在Bash中提取JSON值 [英] Extract JSON value in Bash

问题描述

在Bash中，我将响应数据保存到一个变量中.

In the Bash I saved response data into a variable.

结果如下:

{"token_type":"Bearer","access_token":"022-8baa5324-f57b-445d-c5ec-821c63a5fd35","expires_in":3600,"scope":"any-website.com"}

现在我想将访问令牌的值提取到另一个变量中.

Now I want to extract the value of the access token into an other var.

在Linux中，我以这种方式解决了该问题，并且有效:

In Linux I solved that in this way and it works:

echo "$response_json" | grep -oP '(?<="access_token":")[^"]*'

结果我得到:

022-8baa5324-f57b-445d-c5ec-821c63a5fd35

我的问题是MacOS不再支持grep参数P(Perl表达式).参数E不适用于该表达式.

My problem is that MacOS does not support the grep parameter P (Perl expression) anymore. Parameter E does not work with that expression.

我希望在不安装其他Bash工具的情况下提供解决方案方面的帮助.

I would appreciate any help with a solution without requiring to install additional Bash tools.

推荐答案

每个人都说他们不想安装新工具，但实际上，像grep这样的面向行的工具并不是专门为处理结构化文本而设计的.像JSON.如果要使用JSON，请获取旨在处理JSON的工具.

Everyone says they don't want to install new tools, but really, line-oriented tools like grep simply weren't designed to cope with structured text like JSON. If you are going to work with JSON, get tools designed to process it.

jq 是这样的一种选择:

jq is one such option:

$ echo "$response_json" | jq -r '.access_token'
022-8baa5324-f57b-445d-c5ec-821c63a5fd35

这篇关于在Bash中提取JSON值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！