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]
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:
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
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