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regexstringbashunixcut

How to cut a string from a string

My script gets this string for example:

/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file

let's say I don't know how long the string until the /importance.

I want a new variable that will keep only the /importance/lib1/lib2/lib3/file from the full string.

I tried to use sed 's/.*importance//' but it's giving me the path without the importance....

Here is the command in my code:

find <main_path> -name file | sed 's/.*importance//

I am not familiar with the regex, so I need your help please :)

Sorry my friends I have just wrong about my question, I don't need the output /importance/lib1/lib2/lib3/file but /importance/lib1/lib2/lib3 with no /file in the output.

Can you help me?

Solution

  • I would use awk:

    $ echo "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file" | awk -F"/importance/" '{print FS$2}'
importance/lib1/lib2/lib3/file

    Which is the same as:

    $ awk -F"/importance/" '{print FS$2}' <<< "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file"
importance/lib1/lib2/lib3/file

    That is, we set the field separator to /importance/, so that the first field is what comes before it and the 2nd one is what comes after. To print /importance/ itself, we use FS!

    All together, and to save it into a variable, use:

    var=$(find <main_path> -name file | awk -F"/importance/" '{print FS$2}')

    Update

    I don't need the output /importance/lib1/lib2/lib3/file but /importance/lib1/lib2/lib3 with no /file in the output.

    Then you can use something like dirname to get the path without the name itself:

    $ dirname $(awk -F"/importance/" '{print FS$2}' <<< "/dir1/dir2/dir3.../importance/lib1/lib2/lib3/file")
/importance/lib1/lib2/lib3