问题使用sed修改文件

Question

I have a problem with the following code, if I make the 1.log file but does not modify the files containing PHP eval (..). shows me the following: sed: no input files.

In that I'm wrong?

grep -lr --include=*.php "eval(base64_decode" ./ >> 1.log | xargs sed -i.bak 's/<?php eval(base64_decode[^;]*;/<?php\n/g'

Answer 1

You get the error message because your grep command does not produce any output (which you pass to sed via xargs). You probably wanted this:

grep -lr --include=*.php "eval(base64_decode" ./ | xargs sed -i.bak 's/<?php eval(base64_decode[^;]*;/<?php\n/g'

Alternative with a file list:

grep -lr --include=*.php "eval(base64_decode" ./ > files
sed -i.bak 's/<?php eval(base64_decode[^;]*;/<?php\n/g' $(cat files)

Answer 2

If you want to save the output in the log file and also pipe it to xargs , use tee :

grep -lr --include=*.php "eval(base64_decode" ./ |
    tee -a 1.log |
    xargs sed -i.bak 's/<?php eval(base64_decode[^;]*;/<?php\n/g'

The -a option makes tee append to the file.