How to put a string search with grep command into if statement?

Question

I want to search multiple strings in two files. If one string is found in both files, then make something. If one string is found in only one file, then make another thing.

My commands are the next:

####This is for the affirmative sentence in both files
if grep -qw "$users" "$file1" && grep -qw "$users" "$file2"; then

####This is for the affirmative sentence in only one file, and negative for the other one
if grep -qw "$users" "$file1" ! grep -qw "$users" "$file2"; then

is it a correct the way to deny and affirm the statements? p.d. I'm using KSH shell.

Thank you in advance.

Answer 1

Try this:

if grep -wq -- "$user" "$file1" && grep -wq -- "$user" "$file2" ; then
   echo "string avail in both files"
elif grep -wq -- "$user" "$file1" "$file2"; then
   echo "string avail in only one file"
fi

• grep can search for patterns in multiple files, so no need to use an OR/NOT operator.

Answer 2

Another option:

grep -qw -- "$users" "$file1"; in_file1=$?
grep -qw -- "$users" "$file2"; in_file2=$?

case "${in_file1},${in_file2}" in
    0,0) echo found in both files ;;
    0,*) echo only in file1 ;;
    *,0) echo only in file2 ;;
      *) echo in neither file ;;
esac

Answer 3

n=0

#Or if you have more files to check, you can put your while here. 
grep -qw -- "$users" "$file1" && ((n++))
grep -qw -- "$users" "$file2" && ((n++))

case $n in 
   1) 
       echo "Only one file with the string"
    ;;
   2)
       echo "The two files are with the string"
   ;;
   0)
       echo "No one file with the string"
   ;;
   *)
       echo "Strange..."
   ;;
esac 

Note: ((n++)) is a ksh extension (also supported by zsh and bash). In POSIX sh syntax, you'd need n=$((n + 1)) instead.

Answer 4

If your filenames don't contain newlines, you can avoid multiple invocations of grep by having grep print the names of matching files, and count the results.

 local IFS=$'\n'    # inside a function.  Otherwise use some other way to save/restore IFS
 matches=( $(grep -lw "$users" "$file1" "$file2") )

The number of matches is "${#matches[@]}".

There might be a way to use grep --null -lw here, but I'm not sure how to parse the output. Bash var=( array elements ) don't have a way to use a \0 delimiter instead of \n. Maybe bash's mapfile builtin can do it? But probably not, because you specify the delimiter with -d string.

---

You could count=$(grep -l | wc -l), but then you have two external processes so you might as well just run grep on the two files separately. (The difference between grep vs. wc startup overhead is small compared to fork+exec + dynamic linker stuff to start up a separate process at all).

Also, with wc -l you don't find out which file matched.

---

With the results captured in an array, that might already be what you want, or if there's exactly 1 match, you can check if it was the first input or not.

local IFS=$'\n'    # inside a function.  Otherwise use some other way to save/restore IFS
matches=( $(grep -lw "$users" "$file1" "$file2") )

# print the matching filenames
[[ -n $matches ]] && printf  'match in %s\n'  "${matches[@]}"

# figure out which input position the name came from, if there's exactly 1.
if [[ "${#matches[@]" -eq 1 ]]; then
    if [[ $matches == "$file1" ]];then
        echo "match in file1"
    else
        echo "match in file2"
    fi
fi

$matches is shorthand for ${matches[0]}, the first array element.