We Keep Coding

Replacing quotes in a Java String only on specific places

We have a String as below.
\config\test\[name="sample"]\identifier["2"]\age["3"]
I need to remove the quotes surrounding the numbers. For example, the above string after replacement should look like below.
\config\test\[name="sample"]\identifier[2]\age[3]
Currently I'm trying with the regex as below
String.replaceAll("\"\\\\d\"", "");
This is replacing the numbers also. Please help to find out a regex for this.

You can use replaceAll with this regex \"(\d+)\" so you can replace the matching of \"(\d+)\" with the capturing group (\d+) :
String str = "\\config\\test\\[name=\"sample\"]\\identifier[\"2\"]\\age[\"3\"]";
str = str.replaceAll("\"(\\d+)\"", "$1");
//----------------------^____^------^^
Output
\config\test\[name="sample"]\identifier[2]\age[3]
regex demo
Take a look about Capturing Groups

We can try doing a blanket replacement of the following pattern:
\["(\d+)"\]
And replacing it with this:
\[$1\]
Note that we specifically target quoted numbers only appearing in square brackets. This minimizes the risk of accidentally doing an unintended replacement.
Code:
String input = "\\config\\test\\[name=\"sample\"]\\identifier[\"2\"]\\age[\"3\"]";
input = input.replaceAll("\\[\"(\\d+)\"\\]", "[$1]");
System.out.println(input);
Output:
\config\test\[name="sample"]\identifier[2]\age[3]
Demo here:
Rextester

You can use:
(?:"(?=\d)|(?<=\d)")
and replace it with nothing == ( "" )
fast test:
echo '\config\test\[name="sample"]\identifier["2"]\age["3"]' | perl -lpe 's/(?:"(?=\d)|(?<=\d)")//g'
the output:
\config\test\[name="sample"]\identifier[2]\age[3]
test2:
echo 'identifier["123"]\age["456"]' | perl -lpe 's/(?:"(?=\d)|(?<=\d)")//g'
the output:
identifier[123]\age[456]
NOTE
if you have only a single double quote " it works fine; otherwise you should add quantifier + for both beginning and end "
test3:
echo '"""""1234234"""""' | perl -lpe 's/(?:"+(?=\d)|(?<=\d)"+)//g'
the output:
1234234

How to remove comma after a word pattern in java

Please help me out to get the specific regex to remove comma after a word pattern in java.
Assume, I would like to delete comma after each pattern where the pattern is <Word$TAG>, <Word$TAG>, <Word$TAG>, <Word$TAG>, <Word$TAG> now I want my output to be <Word$TAG> <Word$TAG> <Word$TAG> <Word$TAG> . if I used .replaceAll(), it will replace all commas, but in my <Word$TAG> Word may have a comma(,).
For example, Input.txt is as follows
mms§NNP_ACRON, site§N_NN, pe§PSP, ,,,,,§RD_PUNC, link§N_NN, ....§RD_PUNC, CID§NNP_ACRON, team§N_NN, :)§E
and Output.txt
mms§NNP_ACRON site§N_NN pe§PSP ,,,,,§RD_PUNC link§N_NN ....§RD_PUNC CID§NNP_ACRON team§N_NN :)§E

You could use ", " as search and replace it with " " (space) as below:
one.replace(", ", " ");
If you think, you have "myString, ,,," or multiple spaces in between, then you could use replace all with regex like
one.replaceAll(",\\s+", " ");

(?<=[^,\s]),
Try this.Replace by empty string.See demo.
http://regex101.com/r/lZ5mN8/5

Match the data you want, not the one you don't want.
You probably want ([^ ]+), and keep the bracketed data, separated by whitespace.
You might even want to narrow it down to ([^ ]+§[^ ]+),. Usually, stricter is better.

You could use a positive lookahead assertion to match all the commas which are followed by a space or end of the line anchor.
String s = "mms§NNP_ACRON, site§N_NN, pe§PSP, ,,,,,§RD_PUNC, link§N_NN, ....§RD_PUNC, CID§NNP_ACRON, team§N_NN, :)§E";
System.out.println(s.replaceAll(",(?=\\s|$)",""));
Output:
mms§NNP_ACRON site§N_NN pe§PSP ,,,,,§RD_PUNC link§N_NN ....§RD_PUNC CID§NNP_ACRON team§N_NN :)§E

search and replace string in java using pattern

Given the string
Content ID [9283745997] Content ID [9283005997] There can be text in between Content ID [9283745953] Content ID [9283741197] Content ID [928374500] There can be valid text here which should not be removed.
I want to remove the text starting Content ID followed by [9283745997] any numbers can be present between square brackets. Eventually I want the result string to be
There can be text in between There can be valid text here which should not be removed.
Could anyone please provide a valid regex to capture this recurring text but the numerals within square brackets are unique?
I appreciate your help!
My soulution to this was :
Pattern p = Pattern.compile("(Content ID \\[\\d*\\] )");
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer();
while(m.find()) {
m.appendReplacement(sb, "");
}
m.appendTail(sb);
System.out.println(sb);

