how to check if the first line of file contain a specific string?

Question

I need to write a shell script that find and print all files in a directory which starts with the string: #include. Now, I know how to check if a string is in the file, by using:

for f in `ls`; do
    if grep -q 'MyString' $f; then:
        #DO SOMETHING
    fi

but how can I apply this to the first line? I thought to maybe create a variable of the first line and check if it starts with #include, but I'm not sure how to do this. I tried the read command but I fail to read into a variable.

I'd like to hear other approaches to this problem; maybe awk? Anyway, remember, I need to check if the first line starts with #include, not if it contains that string. That's why I found those questions: How to print file content only if the first line matches a certain pattern? https://stackoverflow.com/questions/5536018/how-to-print-matched-regex-pattern-using-awk they are not completely helping.

Answer 1

It is easy to check if the first line starts with #include in (GNU and AT&T) sed:

sed -n '1{/^#include/p};q'   file

Or simplified (and POSIX compatible):

sed -n '/^#include/p;q'   file

That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.

So, a shell loop for all files (with sed) should be like this:

for file in *
do
    [ "$(sed -n '/^#include/p;q' "$file")" ] && printf '%s\n' "$file"
done

If there are only files (not directories) in the pwd.

If what you need is to print all lines of the file, a solution similar to the first code posted will work (GNU & AT&T version):

sed -n '1{/^#include/!q};p'  file

Or, (BSD compatible POSIXfied version):

sed -ne '1{/^#include/!q;}' -e p  file

Or:

sed -n '1{
           /^#include/!q
         }
         p
       '  file

Answer 2

for file in *; do
  [ -f "$file" ] || continue
  IFS= read -r line < "$file" || [ -n "$line" ] || continue
  case $line in
    ("#include"*) printf '%s\n' "$file"
  esac
done

To print the content of the file instead of its name, replace the printf command with cat < "$file".

If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:

awk '/^#include/{print substr(FILENAME, 3)}; {nextfile}' ./*

Above, we're adding a ./ prefix which we're stripping afterwards in FILENAME to avoid problems with file names containing = characters (or a file called -).

With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.

Or to print the file contents instead of name:

awk 'FNR == 1 {found = /^#include/}; found' ./*

(that one is portable).

Answer 3

for file in *
do
  [ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done

Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.

Answer 4

This question is a perfect example where bash and sed solutions are quite complex, but the task can be much simpler with (GNU) awk:

gawk 'FNR==1 && /^#include/{print FILENAME}{nextfile}' *