I‘m trying to match words using GNU awk command and getting the following error:
echo 'foo bar this that blah' | awk '{gsub("\<regex-word\>", "NEW-WORD");print}'
But getting the following warning on screen and it is not working:
awk: warning: escape sequence `\<' treated as plain `<'
awk: warning: escape sequence `\>' treated as plain `>'
How do I fix this problem under Unix like operating systems?
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I‘ve a file as follows:
This is a test.
One bang two three
Foo dang Bar
001 0xfg 0xA
002 0xA foo bar 0xfG
I'm done
How do I delete all “words” from the above file which ends with a particular letter (say ‘g’) in each line? The output should be as follows:
This is a test.
One two three
Foo Bar
001 0xA
002 0xA foo bar
I'm done
How do I delete regex-based word using sed or awk under Linux / Unix like operating systems?
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I‘m writing a shell script for automation purpose. The output the path of the current working directory is stored in $PWD or it can be obtained using the pwd command. How do I find out find out 3rd field separated by the forward slash (/) delimiter using $PWD under Unix like operating systems?
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How do I read a text file using awk pattern scanning and text processing language under Linux / Unix like operating systems?
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I‘m a new Linux sys admin and I’m unable to find the command to list all users on my RHEL server. What is the command to list users under Linux operating systems?
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I‘d like to skip first two or three fields at the the beginning of a line and print the rest of line. Consider the following input:
This is a test
Giving back more than we take
I want my input file with the following output:
a test
more than we take
How do I printing lines from the nth field using awk under UNIX or Linux operating systems?
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