将grep输出存储在数组中[重复]

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这个问题在这里已有答案：

I need to search a pattern in a directory and save the names of the files which contain it in an array.

我需要搜索目录中的模式，并将包含它的文件的名称保存在数组中。

Searching for pattern:

搜索模式：

grep -HR "pattern" . | cut -d: -f1

This prints me all filenames that contain "pattern".

这将打印出包含“pattern”的所有文件名。

If I try:

如果我尝试：

targets=$(grep  -HR "pattern" . | cut -d: -f1)
length=${#targets[@]}
for ((i = 0; i != length; i++)); do
   echo "target $i: '${targets[i]}'"
done

This prints only one element that contains a string with all filnames.

这只打印一个包含所有filnames字符串的元素。

output: target 0: 'file0 file1 .. fileN'

But I need:

但是我需要：

 output: target 0: 'file0'
 output: target 1: 'file1'
 .....
 output: target N: 'fileN'

How can I achieve the result without doing a boring split operation on targets?

如果不对目标执行无聊的拆分操作，我怎样才能实现结果？

1 个解决方案

You can use:

您可以使用：

targets=($(grep -HRl "pattern" .))

Note use of (...) for array creation in BASH.

注意在BASH中使用（...）进行数组创建。

Also you can use grep -l to get only file names in grep's output (as shown in my command).