使用bash变量将多个标头传递给curl命令 [英] Using a bash variable to pass multiple headers to curl command

问题描述

我想发出带有多个标头的 curl 请求。解决方案是发出以下命令：

I would like to make curl request with multiple headers.The solution would be to make this command :

curl -H "keyheader: value" -H "2ndkeyheader: 2ndvalue" ...

我的目标是仅对所有标头使用一个变量：

My goal is to use only one variable with all the headers like :

headers='-H "keyheader: value" -H "2ndkeyheader: 2ndvalue" '
curl $headers

to发送

curl -H "keyheader: value" -H "2ndkeyheader: 2ndvalue"

当前，问题是：我可以使用'或 声明我的字符串，但是 bash 尝试运行-H 之后的内容参数和答案：

Currently, the problem is : I can use ' or " to declare my string, but bash tries to run what's after "-H" as arguments and then answers :

command unknown

想知道这里出了什么问题。

Would like to know what is going wrong here.

推荐答案

您只需要使用数组和 not 变量以传递引号字符串。

You just need to use an array and not a variable to pass the quoted strings.

declare -a curlArgs=('-H' "keyheader: value" '-H' "2ndkeyheader: 2ndvalue")

现在以这种方式完全传递此数组，数组扩展（带双引号）将处理

and now pass this array fully that way, the array expansion(with double-quotes) takes care of the arguments within double-quotes to be not split while passing.

curl "${curlArgs[@]}"

有关为何将args放入变量失败的更多信息，请参见 BashFAQ / 050-我试图将命令放入变量中，但复杂的情况总是会失败！

For more insight into why putting the args in your variables fail, see BashFAQ/050 - I'm trying to put a command in a variable, but the complex cases always fail!

这篇关于使用bash变量将多个标头传递给curl命令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！