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How can I find a String within a Java program converted to a string?

Basically, I read a java program into my program as a string, and I'm trying to find a way to extract strings from this. I have a loop counting through each character of this program, and this is what happens when it reaches a '"'.
else if (ch == '"')
{
String subString = " ";
index ++;
if (ch != '"')
{
subString += ch;
}
else
{
System.out.println(lineNumber + ", " + TokenType.STRING + ", " + subString);
index ++;
continue;
}
Unfortunately, this isn't working. This is the way I am trying to output the subString.
Essentially, I am looking for a way to add all the characters in between two "s together in order to get a String.

You could use regular expressions:
Pattern regex = Pattern.compile("(?:(?!<')\"(.*?(?<!\\\\)(?:\\\\\\\\)*)\")");
Matcher m = regex.matcher(content);
while (m.find())
System.out.println(m.group(1));
This will capture quoted strings, and takes account of escaped quotes/backslashes.
To break down the pattern:
(?: ... ) = don't capture as a group (the inside is captured instead)
(?!<') = make sure there isn't a single quote before (to avoid '"')
\"( ... )\" = capture what is inside the quotes
.*? = match the minimum of string of any chars
(?<!\\\\) = don't match single backslash before (double-escape = single backslash in content)
(?\\\\\\\\)* = match 0 or even number of backslashes
Together, 5. & 6. only match an even number of backslashes before the quote. This allows string endings like \\", \\\\", but not \" and \\\", which would be part of the string.
Non-regex solution, also taking care of escaped quotes:
List<String> strings = new ArrayList<>();
int start = -1;
int backslashes = 0;
for (int i = 0; i < content.length(); i++) {
char ch = content.charAt(i);
if (ch == '"') {
if (start == -1) {
start = i + 1;
backslashes = 0;
} else if (backslashes % 2 == 1) {
backslashes = 0;
} else {
strings.add(content.substring(start, i));
start = -1;
}
} else if (ch == '\\') backslashes++;
}
strings.forEach(System.out::println);

Java Regular expression to find out the number of matching words

I am learning regular expression.Suppose, If I have two String like abcd & bcdd. To make them equal Strings I have to remove a from first and d from last string. is this possible to count the matched number like bcd=> (3).
Currently, I am doing this
Pattern p= Pattern.compile("["+abcd+"]{2}");
Matcher m= p.matcher("abcd bcdd");
My current solution doesn't provide me the correct result. So, my question
1) Is this possible ?
2) If possible, then how can I achieve that ?
Hope, you will help to increase my regular expression knowledge.

Not sure why you would use regex at all, if all you need is the number of "bcd"s. I've put both a non-regex and regex version here for comparison.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
/**
<P>{#code java BcdRegexXmpl}</P>
**/
public class BcdRegexXmpl {
public static final void main(String[] igno_red) {
String sSentence = "abcd bcdd";
int iBcds = 0;
int iIdx = 0;
while(true) {
int iBcdIdx = sSentence.indexOf("bcd", iIdx);
if(iBcdIdx == -1) {
break;
}
iIdx = iBcdIdx + "bcd".length();
iBcds++;
}
System.out.println("Number of 'bcd's (no regex): " + iBcds);
//Alternatively
iBcds = 0;
//Same regex as #la-comadreja, with word-boundaries
//(for multiple "bcd"-s in a single word, remove the "\\b"-s)
Matcher m = Pattern.compile("\\b\\w*bcd\\w*\\b").matcher(sSentence);
while(m.find()) {
System.out.println("Found at index " + m.start());
iBcds++;
}
System.out.println("Number of 'bcd's (with regex): " + iBcds);
}
}
Output:
[R:\jeffy\programming\sandbox\xbnjava]java BcdRegexXmpl
Number of 'bcd's (no regex): 2
Found at index 0
Found at index 5
Number of 'bcd's (with regex): 2

Your pattern should be:
(a?)(bcd)(d?)
Another possibility is to write it as
\w*bcd\w*
If you want to count the number of "bcd"s in the string:
int bcds = 0;
for (int i = 0; i < str.length() - 2; i++) {
if (str.charAt(i) == 'b' && str.charAt(i+1) == 'c' && str.charAt(i+2) == 'd')
bcds++;
}

A maximally generalizable, concise and readable (and reasonably efficient) non-Regex answer:
int countMatches(String s, String searchStr) {
//Here, s is "abcd bcdd" and searchStr is "bcd"
int matches = 0;
for (int i = 0; i < s.length() - searchStr.length() + 1; i++) {
for (int j = 0; j < searchStr.length(); j++) {
if (s.charAt(i + j) != searchStr.charAt(j)) break;
if (j == searchStr.length() - 1) matches++;
}
}
return matches;
}

