grep输出到数组 [英] grep output into array

问题描述

伙计们，我该如何做

`查找/xyz/abc/music/| grep def`

`find /xyz/abc/music/ |grep def`

我不想将数组存储在任何临时变量中.我们如何直接在此数组上进行操作.

I don't want to store the array in any temporary variable. How can we directly operate on this array.

因此获得该数组的第一个元素

so to get the 1st element of that array

echo $ {$(`find/xyz/abc/music/| grep def`)[0]}请帮帮我

echo ${$(`find /xyz/abc/music/ |grep def`)[0]} Please help me How I can achieve this

推荐答案

如果只需要第一个元素(或者更确切地说是行)，则可以使用 head :

If you just need the first element (or rather line), you can use head:

`find /xyz/abc/music/ |grep def | head -n 1`

如果需要访问任意元素，则可以先存储数组，然后检索元素:

If you need access to arbitrary elements, you can store the array first, and then retrieve the element:

arr=(`find /xyz/abc/music/ |grep def`)
echo ${arr[n]}

但这不会将grep输出的每一行都放入数组的单独元素中.

but this will not put each line of grep output into a separate element of an array.

如果您只关心整行而不是单词，则可以使用 head 和 tail 来完成此任务，如下所示:

If you care for whole lines instead of words, you can use head and tail for this task, like so:

`find /xyz/abc/music/ |grep def | head -n line_number | tail -n 1`

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