bash - Using grep -E with variables

Question

I am trying to do a search with find and pass the results into grep. Grep has to find the files matched against string1 and string2 and string3.

I have the following command:

#/bin/bash
searchpath="/home/myfolder"
string1="abc"
string2="def"
string3="ghi"
find `echo "${searchpath}"` -type f -print0 | xargs -0 grep -l -E '"${string1}".*"${string2}".*"${string3}"'

But the result is blank, but when I do:

find /home/myfolder -type f -print0 | xargs -0 grep -l -E 'abc.*def.*ghi'

I get results. What am I doing wrong?

Answer 1

从行中删除单引号：

find `echo "${searchpath}"` -type f -print0 | xargs -0 grep -l -E '"${string1}".*"${string2}".*"${string3}"'

即，说：

find "${searchpath}" -type f -print0 | xargs -0 grep -l -E "${string1}".*"${string2}".*"${string3}"

会工作。当你用单引号括起来时，shell 将其解释为：

"${string1}".*"${string2}".*"${string3}"

（不扩展变量）

而且，你不用说

`echo "${searchpath}"` 

说

"${searchpath}"

就足够了。

Answer 2

您不需要使用命令替换。你应该使用""引用变量，而不是'：

find "${searchpath}" -type f -print0 | xargs -0 grep -l -E "${string1}.*${string2}.*${string3}"