Develop Reference

How can I get non-matching groups using a Matcher in Java?

I'm trying to write a java regex to catch some groups of words from a String using a Matcher.
Say i got this string: "Hello, we are #happy# to see you today".
I would like to get 2 group of matches, one having
Hello, we are
to see you today
and the other
happy
So far, I was only able to match the word between the #s using this Pattern:
Pattern p = Pattern.compile("#(.+?)#");
I've read about negative lookahead and lookaround, played a bit with it but without success.
I assume I should do some sort of negation of the regex so far, but I couldn't come up with anything.
Any help would be really appreciated, thank you.

From comment:
I may incur in a string where I got more than one instances of words wrapped by #, such as "#Hello# kind #stranger#"
From comment:
I need to apply some different style format to both the text inside and outside.
Since you need to apply different stylings, the code need to process each block of text separately, and needs to know if the text is inside or outside a #..# section.
Note, in the following code, it will silently skip the last #, if there is an odd number of them.
String input = ...
for (Matcher m = Pattern.compile("([^#]+)|#([^#]+)#").matcher(input); m.find(); ) {
if (m.start(1) != -1) {
String outsideText = m.group(1);
System.out.println("Outside: \"" + outsideText + "\"");
} else {
String insideText = m.group(2);
System.out.println("Inside: \"" + insideText + "\"");
}
}
Output for input = "Hello, we are #happy# to see you today"
Outside: "Hello, we are "
Inside: "happy"
Outside: " to see you today"
Output for input = "#Hello# kind #stranger#"
Inside: "Hello"
Outside: " kind "
Inside: "stranger"
Output for input = "This #text# has unpaired # characters"
Outside: "This "
Inside: "text"
Outside: " has unpaired "
Outside: " characters"

The best I could do is splitting in 3 groups, then merging the group 1 and 4 :
(^.*)(\#(.+?)\#)(.*)
Test it here
EDIT: Taking remarks from the comments :
(^[^\#]*)(?:\#(.+?)\#)([^\#]*)
Thanks to #Lino we don't capture the useless group with # anymore, and we capture anything except #, instead of any non whitespace character in the 1st and 2nd groups.
Test it here

Is this solution fine?
Pattern pattern =
Pattern.compile("([^#]+)|#([^#]*)#");
Matcher matcher =
pattern.matcher("Hello, we are #happy# to see you today");
List<String> notBetween = new ArrayList<>(); // not surrounded by #
List<String> between = new ArrayList<>(); // surrounded by #
while (matcher.find()) {
if (Objects.nonNull(matcher.group(1))) notBetween.add(matcher.group(1));
if (Objects.nonNull(matcher.group(2))) between.add(matcher.group(2));
}
System.out.println("Printing group 1");
for (String string :
notBetween) {
System.out.println(string);
}
System.out.println("Printing group 2");
for (String string :
between) {
System.out.println(string);
}

Java Regex ReplaceAll with grouping

I want to surround all tokens in a text with tags in the following manner:
Input: " abc fg asd "
Output:" <token>abc</token> <token>fg</token> <token>asd</token> "
This is the code I tried so far:
String regex = "(\\s)([a-zA-Z]+)(\\s)";
String text = " abc fg asd ";
text = text.replaceAll(regex, "$1<token>$2</token>$3");
System.out.println(text);
Output:" <token>abc</token> fg <token>asd</token> "
Note: for simplicity we can assume that the input starts and ends with whitespaces

Use lookaround:
String regex = "(?<=\\s)([a-zA-Z]+)(?=\\s)";
...
text = text.replaceAll(regex, "<token>$1</token>");

If your tokens are only defined with a character class you don't need to describe what characters are around. So this should suffice since the regex engine walks from left to right and since the quantifier is greedy:
String regex = "[a-zA-Z]+";
text = text.replaceAll(regex, "<token>$0</token>");

// meaning not a space, 1+ times
String result = input.replaceAll("([^\\s]+)", "<token>$1</token>");
this matches everything that isn't a space. Prolly the best fit for what you need. Also it's greedy meaning it will never leave out a character that it shouldn't ( it will never find the string "as" in the string "asd" when there is another character with which it matches)

Unescaped java not matching in regex matcher.find()

I have the following code that basically matches "Match this:" and keeps the first sentence. However, there are sometimes unicode characters that get passed into the text that are causing backtracking on other more complicated regex's. Escaping seem to alleviate the backtracking index out of range exceptions. However, now the regex isn't matching.
What i would like to know is why this regex isn't matching when escaped? If you comment out the escape/unescape java lines everything.
String text = "Keep this\n\n"
+ "Match this:\n\nDelete 📱 this";
text = org.apache.commons.lang.StringEscapeUtils.escapeJava(text);
Pattern PATTERN = Pattern.compile("^Match this:$",
Pattern.MULTILINE);
Matcher m = PATTERN.matcher(text);
if (m.find()) {
text = text.substring(0, m.start()).replaceAll("[\\n]+$", "");
}
text = org.apache.commons.lang.StringEscapeUtils.unescapeJava(text);
System.out.println(text);

