How can i grep only one word from a file, not the whole string, if i dont have fixed position ( cut cannt help me) Ex. 11,40 * * * * /usr/local/bin/apache_test.sh -s i need only script name as variable, but position is always another (crontab) BIG THANKS 4el

Id do it a messy, but it would work grep "word" file | tr "[ ]" "[\n]" | grep "word"

Actually, that was pretty creative. The following awk solution will print each word in the file that contains a slash. It could easily be adjusted to print only the first qualifying word in each line, or whatever: awk '{for (w=1;w<=NF;w++) \ if (match($w,"/")) \ print $w}' myfile

Aye, the example I gave can also be easily modified (forgot to mention that) In thr first bracket put a slash: grep "word" file | tr "[/]" "[\n]" | grep "word" And that would also work fine

cool, but i made a small mistake, i need the whole path to the script:) sorry boys, how can i do this?:) output must be smt like this -> /usr/local/apache_test.sh, a not a single word. Any idea? BIG THANKS (Episode II)

and smt more :) it can be a flag after script that i dont need, so how can have only the path, smt like grep from slash until space. thx

If its going to be the path everytime, all you need to do is use grep to produce that line then awk: awk '{print $6}' To print that path and the scriptname.