How to tell if grep didn't return anything

[realtest]

Hi,

I have a file:

979798707
787862348
766428634

I want to see if all the records in the file are present in the contents of the files of a particular directory.

Basically I want to say if grep doesn't return anything, then report.

For example in /tmp dir I have 4 files and flast 2 values (787862348 and 766428634) are present in the files of /tmp dir, but first one (979798707) is not. I want to echo that in a reporting file.

something like:

while read line
do
# if ! grep -rl $line /tmp
echo $line >> are_not_present
done < "myFile"


How do I achieve " if ! grep -rl $line /tmp"? That is, if the line is found by grep, then grep will print the output, but if grep does'nt find it, it will print nothing. How can I check if grep didn't find it (i.e. printed nothing)?

Thanks,

[lomcevak]

Check the return status ($?)

When grep finds something return status is 0
When grep doesn't find something return status is 1

[stokes]

how about something like:

for line in $(cat file); do
 for file in $(ls /tmp); do
  if [ -z "(grep $line /tmp/$file)" ]; then
   echo "I couldn't find $line in /tmp/$file"; 
  else
   echo "I found $line in /tmp/$file";
  fi
 done
done

this is checking for the existance of each line in "file" in every file in /tmp/ and reporting a yes or no. If you don't want to see the "I couldn't find the file", you could change the -z clause to "! -z".

for line in $(cat file); do
 for file in $(ls /tmp); do
  if [ ! -z "(grep $line /tmp/$file)" ]; then
   echo "I found $line in /tmp/$file";
  fi
 done
done

Edit: I should have explained. testing using -z is testing for a blank string. Check out "man test", which explains all the different clauses you can use in if statements.