Check and extract a number from a String in Java

I'm writing a program where the user enters a String in the following format:
"What is the square of 10?"
I need to check that there is a number in the String
and then extract just the number.
If i use .contains("\\d+") or .contains("[0-9]+"), the program can't find a number in the String, no matter what the input is, but .matches("\\d+")will only work when there is only numbers.
What can I use as a solution for finding and extracting?

try this
str.matches(".*\\d.*");

If you want to extract the first number out of the input string, you can do-
public static String extractNumber(final String str) {
if(str == null || str.isEmpty()) return "";
StringBuilder sb = new StringBuilder();
boolean found = false;
for(char c : str.toCharArray()){
if(Character.isDigit(c)){
sb.append(c);
found = true;
} else if(found){
// If we already found a digit before and this char is not a digit, stop looping
break;
}
}
return sb.toString();
}
Examples:
For input "123abc", the method above will return 123.
For "abc1000def", 1000.
For "555abc45", 555.
For "abc", will return an empty string.

I think it is faster than regex .
public final boolean containsDigit(String s) {
boolean containsDigit = false;
if (s != null && !s.isEmpty()) {
for (char c : s.toCharArray()) {
if (containsDigit = Character.isDigit(c)) {
break;
}
}
}
return containsDigit;
}

s=s.replaceAll("[*a-zA-Z]", "") replaces all alphabets
s=s.replaceAll("[*0-9]", "") replaces all numerics
if you do above two replaces you will get all special charactered string
If you want to extract only integers from a String s=s.replaceAll("[^0-9]", "")
If you want to extract only Alphabets from a String s=s.replaceAll("[^a-zA-Z]", "")
Happy coding :)

The code below is enough for "Check if a String contains numbers in Java"
Pattern p = Pattern.compile("([0-9])");
Matcher m = p.matcher("Here is ur string");
if(m.find()){
System.out.println("Hello "+m.find());
}

I could not find a single pattern correct.
Please follow below guide for a small and sweet solution.
String regex = "(.)*(\\d)(.)*";
Pattern pattern = Pattern.compile(regex);
String msg = "What is the square of 10?";
boolean containsNumber = pattern.matcher(msg).matches();

Pattern p = Pattern.compile("(([A-Z].*[0-9])");
Matcher m = p.matcher("TEST 123");
boolean b = m.find();
System.out.println(b);

The solution I went with looks like this:
Pattern numberPat = Pattern.compile("\\d+");
Matcher matcher1 = numberPat.matcher(line);
Pattern stringPat = Pattern.compile("What is the square of", Pattern.CASE_INSENSITIVE);
Matcher matcher2 = stringPat.matcher(line);
if (matcher1.find() && matcher2.find())
{
int number = Integer.parseInt(matcher1.group());
pw.println(number + " squared = " + (number * number));
}
I'm sure it's not a perfect solution, but it suited my needs. Thank you all for the help. :)

Try the following pattern:
.matches("[a-zA-Z ]*\\d+.*")

Below code snippet will tell whether the String contains digit or not
str.matches(".*\\d.*")
or
str.matches(.*[0-9].*)
For example
String str = "abhinav123";
str.matches(".*\\d.*") or str.matches(.*[0-9].*) will return true
str = "abhinav";
str.matches(".*\\d.*") or str.matches(.*[0-9].*) will return false

As I was redirected here searching for a method to find digits in string in Kotlin language, I'll leave my findings here for other folks wanting a solution specific to Kotlin.
Finding out if a string contains digit:
val hasDigits = sampleString.any { it.isDigit() }
Finding out if a string contains only digits:
val hasOnlyDigits = sampleString.all { it.isDigit() }
Extract digits from string:
val onlyNumberString = sampleString.filter { it.isDigit() }

public String hasNums(String str) {
char[] nums = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };
char[] toChar = new char[str.length()];
for (int i = 0; i < str.length(); i++) {
toChar[i] = str.charAt(i);
for (int j = 0; j < nums.length; j++) {
if (toChar[i] == nums[j]) { return str; }
}
}
return "None";
}

You can try this
String text = "ddd123.0114cc";
String numOnly = text.replaceAll("\\p{Alpha}","");
try {
double numVal = Double.valueOf(numOnly);
System.out.println(text +" contains numbers");
} catch (NumberFormatException e){
System.out.println(text+" not contains numbers");
}

As you don't only want to look for a number but also extract it, you should write a small function doing that for you. Go letter by letter till you spot a digit. Ah, just found the necessary code for you on stackoverflow: find integer in string. Look at the accepted answer.

.matches(".*\\d+.*") only works for numbers but not other symbols like // or * etc.

ASCII is at the start of UNICODE, so you can do something like this:
(x >= 97 && x <= 122) || (x >= 65 && x <= 90) // 97 == 'a' and 65 = 'A'
I'm sure you can figure out the other values...

