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How to make a regular expression match based on a condition?

I'm trying to make a conditional regex, I know that there are other posts on stack overflow but there too specific to the problem.
The Question
How can I create a regular expression that only looks to match something given a certain condition?
An example
An example of this would be if we had a list of a string(this is in java):
String nums = "42 36 23827";
and we only want to match if there are the same amount of x's at the end of the string as there are at the beginning
What we want in this example
In this example, we would want a regex that checks if there are the same amount of regex's at the end as there are in the beginning. The conditional part: If there are x's at the beginning, then check if there are that many at the end, if there are then it is a match.
Another example
An example of this would be if we had a list of numbers (this is in java) in string format:
String nums = "42 36 23827";
and we want to separate each number into a list
String splitSpace = "Regex goes here";
Pattern splitSpaceRegex = Pattern.compile(splitSpace);
Matcher splitSpaceMatcher = splitSpaceRegex.matcher(text);
ArrayList<String> splitEquation = new ArrayList<String>();
while (splitSpaceMatcher.find()) {
if (splitSpaceMatcher.group().length() != 0) {
System.out.println(splitSpaceMatcher.group().trim());
splitEquation.add(splitSpaceMatcher.group().trim());
}
}
How can I make this into an array that looks like this:
["42", "36", "23827"]
You could try making a simple regex like this:
String splitSpace = "\\d+\\s+";
But that exludes the "23827" because there is no space after it.
and we only want to match if there are the same amount ofx`'s at the end of the string as there are at the beginning
What we want in this example
In this example, we would want a regex that checks if it is the end of the string; if it is then we don't need the space, otherwise, we do. As #YCF_L mentioned we could just make a regex that is \\b\\d\\b but I am aiming for something conditional.
Conclusion
So, as a result, the question is, how do we make conditional regular expressions? Thanks for reading and cheers!

There are no conditionals in Java regexes.
I want a regex that checks if there are the same amount of regex's at the end as there are in the beginning. The conditional part: If there are x's at the beginning, then check if there are that many at the end, if there are then it is a match.
This may or may not be solvable. If you want to know if a specific string (or pattern) repeats, that can be done using a back reference; e.g.
^(\d+).+\1$
will match a line consisting of an arbitrary number digits, any number of characters, and the same digits matched at the start. The back reference \1 matches the string matched by group 1.
However if you want the same number of digits at the end as at the start (and that number isn't a constant) then you cannot implement this using a single (Java) regex.
Note that some regex languages / engines do support conditionals; see the Wikipedia Comparison of regular-expression engines page.

I would like to use split which accept regex like so :
String[] split = nums.split("\\s+"); // ["42", "36", "23827"]
If you want to use Pattern with Matcher, then you can use String \b\d+\b with word boundaries.
String regex = "\\b\\d+\\b";
By using word boundaries, you will avoid cases where the number is part of the word, for example "123 a4 5678 9b" you will get just ["123", "4578"]

I do not see the "conditional" in the question. The problem is solvable with a straight forward regular expression: \b\d+\b.
regex101 demo
A fully fledged Java example would look something like this:
import java.util.ArrayList;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class Ideone {
public static void main(String args[]) {
final String sample = "123 45 678 90";
final Pattern pattern = Pattern.compile("\\b\\d+\\b");
final Matcher matcher = pattern.matcher(sample);
final ArrayList<String> results = new ArrayList<>();
while (matcher.find()) {
results.add(matcher.group());
}
System.out.println(results);
}
}
Output: [123, 45, 678, 90]
Ideone demo

Regular expression is not working with single character

I am trying to write regular expression in Java to evaluate two strings mentioned with () separated by ,
Example: (test1,test2)
I have written below code
public static void main(String[] a){
String pattern = "\\([a-zA-Z0-9]+,[a-zA-Z0-9]+.\\)";
String test = "(test1,test2)";
System.out.println(test.matches(pattern));
}
It works as expected and prints true in below cases
String test = "(test1,test2)";
String test = "(t,test2)";
But it is printing false when I send below
String test = "(test1,t)";
It is strange because I am using same expression before and after ,
It returns true for (t,test2) but not for (test1,t)
Please let me know what am I missing in this regular expression. I need it to evaluate and return true for (test1,t)

There's no need for the . (that matches one character) in your regex. Remove . from your regex so it becomes "\\([a-zA-Z0-9]+,[a-zA-Z0-9]+\\)" and it should work.

Use this regex:
String pattern = "^\\(.+,.+\\)";
This will match your required strings.

In the second part of your pattern, you have "[a-zA-Z0-9]+."
If you're trying to match "t", it will see t for the [a-zA-Z0-9]+ part, but it requires another character after that to match the . part.
Revised pattern: "\\([a-zA-Z0-9]+,[a-zA-Z0-9]+\\)"

Delete the dot after the second group[a-zA-Z0-9]
Demo

and even simpler you can use \w for words, you can use instead of [a-zA-Z0-9]
so your regular expression would be like that
\(\w+,\w+\)

In your regular expression '.' is not needed in the latter part.
change is as "\([a-zA-Z0-9]+,[a-zA-Z0-9]+\)" so that it will be returning true for "(test,t)"
String pattern = "\\([a-zA-Z0-9]+,[a-zA-Z0-9]+\\)";
String test = "(te,t)";
System.out.println(test.matches(pattern)); // true

String pattern = "\\([a-zA-Z0-9]+,[a-zA-Z0-9]+\\)";
String test = "(test1,test2)";
String t1 = "(t,test2)";
String t2="(test2,t)";
System.out.println(test.matches(pattern));
System.out.println(t1.matches(pattern));
System.out.println(t2.matches(pattern));
just try this code, it will give you answer you want.
You have written "." at the end after + in your pattern so clear it.

