Print lines that do not match the pattern

Shell Programming and Scripting
1

[LEFT]I need to print the lines that do not match a pattern. I tried using grep -v and sed -n '/pattern/!p', but both of them are not working as I am passing the pattern as variable and it can be null some times.

Example

........ abcd[1234]......
.........abcd[2341]......
.........abcd[1234]......
.........abcd[4567]......
.........abcd[9876]......

would need the output to be not matching pattern 1234, so it would be

.........abcd[2341]......
.........abcd[4567]......
.........abcd[9876]......

[/LEFT]

The value in the brackets is passed as a variable and sometimes it can be null
so the grep or sed is not returning any value. when it is supposed to.

Please help.

2

Check if pattern is defined and then grep for it:

pattern="1234"

[[ ! -z "$pattern" ]] && grep -v "\[$pattern\]" file
3

Hi

Thank you for the quick reply, this option would not work as the structure of the line is different, as below.

For other purposes, in the whole script, I am extracting the value in the brackets and assigning it to a variable. The value in the brackets is alpha numeric with "-" in between ex: abcd[xyz12-45de-5678se].

I am extracting the value based on a condition. so sometimes it can be null. In that case, the grep will look like grep -v ' '

Thanks

4

Your requirement is not so clear. The command should have worked for the sample data that you posted!

I think it will help if you can post few lines from your original input file and desired output in code tags

5

try also:

awk '! ($0 ~ "[[]" 1234 "[]]")' infile