing (that's possibly better for you to learn debugging so you know what your problem is)
However it can be easily done by regular expression:
String s = "FOO()FOOL()SOMEFOO()FOO";
Pattern p = Pattern.compile("\\bFOO\\b");
Matcher m = p.matcher(s);
int count = 0;
while (m.find()) {
count++;
}
// count= 2
The main trick here is \b in the regex. \b means word boundary. In short, if there is a alphanumeric character at the position of \b, it will not match.
http://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html
I think you can solve your problem by finding String IF(.
Try to do same thing in another way .
For example:
inputStrin = IF(hello)IF(hello)....IF(helloIF(hello))....
inputString.getIndexOf("IF(");
That solves your problem?
Click Here Or You can use regular expression also.
I have a string representing a 32 character long barcode made up of "|" and ":".
I want to check the validity of any given string to make sure it is a barcode. One of the tests is to check that the only symbols it contains are the two mentioned above. How can I check that?
I first I was using a delimiter, but I don't think that is the right way to go about this.
public boolean isValidBarCode (String barCode)
{
barCode.useDelimiter ("[|:]");
if (barCode.length() == 32)
{
return true;
}
else
{
return false;
}
I know there are other things I need to check in order to validate it as a barcode, but I'm asking only for the purposes of checking the symbols within the given string.
I'm a beginner programmer, so the help is greatly appreciated!
You can use a regex:
boolean correct = string.matches("[\\:\\|]+");
Explanation for the regex: it checks that the string is constituted of 1 or more characters (that's what the + suffix does) being either : or |. We would normally write [:|]+, but since : and (I think) | are special characters in regexes, they need to be escaped with a backslash. And backslashes must be escaped in a string literal, hence the double backslash.
Or you can simply code a 5 lines algorithm using a loop:
boolean correct = false;
for (int i = 0; i < string.length() && correct; i++) {
char c = string.charAt(i);
if (c != ':' && c != '|') {
correct = false;
}
}
Since you require the barcode to be exactly 32 characters long and consist only of the : and | characters, you should use a combination of length and regex checking:
boolean isCorrect = barCode.matches( "[\\|\\:]*" );
if(isCorrect && barCode.length() == 32) {
//true case
} else {
//false case
}
boolean isBarCode = barCode.matches( "[\\|\\:]*" );
I have a string like "portal100common2055".
I would like to split this into two parts, where the second part should only contain numbers.
"portal200511sbet104" would become "portal200511sbet", "104"
Can you please help me to achieve this?
Like this:
Matcher m = Pattern.compile("^(.*?)(\\d+)$").matcher(args[0]);
if( m.find() ) {
String prefix = m.group(1);
String digits = m.group(2);
System.out.println("Prefix is \""+prefix+"\"");
System.out.println("Trailing digits are \""+digits+"\"");
} else {
System.out.println("Does not match");
}
String[] parts = input.split("(?<=\\D)(?=\\d+$)");
if (parts.length < 2) throw new IllegalArgumentException("Input does not end with numbers: " + input);
String head = parts[0];
String numericTail = parts[1];
This more elegant solution uses the look behind and look ahead features of regex.
Explanation:
(?<=\\D) means at the current point, ensure the preceding characters ends with a non-digit (a non-digit is expressed as \D)
(?=\\d+$) means t the current point, ensure that only digits are found to the end of the input (a digit is expressed as \d)
This will only the true at the desired point you want to divide the input