My test file has text like:
> cat test.txt new dummy("test1", random1).foo("bar1"); new dummy(" test2", random2); new dummy("test3", random3).foo("bar3"); new dummy = dummy( "test4", random4).foo("bar4");
I am trying to match all single lines ending with semicolon (;) and having text “dummy(“. Then I need to extract the string present in the double quotes inside dummy. I have come up with the following command, but it matches only the first and third statement.
> perl -ne 'print if /dummy/ .. /;/' test.txt | grep -oP 'dummy((.|n)*,' dummy("test1", dummy("test3",
With -o flag I expected to extract string between the double quotes inside dummy. But that is also not working. Can you please give me an idea on how to proceed?
Expected output is:
test1 test2 test3 test4
Some of the below answers work for basic file structures. If lines contains more than 1 new line characters, then code breaks. e.g. Input text files with more new line characters:
new dummy("test1", random1).foo("bar1"); new dummy(" test2", random2); new dummy("test3", random3).foo("bar3"); new dummy = dummy( "test4", random4).foo("bar4"); new dummy("test5", random5).foo("bar5"); new dummy("test6", random6).foo( "bar6"); new dummy("test7", random7).foo(" bar7");
I referred to following SO links:
How to give a pattern for new line in grep?
how to grep multiple lines until ; (semicolon)
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Answer
@TLP was pretty close:
perl -0777 -nE 'say for map {s/^s+|s+$//gr} /bdummy(s*"(.+?)"/gs' test.txt
test1 test2
Using
-0777
to slurp the file in as a single string/bdummy(s*"(.+?)"/gs
finds all the quoted string content after “dummy(” (with optional whitespace before the opening quote)- the
s
flag allows.
to match newlines. - any string containing escaped double quotes will break this regex
- the
map {s/^s+|s+$//gr}
trims leading/trailing whitespace from each string.