UNIX for Dummies Questions & Answers

Hi,

I was able to succesfully search/find the files I needed using the command : find . -exec grep string {} /dev/null \; however,
I would like to have the result written in a log file.

Can someone please show me how to do this?

Thanks in advance.
Bong

Just redirect the output to a log file using >

e.g.

find . -exec grep string {} \; > ./mylog

Matt.

Hi Bong,

In addition again :

for speeding up your find command try to be more specific :

find . -type f -exec grep string {} \;

If you specify to only grep files, it speeds up definitly. Also redirecting output is one of the standards you have to understand. So therefor a little story

When specify-ing > you say that all output generated should be redirected. By specify-ing > you actualy say 1>. As you are telling only standard output to be redirected.

You have :
0 --> Standard input
1 --> Standard output
2 --> Standard error

So to make things a bit more complicated :

ls -l /tmp 1> /tmp/output 2>/dev/null

Standard output is written to a file and standard Error is being thrown away (/dev/null is a kind of paper.bin)

ls -l 1>> /tmp/output 2>&1

Standard output is appended to a file and standard error is redirected to standard output.

Good luck.

Regs David

In addition to davidg's suggestion for find: If you will be finding a lot of files and then grepping them, you can save a good bit more time by feeding more file names to grep at once using xargs, thus:

find . -type f | xargs grep string

This has the side benefit of printing the file names with each file in which your string is found. If there are too many files to put on one command line (for the grep part), xargs will split the names according to your system's command line rules (which vary frustratingly and provide another motivation for using xargs). If there is only one file in the last "group" that xargs feeds to grep, however, its filename won't be printed (unless your grep has an option to force the filename).