Java Pattern REGEX Where Not Matching

Question

I'm trying to use the Java Pattern and Matcher to apply input checks. I have it working in a really basic format which I am happy with so far. It applies a REGEX to an argument and then loops through the matching characters.

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class RegexUtil {

   public static void main(String[] args) {

      String    argument;
      Pattern   pattern;
      Matcher   matcher;

      argument = "#a1^b2";
      pattern = Pattern.compile("[a-zA-Z]|[0-9]|\\s");
      matcher = pattern.matcher(argument);

      // find all matching characters
      while(matcher.find()) {
         System.out.println(matcher.group());
      }

   }

}

This is fine for extracting all the good characters, I get the output

a
1
b
2

Now I wanted to know if it's possible to do the same for any characters that don't match the REGEX so I get the output

#
^

Or better yet loop through it and get TRUE or FALSE flags for each index of the argument

false
true
true
false
true
true

I only know how to loop through with matcher.find(), any help would be greatly appreciated

[^a-zA-Z0-9] will give you all characters not matching the range so: # ^ — Jérôme
– Jérôme, Commented Mar 9, 2018 at 13:32

Wiktor Stribiżew · Accepted Answer · 2018-03-09 13:36:32Z

3

You may add a |(.) alternative to your pattern (to match any char but a line break char) and check if Group 1 matched upon each match. If yes, output false, else, output true:

String argument = "#a1^b2";
Pattern pattern = Pattern.compile("[a-zA-Z]|[0-9]|\\s|(.)"); // or "[a-zA-Z0-9\\s]|(.)"
Matcher matcher = pattern.matcher(argument);

while(matcher.find()) {                           // find all matching characters
    System.out.println(matcher.group(1) == null);

See the Java demo, output:

false
true
true
false
true
true

Note you do not need to use a Pattern.DOTALL here, because \s in your "whitelist" part of the pattern matches line breaks.

edited Mar 9, 2018 at 13:36

answered Mar 9, 2018 at 13:31

Wiktor Stribiżew

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7 Comments

azro Over a year ago

Check condition to print boolean can be simplified as basic while (matcher.find()) { System.out.println(matcher.group(1) == null); }

Wiktor Stribiżew Over a year ago

@azro Yes, I just thought that a custom string might be necessary, if true or false are the only values needed then sure.

Wiktor Stribiżew Over a year ago

@Trent Because Group 1 will hold them. The regex is basically tokenizing a string into 2 types of chars, those you want to be labelled as true and those that must be false.

Wiktor Stribiżew Over a year ago

@Trent Sorry, I do not get you. If you just want to print the match print it with matcher.group(0).

TheLovelySausage Over a year ago

It's ok, I'm to tinker a bit but this answer is perfect for what I need, thanks a bunch

|

MGothe · Accepted Answer · 2018-03-09 13:36:56Z

1

Why not simply removing all matching chars from your string, so you get only the non matching ones back:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class RegexUtil {

   public static void main(String[] args) {

      String    argument;
      Pattern   pattern;
      Matcher   matcher;

      argument = "#a1^b2";
      pattern = Pattern.compile("[a-zA-Z]|[0-9]|\\s");
      matcher = pattern.matcher(argument);

      // find all matching characters
      while(matcher.find()) {
         System.out.println(matcher.group());
         argument = argument.replace(matcher.group(), "");
      }

      System.out.println("argument: " + argument);

   }

}

answered Mar 9, 2018 at 13:36

MGothe

113 bronze badges

Comments

azro · Accepted Answer · 2018-03-09 13:35:26Z

0

You have to iterate over each char of the String and check one by one :

//for-each loop, shorter way
for (char c : argument.toCharArray()){
    System.out.println(pattern.matcher(c + "").matches());
}

or

//classic for-i loop, with details
for (int i = 0; i < argument.length(); i++) {
    String charAtI = argument.charAt(i) + "";
    boolean doesMatch = pattern.matcher(charAtI).matches();
    System.out.println(doesMatch);
}

Also, when you don't require it, you can do declaration and give a value at same time :

String argument = "#a1^b2";
Pattern pattern = Pattern.compile("[a-zA-Z]|[0-9]|\\s");

edited Mar 9, 2018 at 13:35

answered Mar 9, 2018 at 13:29

azro

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Comments

markusk · Accepted Answer · 2018-03-09 13:44:08Z

0

Track positions, and for each match, print the characters between last match and current match.

int pos = 0;
while (matcher.find()) {
    for (int i = pos; i < matcher.start(); i++) {
        System.out.println(argument.charAt(i));
    }
    pos = matcher.end();
}
// Print any trailing characters after last match.
for (int i = pos; i < argument.length(); i++) {
    System.out.println(argument.charAt(i));
}

answered Mar 9, 2018 at 13:44

markusk

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Comments

Anto Livish A · Accepted Answer · 2018-03-09 14:04:58Z

0

One solution is

    String  argument;
    Pattern pattern;
    Matcher matcher;

    argument = "#a1^b2";
    List<String> charList = Arrays.asList(argument.split(""));
    pattern = Pattern.compile("[a-zA-Z]|[0-9]|\\s");
    matcher = pattern.matcher(argument);
    ArrayList<String> unmatchedCharList = new ArrayList<>();
    // find all matching
    while(matcher.find()) {
        unmatchedCharList.add(matcher.group());
    }
    for(String charr : charList)
    {
        System.out.println(unmatchedCharList.contains(charr ));
    }

Output

false true true false true true

answered Mar 9, 2018 at 14:04

Anto Livish A

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