What i would like to know is why this regex isn't matching when escaped?
When you escape string like "foo\nbar" which printed is similar to
foo
bar
you are getting "foo\\nbar" which printed looks like
foo\nbar
It happens because StringEscapeUtils.escapeJava escapes also \n and is replacing it with \\n, so it is no longer line separator but simple literal, so it can't be matched with ^ or $.
Possible solution could be replacing back "\\n" with "\n" after StringEscapeUtils.escapeJava. You will need to be careful here, not to "unescapee" real "\\n" which after replacing would give you "\\\\n" which printed would look like \\n. So maybe use
text = org.apache.commons.lang3.StringEscapeUtils.escapeJava(text);
text = text.replaceAll("(?<!\\\\)\\\\n", "\n");// escape `\n`
// if it is not preceded with `\`
//do your job
//and now you can unescape your text (\n will stay \n)
text = org.apache.commons.lang3.StringEscapeUtils.unescapeJava(text);
Another option could be creating your own implementation similar to StringEscapeUtils.escapeJava. If you take a look at this method body you will see
return ESCAPE_JAVA.translate(input);
Where ESCAPE_JAVA is
CharSequenceTranslator ESCAPE_JAVA =
new LookupTranslator(
new String[][] {
{"\"", "\\\""},
{"\\", "\\\\"},
}).with(
new LookupTranslator(EntityArrays.JAVA_CTRL_CHARS_ESCAPE())
).with(
UnicodeEscaper.outsideOf(32, 0x7f)
);
and EntityArrays.JAVA_CTRL_CHARS_ESCAPE() returns clone of
String[][] JAVA_CTRL_CHARS_ESCAPE = {
{"\b", "\\b"},
{"\n", "\\n"},
{"\t", "\\t"},
{"\f", "\\f"},
{"\r", "\\r"}
};
array. So if you provide here your own table which will tell explicitly that \n should be left as it is (so it should be replaced with itself \n) your code will ignore it.
So this is how your own implementation can look like
private static CharSequenceTranslator translatorIgnoringLineSeparators =
new LookupTranslator(
new String[][] {
{ "\"", "\\\"" },
{ "\\", "\\\\" },
}).with(
new LookupTranslator(new String[][] {
{ "\b", "\\b" },
{ "\n", "\n" },//this will handle `\n` and will not change it
{ "\r", "\r" },//this will handle `\r` and will not change it
{ "\t", "\\t" },
{ "\f", "\\f" },
})).with(UnicodeEscaper.outsideOf(32, 0x7f));
public static String myJavaEscaper(CharSequence input) {
return translatorIgnoringLineSeparators.translate(input);
}
This method will prevent escaping \r and \n.

how to suppress single and multi-line comments using a regexp?

I need do find all multiline comments in a string and replace them with a space (if the comment is in one line) or with a \n (if the comment is on more than one line).
for example:
int/* one line comment */a;
should be changed to:
int a;
and this:
int/*
more
than one
line comment*/a;
should be changed to:
int
a;
I have one String with all the text and I used this command:
file = file.replaceAll("(/\\*([^*]|(\\*+[^*/]))*\\*+/)"," ");
where file is the string.
The problem is it finds all multiline comment and I want to separate it to 2 cases.
How can I do it?

This can be solved using Matcher.appendReplacement and Matcher.appendTail.
String file = "hello /* line 1 \n line 2 \n line 3 */"
+ "there /* line 4 */ world";
StringBuffer sb = new StringBuffer();
Matcher m = Pattern.compile("(?m)/\\*([^*]|(\\*+[^*/]))*\\*+/").matcher(file);
while (m.find()) {
// Find a comment
String toReplace = m.group();
// Figure out what to replace it with
String replacement = toReplace.contains("\n") ? "\n" : "";
// Perform the replacement.
m.appendReplacement(sb, replacement);
}
m.appendTail(sb);
System.out.println(sb);
Output:
hello
there world
Note: If you want to preserve correct line number / columns for all text that is not inside comments (good if you want to refer back to the source code in error messages etc) I would recommend doing
String replacement = toReplace.replaceAll("\\S", " ");
which replaces all non-whitespace with white space. This way \n is preserved, and
"/* abc */"
is replaced by
" "

Java - New Line Character Issue

I have a code like,
String str = " " ;
while( cond ) {
str = str + "\n" ;
}
Now, I don't know why at the time of printing, the output string is not printing the newline character. However, when I add any other character like ( str = str + "c"), it is printing properly. Can anybody help me, how to solve this problem and why this happening ?

The newline character is considered a control character, which doesn't print a special character to the screen by default.
As an example, try this:
String str = "Hi";
while (cond) {
str += "\n"; // Syntactically equivalent to your code
}
str += "Bye";
System.out.println(str);

Looks like you are trying to run the above code on Windows. Well the line separator or new line is different on Windows ( '\r\n' ) and Unix flavors ('\n').
So, instead of hard coding and using '\n' as new line. Try getting new line from the system like:
String newLine = System.getProperty("line.separator");
String str = " " ;
while( cond ) {
str = str + newLine ;
}

If you really want \n, to get printed, do it like this.
String first = "C:/Mine/Java" + "\\n";
System.out.println(first);
OUTPUT is as follows :
For a good reference as to why is this happening, visit JAVA Tutorials
As referred in that TUTORIAL : A character preceded by a backslash is an escape sequence, and has a special meaning to the compiler. When an escape sequence is encountered in a print statement, the compiler interprets it accordingly
Hope this might help.
Regards

Based on your sample, the only reason it would not show a new line character is that cond is never true and thus the while loop never runs...