Regular expression not matching on first and last word of string

I am trying to write a java program that will look for a specific words in a string. I have it working for the most part but it doesnt seem to match if the word to match is the first or last word in the string. Here is an example:
"trying to find the first word".matches(".*[^a-z]find[^a-z].*") //returns true
"trying to find the first word".matches(".*[^a-z]trying[^a-z].*") //returns false
"trying to find the first word".matches(".*[^a-z]word[^a-z].*") //returns false
Any idea how to make this match on any word in the string?
Thanks in advance,
Craig

The problem is your character class before and after the words [^a-z]- I think that what you actually want is a word boundary character \b (as per ColinD's comment) as opposed to not a character in the a-z range. As pointed out in the comments (thanks) you'll also needs to handle the start and end of string cases.
So try, eg:
"(?:^|.*\b)trying(?:\b.*|$)"

You can use the optional (?) , check below link and test more cases if this give proper output:
https://regex101.com/r/oP5zB8/1
(.*[^a-z]?trying[^a-z]?.*)

I think (^|^.*[^a-z])trying([^a-z].*$|$) just fits your need.
Or (?:^|^.*[^a-z])trying(?:[^a-z].*$|$) for non capturing parentheses.

You can try following program to check the existence on start and end of any string:
package com.ajsodhi.utilities;
import java.util.regex.Pattern;
public class RegExStartEndWordCheck {
public static final String stringToMatch = "StartingsomeWordsEndWord";
public static void main(String[] args) {
String regEx = "Starting[A-Za-z0-9]{0,}EndWord";
Pattern patternOriginalSign = Pattern.compile(regEx, Pattern.CASE_INSENSITIVE);
boolean OriginalStringMatchesPattern = patternOriginalSign.matcher(stringToMatch).matches();
System.out.println(OriginalStringMatchesPattern);
}
}

you should use the boundary \b that's specify a beginning or a ending of a word instead of [^a-z] which is not so logic.
Just something like
".*\\bfind\\b.*"

Match String ending with (regex) java

I am following the suggestions on the page, check if string ends with certain pattern
I am trying to display a string that is
Starts with anything
Has the letters ".mp4" in it
Ends explicitly with ', (apostrophe followed by comma)
Here is my Java code:
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.regex.*;
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
String str = " _file='ANyTypEofSTR1ngHere_133444556_266545797_10798866.mp4',";
Pattern p = Pattern.compile(".*.mp4[',]$");
Matcher m = p.matcher(str);
if(m.find())
System.out.println("yes");
else
System.out.println("no");
}
}
It prints "no". How should I declare my RegEx?

There are several issues in your regex:
"Has the letters .mp4 in it" means somewhere, not necessarily just in front of ',, so another .* should be inserted.
. matches any character. Use \. to match .
[,'] is a character group, i.e. exactly one of the characters in the brackets has to occur.
You can use the following regex instead:
Pattern p = Pattern.compile(".*\\.mp4.*',$");

Your character set [',] is checking whether the string ends with ' or , a single time.
If you want to match those character one or more times, use [',]+. However, you probably don't want to use a character set in this case since you said order is important.
To match an apostrophe followed by comma, just use:
.*\\.mp4',$
Also, since . has special meaning, you need to escape it in '.mp4'.

java regular expression

Can anyone please help me do the following in a java regular expression?
I need to read 3 characters from the 5th position from a given String ignoring whatever is found before and after.
Example : testXXXtest
Expected result : XXX

You don't need regex at all.
Just use substring: yourString.substring(4,7)
Since you do need to use regex, you can do it like this:
Pattern pattern = Pattern.compile(".{4}(.{3}).*");
Matcher matcher = pattern.matcher("testXXXtest");
matcher.matches();
String whatYouNeed = matcher.group(1);
What does it mean, step by step:
.{4} - any four characters
( - start capturing group, i.e. what you need
.{3} - any three characters
) - end capturing group, you got it now
.* followed by 0 or more arbitrary characters.
matcher.group(1) - get the 1st (only) capturing group.

You should be able to use the substring() method to accomplish this:
string example = "testXXXtest";
string result = example.substring(4,7);

This might help: Groups and capturing in java.util.regex.Pattern.
Here is an example:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Example {
public static void main(String[] args) {
String text = "This is a testWithSomeDataInBetweentest.";
Pattern p = Pattern.compile("test([A-Za-z0-9]*)test");
Matcher m = p.matcher(text);
if (m.find()) {
System.out.println("Matched: " + m.group(1));
} else {
System.out.println("No match.");
}
}
}
This prints:
Matched: WithSomeDataInBetween
If you don't want to match the entire pattern rather to the input string (rather than to seek a substring that would match), you can use matches() instead of find(). You can continue searching for more matching substrings with subsequent calls with find().
Also, your question did not specify what are admissible characters and length of the string between two "test" strings. I assumed any length is OK including zero and that we seek a substring composed of small and capital letters as well as digits.

You can use substring for this, you don't need a regex.
yourString.substring(4,7);
I'm sure you could use a regex too, but why if you don't need it. Of course you should protect this code against null and strings that are too short.

Use the String.replaceAll() Class Method
If you don't need to be performance optimized, you can try the String.replaceAll() class method for a cleaner option:
String sDataLine = "testXXXtest";
String sWhatYouNeed = sDataLine.replaceAll( ".{4}(.{3}).*", "$1" );
References
https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html
http://www.vogella.com/tutorials/JavaRegularExpressions/article.html#using-regular-expressions-with-string-methods