By default grep command gives the filename and the line contains the search word. What command should i use to display only the filenames?

I have another question. What command should i use to display a count of all of filenames in my current working directory that start with the string "Foo" and end with an odd digit?

Thank you very much if you help me.

For the first question:

grep -l 'pattern' *

For the second, try something like this:

ls -1 | grep '^Foo.*[13579]' | wc -l

Thank you for your help!

Could you explain the command "ls -1 | grep '^Foo.*[13579]' | wc -l", step by step?

What is the different between ls -1 and ls?

man ls----always check the man pages first.....

ls -l (long listing)
piped to grep
grep looks for lines beginning with "Foo" followed by any character, and then one of the characters 1,3,5,7,9

then piped to wc (word count)

To see if I did this right--and to learn much more, got to tldp.org and get "Bash guide for Beginners" by Machtelt Garrels.

'ls -1' (that's the number one) gives you a listing of all the files in a single column, good for counting. A regular 'ls' will output the listing with spaces between the filenames, but the listing goes across so it's (slightly) harder to count the number of files.

the 'wc -l' command counts lines fed to it, as opposed to just counting words. You could use the regular 'ls' command and pipe the output to 'wc' with no -l option, but it would miscount files that had spaces in their names. If none of your files had spaces then you could use the command like this:

ls | grep '^Foo.*[13579]' | wc

However, the output of 'wc' without the -l option includes the number of lines, bytes, and words, so you'd have to know what to look at to pull out the value you need. The original method is more direct and could be fed to something else that just needed the line count.