使用grep查找匹配项超过1个的文件

Question

I'm wondering if there's a way to list files with more than 1 match using grep. Here's what i'm doing. I'm trying to go through a client's website and replace the last match with a string. so at the bottom of every page, he has this line:

<p align="center"><font face="Arial" color="#808080" size="2">

which is followed by a footer. I need to replace it (using sed) with this line:

<?php include 'footer.php'; ?>

but i want to isolate the files that only have 1 instance of that first line, so i can replace them. Another solution to my problem would be to use sed on only the last match, or use sed on the last 50 lines of a file. any ideas for me?

Answer 1

You can actually use grep -c to get a count of matches in a file:

grep -c '<p align="center"><font face="Arial" color="#808080" size="2">' file

To store its output

cnt=$(grep -c '<p align="center"><font face="Arial" color="#808080" size="2">' file)

Answer 2

I would use this:

for a in *; do one=\`grep exit $a | wc -l\`; if [ $one = 1 ]; then echo replace in $a; fi; done

and in place of echo , do your replacement.