How to get float value from string

I have a string like
> 12.4N-m/kg.
From the above string I need to get a value 12.4.
When I use replace all function str.replaceAll("[^.0-9]", "").
This doesn't work when then string has two dots.
The location of float value may differ.

First discard all non flot characters and then covert to Float like this:
float f = Float.valueOf("> 12.4N-m/kg.".replaceAll("[^\\d.]+|\\.(?!\\d)", ""));
// now f = 12.4

Assuming your input always has a space before the number and an N after it:
String t = "> 12.4N-m/kg.";
Pattern p = Pattern.compile("^.*\\s(\\d+\\.\\d)N.*$");
Matcher matcher = p.matcher(t);
if (matcher.matches()) {
System.out.println(Float.valueOf(matcher.group(1)));
}

Try to use this:
Float.valueOf(str.substring(0,4));

following code will work with assumption that input string always starts with "> " and it has proper float prefixed.
int i=2;
while(Character.isDigit(str.charAt(i)) || str.charAt(i) == '.')
i++;
float answer = Float.valueOf(str.substring(2,i));

Try to use this regular expression
^[-+]?[0-9]*\.?[0-9]+$

I think the previous answers leave out two points:
There are more complicated numbers than this.
There might be a digit in the unit which souldn't end up in the float.
Because of the second point I don't think replacing everything that is a non-digit is a good idea. One should rather search for the first number in the string:
Matcher m = p.matcher(str);
System.out.println("Input: "+ str);
if (m.find()) {
System.out.println("Found: "+ m.group());
try {
System.out.println("Number: "+ Float.parseFloat(m.group()));
} catch (Exception exc) {
exc.printStackTrace();
}
}
Alternatively, you could do something like
int i, j;
for (i = 0; i < str.length(); ++i) {
if (mightBePartOfNumber(str.charAt(i))) {
break;
}
}
for (j = i; j < str.length(); ++j) {
if (!mightBePartOfNumber(str.charAt(j))) {
break;
}
}
String substr = str.substring(i, j);
System.out.println("Found: "+ substr);
try {
System.out.println("Number: "+ Float.parseFloat(substr));
} catch (Exception exc) {
exc.printStackTrace();
}
with a helper
private static boolean mightBePartOfNumber(char c) {
return ('0' <= c && c <= '9') || c == '+' || c == '-' || c == '.' || c == 'e' || c == 'E';
}

I have tried the above options but not worked for me , Please try below pattern
Pattern pattern = Pattern.compile("\\d+(?:\\.\\d+)?");

Find count of digits in string variable

I have a string which sometimes gives character value and sometimes gives integer value. I want to get the count of number of digits in that string.
For example, if string contains "2485083572085748" then total number of digits is 16.
Please help me with this.

A cleaner solution using Regular Expressions:
// matches all non-digits, replaces it with "" and returns the length.
s.replaceAll("\\D", "").length()

String s = "2485083572085748";
int count = 0;
for (int i = 0, len = s.length(); i < len; i++) {
if (Character.isDigit(s.charAt(i))) {
count++;
}
}

Just to refresh this thread with stream option of counting digits in a string:
"2485083572085748".chars()
.filter(Character::isDigit)
.count();

If your string gets to big and full of other stuff than digits you should try to do it with regular expressions. Code below would do that to you:
String str = "asdasd 01829898 dasds ds8898";
Pattern p = Pattern.compile("\d"); // "\d" is for digits in regex
Matcher m = p.matcher(str);
int count = 0;
while(m.find()){
count++;
}
check out java regex lessons for more.
cheers!

Loop each character and count it.
String s = "2485083572085748";
int counter = 0;
for(char c : s.toCharArray()) {
if( c >= '0' && c<= '9') {
++counter;
}
}
System.out.println(counter);

public static int getCount(String number) {
int flag = 0;
for (int i = 0; i < number.length(); i++) {
if (Character.isDigit(number.charAt(i))) {
flag++;
}
}
return flag;
}

in JavaScript:
str = "2485083572085748"; //using the string in the question
let nondigits = /\D/g; //regex for all non-digits
let digitCount = str.replaceAll(nondigits, "").length;
//counts the digits after removing all non-digits
console.log(digitCount); //see in console
Thanks --> https://stackoverflow.com/users/1396264/vedant for the Java version above. It helped me too.

int count = 0;
for(char c: str.toCharArray()) {
if(Character.isDigit(c)) {
count++;
}
}
Also see
Javadoc

Something like:
using System.Text.RegularExpressions;
Regex r = new Regex( "[0-9]" );
Console.WriteLine( "Matches " + r.Matches("if string contains 2485083572085748 then" ).Count );