搜索文件中带有特定文本的行并输出文件名（文件命令和grep）

Question

I know this question may be simple to some of you, but I have tried several combinations and googled a lot, without success.

Problem: I have a bunch of files with a given file name, but in different directories.

For example, I have a file called 'THEFILE.txt' in directories a, b, c, d. I am in a directory that has these as subdirectories. In each of 'THEFILE.txts' I am looking for lines with the following pattern :'Has this property blah blah blah _apple'. So what I know for sure about the line is that it starts with 'Has this property ' and ends with '_apple'.

I tried:

1. find . -name 'THEFILE.txt' -exec grep -l 'Has this property' {} \\; This works, but I get each and every line with 'Has this property'. I only want ones with _apple at the end

2. So I tried: find . -name 'THEFILE.txt' -exec grep -l 'Has this property*_apple' {} \\; //Does not work, and from my google searches, I don't expect it to.

3. So, next I tried: find . -name 'THEFILE.txt' -exec grep -l 'Has this property[!-~]*_apple' {} \\; //DOES NOT WORK

4. find . -name 'THEFILE.txt' -exec grep 'Has this property' {} \\; | grep '_apple$' //This outputs all matching lines, but not the file names

5. find . -name 'THEFILE.txt' -exec grep 'Has this property' {} \\; | grep -l '_apple$' //Says file is stdin

Expected output: (say files a and c have desired lines)

./a/THEFILE.txt
./c/THEFILE.txt

Answer 1

Your attempt 2. was almost there. With a little adjustment:

find . -name THEFILE.txt -exec grep -q '^Has this property.*_apple$' '{}' ';' -print

it is more precise than recursive grepping and simpler (no pipeline).

The reason why the above works (as opposed to grep -l ), is that the -exec action evaluates to whatever the exit status of its command was. grep will exit with 0 status (which is true ) if it finds what it looked for, that will make -exec yield true , and that in turn will cause the next action ( -print ) to be taken too.

Answer 2

您可以只使用grep -r，如果只需要文件名，则将其通过管道传输到awk中。

grep -r "_apple$" . | awk -F: '{print $1}'