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java - 使用 POST 调用使用 REST Web 服务返回 415

转载 作者:行者123 更新时间:2023-11-30 08:41:37 24 4
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我对基于 REST 的 Web 服务还很陌生。我正在尝试调用我创建的小型 REST WS,其开头如下所示

    package webServices;

import java.net.UnknownHostException;

import javax.ws.rs.Consumes;
import javax.ws.rs.FormParam;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

import org.json.JSONException;
import org.json.JSONObject;

import com.mongodb.BasicDBObject;
import com.mongodb.DB;
import com.mongodb.DBCollection;
import com.mongodb.DBCursor;
import com.mongodb.MongoClient;



@Path("/login")
public class LoginService {



@Path("/isUp")
@GET
@Produces({ MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON })

public String checkServiceStatus(){

return "up and running";

}

@Path("/authenticate")
@POST
@Produces({ MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON, MediaType.TEXT_PLAIN })
@Consumes(MediaType.APPLICATION_FORM_URLENCODED)
public String authenticateUser(@FormParam("user") String user, @FormParam("password") String pwd){


DB db;
DBCollection coll;

MongoClient mongoClient;
String loginResponse="user does not exist";
try {
mongoClient = new MongoClient( "localhost" , 27017 );

db = mongoClient.getDB( "Hackathon" );
coll = db.getCollection("users");

BasicDBObject filter = new BasicDBObject();
filter.put("user", user);
BasicDBObject selectField = new BasicDBObject();
selectField.put("password", 1);
selectField.put("_id", 0);

DBCursor cursor = coll.find(filter, selectField);
String jsonString = cursor.next().toString();
JSONObject json = new JSONObject(jsonString);
String password = json.getString(user);
System.out.println("password "+password);

if(password.equals(pwd)){
loginResponse="success";
System.out.println("success");
}else{
loginResponse="failure";
System.out.println("failure");
}


} catch (UnknownHostException e) {

e.printStackTrace();
} catch (JSONException e) {

e.printStackTrace();
}


return loginResponse;
}

}

每当我使用表单数据从 Chrome postman 调用相同的 POST 服务时

http://localhost:8080/HackDataEngine/login/authenticate
Content-Type application/json
user admin

password admin

POSTMAN call screenshot

我得到低于响应

<html>
<head>
<title>Apache Tomcat/7.0.67 - Error report</title>
<style>
<!--H1 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:22px;} H2 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:16px;} H3 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:14px;} BODY {font-family:Tahoma,Arial,sans-serif;color:black;background-color:white;} B {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;} P {font-family:Tahoma,Arial,sans-serif;background:white;color:black;font-size:12px;}A {color : black;}A.name {color : black;}HR {color : #525D76;}-->
</style>
</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/7.0.67</h3>
</body>
</html>

最佳答案

您需要在 Postman 中正确设置标题

在您的请求选项卡上,按 Headers 并像这样设置一个新变量。

  • Content-Type -> application/json

关于java - 使用 POST 调用使用 REST Web 服务返回 415,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34997270/

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