如何解析变量中的Json [重复]答案

【问题标题】：How to parse the Json which is in the variable [duplicate]如何解析变量中的Json [重复]
【发布时间】：2019-12-18 10:57:58
【问题描述】：

json={"PlatformID":1024,"SystemId":11640,"SystemName":"010.10.10.010","DomainName":null,"AccountId":15631,"AccountName":"merg1","AccountNameFull":"merg1","ApplicationID":null,"ApplicationDisplayName":null,"MaximumReleaseDuration":120,"MaxReleaseDurationDays":0,"MaxReleaseDurationHours":2,"MaxReleaseDurationMinutes":0,"InstanceName":"","DefaultReleaseDuration":120,"DefaultReleaseDurationDays":0,"DefaultReleaseDurationHours":2,"DefaultReleaseDurationMinutes":0,"LastChangeDate":"2019-08-21T10:53:25.237","NextChangeDate":null,"IsChanging":false,"IsISAAccess":false,"PreferredNodeID":"3ef3e7c7-5851-451b-b1a4-c62556b588ce"}

grep SystemId | awk -F ':' '{print $2}' $json

读取时出现错误 - 文件名太长。

谁能帮助我如何获取 SystemID。

【问题讨论】：

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Parsing JSON with Unix tools、How to grep the specific field value from json string using shell script、Extract JSON object using Bash shell?、How to parse JSON with shell scripting on Linux? 等可能重复
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标签： json linux shell

【解决方案1】：

看起来$json 的内容被解释为文件名并被报告为长。

试试：echo $json | awk -F ':' '{print $3}' | cut -d ',' -f1

【讨论】：