Question

我试图从url querystring获取contentid值。

例如：http://localhost:8080/app?contentid=10

我想得到contentid的值是：我的弹簧控制器上面的url是10，我怎么能得到这个？

请注意，如果我在那里传递任何值（例如10,50,100,200，......等等 - 我可以输入或传递任何数字到url），所以无论我传递给{{1那个值应该进入我的控制器（contentid）。

目前，点击“提交”按钮后，我收到以下错误：

System.out.println(contentid);

我在STS控制台下面了。

http://localhost:8080/app?contentid=

Whitelabel Error Page
This application has no explicit mapping for /error, so you are seeing this as a fallback.

There was an unexpected error (type=Bad Request, status=400).
Validation failed for object='testData'. Error count: 1

控制器：

2018-02-05 13:09:45.186  INFO 8428 --- [nio-8080-exec-1] o.a.c.c.C.[Tomcat].[localhost].[/]       : Initializing Spring FrameworkServlet 'dispatcherServlet'
2018-02-05 13:09:45.187  INFO 8428 --- [nio-8080-exec-1] o.s.web.servlet.DispatcherServlet        : FrameworkServlet 'dispatcherServlet': initialization started
2018-02-05 13:09:45.231  INFO 8428 --- [nio-8080-exec-1] o.s.web.servlet.DispatcherServlet        : FrameworkServlet 'dispatcherServlet': initialization completed in 44 ms

TestData.java：

package com.example.demo;

import java.util.concurrent.ThreadLocalRandom;
import javax.servlet.http.HttpServletRequest;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestParam;

@Controller
public class MyController {

    @GetMapping("/app")
    public String getApp(Model model) {
        return "app";
    }

    @PostMapping("/app")
    public String postApp(@RequestParam(required=false) Integer contentid, @ModelAttribute TestData testData, Model model, HttpServletRequest request) {
        System.out.println(contentid);
        if(contentid == null) {
            contentid = ThreadLocalRandom.current().nextInt(0, 100 + 1);
        }
        model.addAttribute("contentid",contentid);
        return "app";
    }
}

app.html：

package com.example.demo;
import java.io.Serializable; 
public class TestData implements Serializable { 
private static final long serialVersionUID = 1L; 
private String firstname; 
private String secondname; 
private int contentid; 

public TestData() { 
} 

//setters and getters

public TestData(String firstname, String secondname, int contentid) { 
this.firstname = firstname; 
this.secondname = secondname; 
this.contentid = contentid; 
} 

@Override 
public String toString() { 
return String.format("TestData[firstname=%s, secondname=%s, contentid=%d]", firstname, secondname, contentid); 
} 

}

的pom.xml：

<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml"
    xmlns:th="http://www.thymeleaf.org"
    xmlns:sec="http://www.thymeleaf.org/thymeleaf-extras-springsecurity4">
<head>

<title></title>

</head>
<body>

<th:block th:if="${contentid != null}">
<div th:text="${'contentId: ' + contentid}"></div>
</th:block>

    <form action="#" th:action="@{/app(contentid=${contentid})}" th:object="${TestData}"
        method="post">
        <div class="form-group">
            <label for="firstname">First Name: </label>
            <input type="text" class="form-control" id="firstname"
                name="firstname" />
        </div>
        <div class="form-group">
            <label for="secondname">Second Name:</label>
            <input type="text" class="form-control" id="secondname"
                name="secondname" />
        </div>
        <button type="submit" value="Submit">Submit</button>
    </form>
</body>
</html>

Answer 1

TestData.java存在问题。 int contentId不能为空。将其更改为Integer contentId。

如何使用Spring启动从url传递querystring值

1 个答案: