How to do "grep" on each file in directory
 

Hello,

Can someone show me how I can do the following in a bash script?
1) Load all *.log files into an array variable
2) Iterate through the array, and do a "grep" on each of the file to see if it contains a "word".

Why not simply

grep PATTERN *.log

?

I am with Sergei, I feel you are over thinking your solution.

C'mon guys, bash now supports associative arrays - wouldn't be too difficult to do ....:p

This sounds like a homework question. If you want further help, you should post what you have tried. Look in the bashref manual and read through section 6.7 for information about using array variables.

No this is not a homework question. Unfortunately I can't recall the reason why I was thinking of doing it this way. I have moved on to other projects, and will need to re-visit this at another time. Will post here what I've tried previously.


However, I think it's because of the reason I want to see what it had grepped on what file. So I think I've echoed the file name first, with `grep <word> $filename[count]` ...... Can't remember the exact syntax. But the grep result overwrites part of the echoed file name.

(Re)read 'man grep' - 'grep' has a command line switch allowing to see the file in which a match occurs.

And consider 'find' with '-exec'.

grep -r "word" /directory

- or -

for i in `ls /directory`; do grep "word" $i; done