BASH：将 curl 参数作为 $1 变量传递

Question

我有一个具有以下功能的 bash 脚本：

function curl_the_URL_and_check_status(){

   status=$(curl $1 | grep "X-Cache-Status:" | cut -d " " -f 2)

   if [[ "$status" != *"MISS"* ]]; then
       echo "
   cURL returned non MISS status. Something is not correct. Exiting with status 1
   check the status by entering curl $1"
       exit 1
   fi
}

传递参数：

## My comment :: .... some more output ....
+++ grep -o -P '(?<=\/\/).*?(?=\/)'
++ host=php-mindaugasb.c9.io
+++ echo http://php-mindaugasb.c9.io/Testing/JS/displayName.js
+++ perl -pe 's|(?<=://).+?(?=/)|localhost:805|'
++ modified_URL=http://localhost:805/Testing/JS/displayName.js

## My comment ::  below is the parameter passed to the function as $1
++ cURL_string='-H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js'

我把它传递给 curl：

++ echo -H '"Host:' 'php-mindaugasb.c9.io"' -I -s http://localhost:805/Testing/JS/displayName.js

从控制台尝试哪个不起作用（引发网关超时错误）。

所以我的卷曲看起来像这样：

curl -H '"Host:' 'php-mindaugasb.c9.io"' -I -s http://localhost:805/Testing/JS/displayName.js

我需要它看起来像这样（从控制台测试时有效）：

curl -H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js

我该如何做到这一点？

尝试使用 `curl $1`, curl "$1" ...

谢谢

附录

我这样调用函数：

 # another function that constructs correct CURL string
 cURL_string="\"-H Host: $host\" -I -s $modified_URL"

 # global scope - this is where the curl is called
 curl_params=$(get_prepared_string_for_cURL $1)
 curl_the_URL_and_check_status $curl_params

（更新日期：2015 年 1 月 14 日）

这是我使用数组方法得到的结果：

cURL_string=(-H \"Host: $host\" -I -s $modified_URL)

案例：

curl "${curl_params[@]}" ==> curl '-H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js'

curl：没有指定 URL！

curl ${curl_params[@]} ==> curl -H '"Host:' 'php-mindaugasb.c9.io"' -I -s http://localhost:805/Testing/JS/displayName.js

我需要

curl -H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js

get_prepared_string_for_cURL

function get_prepared_string_for_cURL(){

    # get the host from URL, to use with in curl with the --Host flag
    host=$(echo $1 | grep -o -P '(?<=\/\/).*?(?=\/)')

    # replace the host part with the "localhost:805" to request the resource
    # from the nginx virtual host (server block) dedicated for proxy cache
    modified_URL=$(echo $1 | perl -pe 's|(?<=://).+?(?=/)|localhost:805|')

    # construct cURL string
    cURL_string=(-H Host: $host -I -s $modified_URL)

    # echo "$cURL_string"

    echo "${cURL_string[@]}"
}

Answer 1

shell 解析引号之前 替换变量引用（例如$1），所以如果$1 的值中有引号，当它们到位时为时已晚他们做任何有用的事情。与其将 curl 参数作为嵌入引号的单个参数传递，不如将其作为一系列参数传递并使用 "$@" 来扩展它：

function curl_the_URL_and_check_status(){

   status=$(curl "$@" | grep "X-Cache-Status:" | cut -d " " -f 2)
[...]

...然后用类似这样的方式调用它：

curl_the_URL_and_check_status -H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js

代替：

curl_the_URL_and_check_status '-H "Host: php-mindaugasb.c9.io" -I -s http://localhost:805/Testing/JS/displayName.js'

但看起来您也在变量中构建参数列表，这会导致完全相同的问题 - 没有好的方法可以获取普通变量并根据嵌入的引号将其拆分为参数。同样，有一个解决方案：使用一个数组，每个参数都是数组的一个元素。然后，将数组引用为"${arrayname[@]}"，这样每个元素都会被视为一个单独的参数。

cURL_args=(-H "Host: php-mindaugasb.c9.io" -I -s "$modified_URL")
curl_the_URL_and_check_status "${cURL_args[@]}"

Answer 2

你可以使用eval先解释命令字符串中的变量，然后再执行它

更新：

eval [arg ...]
          The args are read and concatenated together into a single command.  This command is then read and  exe‐
          cuted  by  the  shell,  and its exit status is returned as the value of eval.  If there are no args, or
          only null arguments, eval returns 0.

这样你就可以在运行后构建一个类似command="curl -H \"Content-Type: whatever\" $1 $2" 的命令字符串

eval ${command}

eval 将首先读取并解释所有引用、转义和变量，然后运行解释的命令。希望这可以帮助。问候。