grep - 限制读取的文件数

Question

I have a directory with over 100,000 files. I want to know if the string "str1" exists as part of the content of any of these files.

The command: grep -l 'str1' * takes too long as it reads all of the files.

How can I ask grep to stop reading any further files if it finds a match? Any one-liner?

Note: I have tried grep -l 'str1' * | head grep -l 'str1' * | head but the command takes just as much time as the previous one.

Answer 1

Naming 100,000 filenames in your command args is going to cause a problem. It probably exceeds the size of a shell command-line.

But you don't have to name all the files if you use the recursive option with just the name of the directory the files are in (which is . if you want to search files in the current directory):

grep -l -r 'str1' . | head -1

Answer 2

Use grep -m 1 so that grep stops after finding the first match in a file. It is extremely efficient for large text files.

grep -m 1 str1 * /dev/null | head -1

If there is a single file, then /dev/null above ensures that grep does print out the file name in the output.

If you want to stop after finding the first match in any file:

for file in *; do
  if grep -q -m 1 str1 "$file"; then
    echo "$file"
    break
  fi
done

The for loop also saves you from the too many arguments issue when you have a directory with a large number of files.