bash 使用来自 shell 变量的 JSON 内容运行 curl 命令
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Running curl command with JSON content from a shell variable
提问by Borsn
I have the script below...
我有下面的脚本...
And I cannot seem to get the variables to work
而且我似乎无法让变量起作用
#!/bin/bash
info = 'Help...?'
object='{"attachments": [{"title": "ti1","text": $info }]}'
curl -X POST -H 'Content-type: application/json' --data '$object' https://hooks.slack.com/services/xxxx
exit 0
Even --data '$object'doesn't work without $info...as Slack API couldn't read my request.
即使--data '$object'没有$info……也无法工作，因为 Slack API 无法读取我的请求。
How do I fix this?
我该如何解决？
回答by Inian
Use double-quotes when passing shell variables and remove extra spaces in variable assignments.
传递 shell 变量时使用双引号并删除变量赋值中的额外空格。
curl -X POST -H 'Content-type: application/json' --data "$object"
# ^^^^^^^^^^
Use nested quotes to preserve the value inside JSON syntax
使用嵌套引号保留 JSON 语法中的值
info='Help...?'
object='{"attachments": [{"title": "ti1","text": "'"$info"'" }]}'
