如何将grep字符串输出到bash中的数组元素 [英] How to get grep string output to array elements in bash

问题描述

我目前有验证码

descarray=($(grep -oP "(?<=description\"\:)(.*?)(?=\}})" descfile.json))

但是当我尝试这样做时，我可以正确地找到匹配项，但是由于它是一个带空格的字符串，因此会将每个单词分隔为数组中的元素.

but when I try this I get the match correctly but since it is a string with whitespace, it separates each word as element in array.

匹配的字符串示例为:

"*No_Request_Validation* issue exists @ some other information here""another example goes here"

但是我会得到的是

"*No_Request_Validation*
issue
exists
@
some
...

每个必需元素的开头和结尾都有引号，我想将其与引号分开. 例如:

There are quotes at the start and at the end of each required elements and I would like to separate them with it. for example:

descarray[0]: "*No_Request_Validation* issue exists @ some other information here"
descarray[1]: "another example goes here"

推荐答案

您遇到了单词拆分的问题，该单词拆分IFS上的标记，默认情况下包括换行符，制表符和空格.要将grep的输出读取到由换行符分隔的数组中，请考虑mapfile:

You're running up against wordsplitting, which splits tokens on IFS, which includes newlines, tabs and spaces by default. To read the output of grep into an array split by newlines, consider mapfile:

mapfile -t descarray < <(grep -oP "(?<=description\"\:)(.*?)(?=\}})" descfile.json))

例如，

$ mapfile -t foo <<< '1 2 3
4 5 6'
$ echo "${#foo[@]}"  # we should see two members in the array
2
$ echo "${foo[1]}"  # The second member should be '4 5 6'
4 5 6

(请注意，使用流程替代代替管道.这对于防止隐式操作很重要.从您的descarray变量中删除子外壳.)

(Note the use of process substitution instead of a pipe. This is important to prevent an implicit subshell from eating your descarray variable.)

您可以使用help mapfile或 Bash参考手册.

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