如何使用 cURL POST JSON 数据? [英] How do I POST JSON data with cURL?

问题描述

我使用 Ubuntu 并在其上安装了 cURL.我想用 cURL 测试我的 Spring REST 应用程序.我在 Java 端编写了我的 POST 代码.但是，我想用 cURL 测试它.我正在尝试发布 JSON 数据.示例数据是这样的:

{"value":"30","type":"Tip 3","targetModule":"Target 3","configurationGroup":null,"name":"Configuration Deneme 3","description":null,"identity":"Configuration Deneme 3","version":0,"systemId":3,"active":true}

我使用这个命令:

curl -i -H "接受:应用程序/json" -H "X-HTTP-Method-Override: PUT" -X POST -d "value":"30","type":"Tip 3","targetModule":"Target 3","configurationGroup":null,"name":"Configuration Deneme 3","description":null,"identity":"Configuration Deneme 3","version":0,"systemId":3,"active":true http://localhost:8080/xx/xxx/xxxx

它返回这个错误:

HTTP/1.1 415 不支持的媒体类型服务器:Apache-Coyote/1.1内容类型:文本/html；字符集=utf-8内容长度:1051日期:2011 年 8 月 24 日，星期三 08:50:17 GMT

错误描述是这样的:

<块引用>

服务器拒绝了这个请求，因为请求实体的格式不被请求方法的请求资源支持().

Tomcat 日志:POST/ui/webapp/conf/clear HTTP/1.1" 415 1051

cURL 命令的正确格式是什么?

这是我的 Java 端 PUT 代码(我已经测试了 GET 和 DELETE 并且它们可以工作):

@RequestMapping(method = RequestMethod.PUT)public Configuration updateConfiguration(HttpServletResponse response, @RequestBody Configuration configuration) {//考虑@Valid标签configuration.setName("PUT 工作");//todo 如果发生错误 response.sendError(HttpServletResponse.SC_NOT_FOUND);返回配置；}

解决方案

您需要将 content-type 设置为 application/json.但是 -d(或 --data) 发送内容类型 application/x-www-form-urlencoded，Spring 不接受该内容.

查看 curl 手册页，我认为您可以使用 -H(或 --header):

-H "Content-Type: application/json";

完整示例:

curl --header "Content-Type: application/json";--请求 POST --data '{"username":"xyz","password":"xyz"}' http://localhost:3000/api/login

(-H是--header的缩写，-d是--data的缩写)

请注意，如果您使用 -d，-request POST 是可选，正如 -d 标志所暗示的那样一个 POST 请求.

---

在 Windows 上，情况略有不同.见评论线程.

I use Ubuntu and installed cURL on it. I want to test my Spring REST application with cURL. I wrote my POST code at the Java side. However, I want to test it with cURL. I am trying to post a JSON data. Example data is like this:

{"value":"30","type":"Tip 3","targetModule":"Target 3","configurationGroup":null,"name":"Configuration Deneme 3","description":null,"identity":"Configuration Deneme 3","version":0,"systemId":3,"active":true}

I use this command:

curl -i 
    -H "Accept: application/json" 
    -H "X-HTTP-Method-Override: PUT" 
    -X POST -d "value":"30","type":"Tip 3","targetModule":"Target 3","configurationGroup":null,"name":"Configuration Deneme 3","description":null,"identity":"Configuration Deneme 3","version":0,"systemId":3,"active":true 
    http://localhost:8080/xx/xxx/xxxx

It returns this error:

HTTP/1.1 415 Unsupported Media Type
Server: Apache-Coyote/1.1
Content-Type: text/html;charset=utf-8
Content-Length: 1051
Date: Wed, 24 Aug 2011 08:50:17 GMT

The error description is this:

The server refused this request because the request entity is in a format not supported by the requested resource for the requested method ().

Tomcat log: "POST /ui/webapp/conf/clear HTTP/1.1" 415 1051

What is the right format of the cURL command?

This is my Java side PUT code (I have tested GET and DELETE and they work):

@RequestMapping(method = RequestMethod.PUT)
public Configuration updateConfiguration(HttpServletResponse response, @RequestBody Configuration configuration) { //consider @Valid tag
    configuration.setName("PUT worked");
    //todo If error occurs response.sendError(HttpServletResponse.SC_NOT_FOUND);
    return configuration;
}

解决方案

You need to set your content-type to application/json. But -d (or --data) sends the Content-Type application/x-www-form-urlencoded, which is not accepted on Spring's side.

Looking at the curl man page, I think you can use -H (or --header):

-H "Content-Type: application/json"

Full example:

curl --header "Content-Type: application/json" 
  --request POST 
  --data '{"username":"xyz","password":"xyz"}' 
  http://localhost:3000/api/login

(-H is short for --header, -d for --data)

Note that -request POST is optional if you use -d, as the -d flag implies a POST request.

---

On Windows, things are slightly different. See the comment thread.

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