404 没有错误 jQuery AJAX 呼叫控制器 Spring

我是新的b。 Spring. 我正在生成 JSON, 如下所示：

[

 {

 "customer" : "16", "project" : "19",

 "phase" : "47", "approver" : "5",

 "date1" : "", "date2" : "",

 "date3" : "", "date4" : "",

 "date5" : "", "date6" : "",

 "date7" : "", "activity" : "1"

 },



 {

 "customer" : "17", "project" : "20",

 "phase" : "48", "approver" : "3",

 "date1" : "", "date2" : "",

 "date3" : "", "date4" : "",

 "date5" : "", "date6" : "",

 "date7" : "", "activity" : "1"

 }

]

我通过了这一点 JSON 它的控制器 Spring:

$.ajax/{

 type: 'post',

 url: 'NewTimesheet',

 dataType : 'json',

 data: JSON.stringify/jsonObj/,



 success: function/data/ {

 console.log/data/;

 }

}/;

我将请求与控制器进行比较，如下所示：

@RequestMapping/value="NewTimesheet", headers = { "Content-type=application/json" }/

@ResponseBody

public String addNewTimesheet/@RequestBody List<timesheet> timesheet,

 HttpSession session/ {



 logger.info/"timesheet list size is" + timesheet.size///;

 return "success";

}

Timesheet

班级：

public class Timesheet {

 private String project;

 private String phase;

 private String approver;

 private String customer;

 private String date1;

 private String date2;

 private String date3;

 private String date4;

 private String date5;

 private String date6;

 private String date7;

 private String activity;



 //Getters and setters

}

但我的请求没有结论。我的控制台显示如下：

WARN
org.springframework.web.servlet.PageNotFound.handleNoSuchRequestHandlingMethod:142
- 找不到Servlet请求的相应处理程序：路径 '/NewTimesheet', 方法 'POST', 选项
map['[{"customer":"16","project":"19","phase":"47","approver":"5","date1":"","date2":"","date3":"","date4":"","date5":"","date6":"","date7":"","activity":"1"},{"customer":"17","project":"20","phase":"48","approver":"3","date1":"","date2":"","date3":"","date4":"","date5":"","date6":"","date7":"","activity":"1"}]'
-> array['']]

如何比较我的 JSON 与控制器？任何帮助都会非常感激！
</timesheet>

已邀请:

5 个回复

冰洋

赞同来自:

您需要将类作为控制器注释，添加 RequestMapping 在你的班级和方法中 HTTP, 你在你的方法中打电话。

@Controller

@RequestMapping/"/NewTimesheet"/

public class MyClassName {



 @RequestMapping/value={ "", "/" }, method = RequestMethod.POST, headers = { "Content-type=application/json" }/

 @ResponseBody

 public String addNewTimesheet/@RequestBody List<timesheet> timesheet,HttpSession session/{

 logger.info/"timesheet list size is"+timesheet.size///;

 return "success";

 }

}

</timesheet>

窦买办

赞同来自:

一对可能导致您出现问题的事情：

确保您拥有杰克逊的所有必要依赖项。为了 Maven 这将是：

<dependency>

<groupid>com.fasterxml.jackson.core</groupid>

<artifactid>jackson-core</artifactid>

<version>2.2.2</version>

</dependency>

<dependency>

<groupid>com.fasterxml.jackson.core</groupid>

<artifactid>jackson-databind</artifactid>

<version>2.2.2</version>

</dependency>

你不需要涂抹你的对象 JavaScript, 这是隐含的。您还必须调用您所呈现的变量，因为它必须匹配域对象：

$.ajax/{

 method : 'post',

 url : 'NewTimesheet',

 dataType : 'json',

 data:{ 'timesheet': jsonObj },

 success : function/data/ {

 console.log/data/;



 }

}/;

您的控制器必须配置为明确处理 POST. 不需要在标题中安装所接收的内容类型。此外，我相信您需要将域的对象与数组进行比较，而不是列表：

@RequestMapping/value="NewTimesheet", method = RequestMethod.POST/

public @ResponseBody String addNewTimesheet/@RequestParam/"timesheet"/ Timesheet[] timesheet,HttpSession session/{

 logger.info/"timesheet list size is"+timesheet.length/;

 return "success";

}

如果您使用相对最近的版本 Spring MVC, 然后处理请求和创建 JSON 不需要额外的配置。你的申请 AJAX 表示答案 JSON, 哪一个 Spring 识别和委托杰克逊序列化输入和输出。

二哥

赞同来自:

有必要确定

method=post

. 我也加了

produces = "application/json"

@RequestMapping/value="NewTimesheet", method = RequestMethod.POST, produces = "application/json"/

 @ResponseBody

 public String addNewTimesheet/@RequestBody List<timesheet> timesheet,HttpSession session/{

 logger.info/"timesheet list size is"+timesheet.size///;

 return "success";

 }

</timesheet>