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I have a .txt file, which contains different words like blablabla:thatswhationlywant.

The blablabla words are always different. So I need to print out only all characters after the ":" - how can I do it?

Thanks a lot!

You could use grep like so:

grep -o ':.*' file.txt

but that will include the :

I would probably do it with cut as it will be more performant

cut -d: -f2 file.txt

Grep text only only after [word]:, You can cut the first phrase [something]: and get the rest using. grep and sed: $ grep -w "\[onemore\]:" words.txt | sed 's/\[onemore\]://g' i got mad. The construct for positive lookbehind is (?<=text) a pair of parentheses, with the opening parenthesis followed by a question mark, “less than” symbol, and an equals sign. 1. Print next word after pattern match using grep 1.1 Using lookbehind. Let's look at some examples to print next word after pattern match.

use cut tool to filter the output of grep. grep ":" file.txt | cut -d ":" -f 2 will only show the text after the character ':' have a look at the manual page for cut for more infomations

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2 methods to grep & print next word after pattern match in Linux , Many (if not all) Unix tools provide man pages during install. On Red Hat Enterprise Linux-based systems, we can run the following to list grep  grep -E -o ".{0,5}test_pattern.{0,5}" test.txt This will match up to 5 characters before and after your pattern. The -o switch tells grep to only show the match and -E to use an extended regular expression. Make sure to put the quotes around your expression, else it might be interpreted by the shell.

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grep to show lines only after pattern, Though grep expects to do the matching on text, it has no limits on input line length other Although this is straightforward when letters differ in case only via​  Linux version : Oracle Linux 6.5 Shell : bash In the the below text file (someString.text), I want to grep all lines with .sh in it. ie. Only the lines and should be returned.

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