While read line with grep

Question

I am trying to report lines found using grep and while.

I know you can use the following to compare a list of strings from inputs.txt and find them in your target file like so:

grep -f inputs.txt file_to_check

What I want is to read each line of the inputted strings and grep them individual in a loop.

So I have tried the following methods:

cat inputs.txt | while read line; do if grep "$line" filename_to_check; then echo "found"; else echo "not found"; fi; done

This returns nothing when I redirect the output to a file.

while read line
do
if grep "$line" file_to_check
  then echo "found"
else 
  echo "not found"
fi
done < inputs.txt

Same as the first one but from what I found is better to do.

I know it iterates line by line because I can replace grep with echo $line and it prints each line; but either method doesn't return anything like grep -f above, instead it shows:

not found
not found
not found
.
. etc.

So what I'm looking for is something where it will iterate through each line and check it via grep using an if statement to determine if grep has actually found it or not. I know I may not have all proper logic but the output for what I want should look something like:

Found *matching line in file_to_check* 
Found *matching line in file_to_check*
Not Found $line *(string that does not match)*
.
. etc.

Solution 1

You can also use && and || operators :

while read line; do
         grep -q "$line" file_to_check  && echo "$line found in file_to_check" || echo "$line not found in file_to_check"
done < inputfile > result.txt

The -q parameter of the grep just outputs a status code :

• if $line is found, it outpouts 0 (True) the command after && will be evaluated
• if not found, it outputs 1 (False) the command after || will evaluated

Solution 2

You can rewrite your final solution into

# Do not need this thanks to tr: file=$(dos2unix inputs.txt)

# Use -r so a line with backslashes will be showed like you want
while read -r line
do 
   # Not empty? Check with test -n
   if [ -n "$(grep "${line}" filename)" ]; then 
      echo "found: ${line}"
   else
      echo "not found: ${line}"
   fi 
done < <(tr -d "\r" < "${file}")

Comments

[Joshua]

Thanks I found the file special carriage returns that grep did not know how to handle. So I copied the content into another test file via VIM and it reported what I wanted. @Skyler, your example helped but it gave some weird output and even test it with my new file that worked after testing. The out put using: while read line; do if [ $(grep "$line" file_to_check) != "" ]; then echo "found: $line" else echo "not found: $line" fi done < inputs.txt gave bash: [: !=: unary operator expected not found: item1 bash: [: !=: unary operator expected not found: item2 etc.

[Skyler]

My apologies @Joshua, the $(command) should have been surrouded by double quotes, answer updated.

[Barmar]

@Joshua If the file contains carriage returns, use dos2unix to convert it to normal Unix newlines.

[Joshua]

Thanks, that was my bad on the double quotes there. I think I have solved my solution and will look into dos2unix. I started using cat inputs.txt | tr -d '\r' > newinputs.txt which worked well too.

[William Pursell]

Do not do if [ "$(grep "$line" filename)" != "" ]. Just do if grep -q "$line" filename.