Question

我有一个简单的Java REST Web服务 -

@GET
    @Path("/get1")
    @Produces(MediaType.APPLICATION_JSON)
    @Consumes(MediaType.APPLICATION_JSON)
    public Status getstudent(Track track) {
        System.out.println("GET1 title = " + track.getTitle());
        System.out.println("GET1 singer = " + track.getSinger());
        Status status = new Status();
        status.setStatus_flag("success");
        return status;
    }

轨道

public class Track {

    String title;
    String singer;

    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    public String getSinger() {
        return singer;
    }

    public void setSinger(String singer) {
        this.singer = singer;
    }

    @Override
    public String toString() {
        return "Track [title=" + title + ", singer=" + singer + "]";
    }

}

Python请求模块： -

import requests
title = {"title":"Best Songs","singer":"lucky"}
r = requests.get("http://localhost:8080/StudentService/rest/insert/get1",data=title)
print r.content

错误

<html><head></head><body>
<h1>HTTP Status 415 - Unsupported Media Type</h1>
<HR size="1" noshade="noshade">
<p><b>type</b> Status report</p>
<p><b>message</b> <u>Unsupported Media Type</u></p>
<p><b>description</b> 
    <u>The server refused this request because the request entity is 
       in a format not supported by the requested resource for the requested 
       method.
    </u>
</p>
<HR size="1" noshade="noshade"><h3>Apache Tomcat/6.0.41</h3>
</body></html>

有没有办法创建实体对象并发送它？

Answer 1

您需要使用params发送网址参数;您正在尝试发送请求 body 。

您需要更改服务以接受表单参数：

@GET
@Path("/get1")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_FORM_URLENCODED)
public Status getstudent(
        @FormParam("title") String title,
        @FormParam("singer") String sing) {
    System.out.println("GET1 title = " + title);
    System.out.println("GET1 singer = " + singer);
    Status status = new Status();
    status.setStatus_flag("success");
    return status;
}

现在您可以发送URL编码的参数：

params = {"title": "Best Songs", "singer": "lucky"}
r = requests.get(
    "http://localhost:8080/StudentService/rest/insert/get1",
    params=params)

Answer 2

只需在标题中添加内容类型和接受即可。像：

headers = {"Content-Type": "application/json", "Accept": "application/json"}
r = requests.get("http://localhost:8080/StudentService/rest/insert/get1", data=title, headers=headers)

使用Python请求调用Java Web服务

2 个答案: