如何通过一个变量在shell脚本一个curl命令 [英] How to pass a variable in a curl command in shell scripting

问题描述

我有一个curl命令：

I have a curl command:

curl -u ${USER_ID}:${PASSWORD} -X GET 'http://lppma670.gso.aexp.com:8080/rest/job-execution/job-details/${job_id}'

变量作业ID 中有一个值，也就是说，1160当我在shell中执行curl命令它给了我以下错误：

The variable job_id has a value in it, say, 1160. When I execute the curl command in shell it gives me the following error:

{"message":"Sorry. An unexpected error occured.", "stacktrace":"Bad Request. The request could not be understood by the server due to malformed syntax."}

如果我通过直接数字1160中的命令，如下图所示，curl命令的作品。

If I pass the number '1160' directly in the command, as shown below, the curl command works.

curl -u ${USER_ID}:${PASSWORD} -X GET 'http://lppma670.gso.aexp.com:8080/rest/job-execution/job-details/1160'

有人可以帮助我吗？我想能够通过在curl命令变量的值。

Can someone please help me out? I wanna be able to pass the value of the variable in the curl command.

推荐答案

在使用变量的外壳，你只能使用双引号，而不是单引号：单引号里面的变量没有扩大。
学习之间的区别和`。请参见 http://mywiki.wooledge.org/Quotes 并的http://wiki.bash-hackers.org/syntax/words

When using variables in shell, you can only use doubles quotes, not single quotes : the variables inside single quotes are not expanded. Learn the difference between ' and " and `. See http://mywiki.wooledge.org/Quotes and http://wiki.bash-hackers.org/syntax/words

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