cURL return HTML error 404 instead of JSON when called within bash script requesting php API

Question

when I use cURL directly in terminal like this curl -S -v -d codemeli=12345 https://example.com/api/score-api.php everything OK! when I use cURL directly in terminal like this curl -S -v -d codemeli=12345 https://example.com/api/score-api.php everything OK!
But when I called it inside .sh file it keeps returning HTTP/2 404但是当我在.sh文件中调用它时，它一直返回HTTP/2 404

here is the Bash script:这是Bash脚本：

#! /usr/bin/bash
if [ $# = 2 ]; then
    set -- "https://example.com/api/"${1}"-api.php"
#   t1=`date +%s`
    curl -S -v -d "$2" "$1"
else
    echo "you should specify only two arguments"
fi

Line 3 trying to update variable $1 and add usual URL before & after it.第 3 行尝试更新变量$1并在其前后添加通常的 URL 。

Output difference: Output区别：

direct in terminal:直接在终端：
- verbose log:详细日志：

> POST /api/score-api.php HTTP/2
> Host: example.com
> Content-Length: 14
 * We are completely uploaded and fine
< HTTP/2 200

various line deleted in output!输出中删除了各种行！

output content: output内容：

{"code":false,"text":"No student exist with this NationalID"}

within Bash: Bash 内：
- verbose log:详细日志：

> POST /api/score-api.php HTTP/2
> Host: example.com
> Content-Length: 0
< HTTP/2 404
* HTTP error before end of send, stop sending

Second line ( We are completely uploaded and fine ) gone!第二行（ We are completely uploaded and fine ）不见了！

output content: output内容：

<!DOCTYPE html>
<html style="height:100%">
<head>
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no" >
<title> 404 Not Found
</title></head>
<body style="color: #444; margin:0;font: normal 14px/20px Arial, Helvetica, sans-serif; height:100%; background-color: #fff;">
<div style="height:auto; min-height:100%; ">     <div style="text-align: center; width:800px; margin-left: -400px; position:absolute; top: 30%; left:50%;">
        <h1 style="margin:0; font-size:150px; line-height:150px; font-weight:bold;">404</h1>
<h2 style="margin-top:20px;font-size: 30px;">Not Found
</h2>
<p>The resource requested could not be found on this server!</p>
</div></div></body></html>

Is there problem of POST upload or my bash script wrong? POST上传有问题还是我的bash脚本有问题？

Answer 1

With:和：

set -- "https://example.com/api/"${1}"-api.php"

You replace all the arguments list with only one argument (the URL).您只需用一个参数（URL）替换所有 arguments 列表。 Then argument $2 becomes empty as it calls curl :然后参数$2在调用curl时变为空：

curl -S -v -d "$2" "$1"

You don't need to use set -- with bash as it has arrays.您不需要使用set -- bash，因为它有 arrays。 It is not needed even for your usage since it does not deal with array data at all.即使您使用它也不需要它，因为它根本不处理数组数据。 Anyway, here is your code fixed while still using set --无论如何，这是您在使用set --

#! /usr/bin/bash
if [ $# = 2 ]; then
    set -- "https://example.com/api/"${1}"-api.php" "$2"
#   t1=`date +%s`
    curl -S -v -d "$2" "$1"
else
    echo "you should specify only two arguments"
fi

Here is equivalent code without using set -- and making arguments immutable, witch is a better coding practice:这是不使用set --并使 arguments 不可变，这是一种更好的编码实践：

#! /usr/bin/bash
if [ $# = 2 ]; then
    url="https://example.com/api/$1-api.php"
#   t1=`date +%s`
    data="$2"
    curl -S -v -d "$data" "$url"
else
    echo "you should specify only two arguments"
fi

cURL return HTML error 404 instead of JSON when called within bash script requesting php API

问题描述

Output difference: Output区别：

1 个解决方案

解决方案1
2 已采纳 2020-07-22 15:18:11

cURL return HTML error 404 instead of JSON when called within bash script requesting php API

问题描述

Output difference: Output区别：

1 个解决方案

解决方案1 2 已采纳 2020-07-22 15:18:11

解决方案1
2 已采纳 2020-07-22 15:18:11