So basically you are trying to remove each of Content ID [one or more digits].
To do this you can use replaceAll("regex","replacement") method of String class. As replacement you can use empty String "".
Only problem that stays is what regex should you use.
to match Content ID just write it normally as "Content ID "
to match [ or ] you will have to add \ before each of them because they are regex metacharacters and you need to escape them (in Java you will need to write \ as "\\")
to represent one digit (character from range 0-9) regex uses \d (again in Java you will need to write \ as "\\" which will result in "\\d")
to say "one or more of previously described element" just add + after definition of such element. For example if you want to match one or more letters a you can write it as a+.
Now you should be able to create correct regex. If you will have some questions feel free to ask them in comments.

Try this one:
(Content ID \[[0-9]+\])
You can test it here: http://regexpal.com/

I would use the regex
Content ID \[\d+\] ?
Implement it like this:
str.replaceAll("Content ID \\[\\d+\\] ?", "");
You can find an explanation and demonstration here: http://regex101.com/r/qD5rJ6

A Little RegEx Help Please - Part 2

Here's the input string:
loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml', '/videos/video-splash-image.gif')
With this RegExp: (?<=')[^']+.xml(?=')
... we get this:
http://www.something.com/videos/JohnsAwesomeCaption.xml
... which is exactly what I wanted. BUT this time I'd like to select the complete string EXCEPT for the above. Basically an inverse selection. The output should look like this:
loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', '', '/videos/video-splash-image.gif')
Thanks!

Do a match of your pattern, get the substring that matched, then do a replace of that exact substring with an empty string (or '', or whatever). If your language has the option to return the indices that matched rather than the text that matched it's even easier because you can just remove the text range identified by the indices.
Some languages provide a "regusb" (regular expression substitute) command that lets you replace a regular expression with another string, which lets you do the match-and-replace all in one step. Different languages may call that command by different names.

Try this if you have to do the task in Regex:
Replace
(.+)(?<=')[^']+\.xml(?=')(.+)
with
$1$2

Not sure which language you are using but in Java you can do:
str = "loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml', '/videos/video-splash-image.gif')";
System.out.println(str.replaceFirst("(?<=')[^']+\\.xml(?=')", ""));
Or in php use this code:
$str = "loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml', '/videos/video-splash-image.gif')";
echo preg_replace("~(?<=')[^']+\.xml(?=')~", "", $str);
OUTPUT (from both)
loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', '', '/videos/video-splash-image.gif')
EDIT
Use following Java code to break the string into 2 parts as per your comments:
Pattern p = Pattern.compile("^(.*)(?<=')[^']+\\.xml(?=')(.*)$");
Matcher m = p.matcher(str);
if (m.find())
System.out.println("Output: " + m.group(1) + "<br /><br />" + m.group(2));
OUTPUT
Matched: loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', '<br /><br />', '/videos/video-splash-image.gif')

How to split this string using Java Regular Expressions

I want to split the string
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
to
name
employeeno
dob
joindate
I wrote the following java code for this but it is printing only name other matches are not printing.
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
Pattern pattern = Pattern.compile("\\[.+\\]+?,?\\s*" );
String[] split = pattern.split(fields);
for (String string : split) {
System.out.println(string);
}
What am I doing wrong here?
Thank you

This part:
\\[.+\\]
matches the first [, the .+ then gobbles up the entire string (if no line breaks are in the string) and then the \\] will match the last ].
You need to make the .+ reluctant by placing a ? after it:
Pattern pattern = Pattern.compile("\\[.+?\\]+?,?\\s*");
And shouldn't \\]+? just be \\] ?

The error is that you are matching greedily. You can change it to a non-greedy match:
Pattern.compile("\\[.+?\\],?\\s*")
^

There's an online regular expression tester at http://gskinner.com/RegExr/?2sa45 that will help you a lot when you try to understand regular expressions and how they are applied to a given input.

WOuld it be better to use Negated Character Classes to match the square brackets? \[(\w+\s)+\w+[^\]]\]
You could also see a good example how does using a negated character class work internally (without backtracking)?