Spring HTTP Status 400 - 客户端发送的请求在语法上不正确

Question

我有一个表单,用户可以使用该表单将文件上载到本地驱动器,并将一些数据保存到数据库.但是这个错误信息出现了,我想知道如何处理它.

   HTTP Status 400 - The request sent by the client was syntactically incorrect.

控制器

@RequestMapping(value = "/main/user/setter/addpage", method = 

RequestMethod.POST, params = "save")
public String saveProcess(@ModelAttribute("module") Module module, 

@RequestParam("userId") Integer userId, 
@RequestParam("name") String name,
@RequestParam("file") MultipartFile file, 
BindingResult result, HttpSession session) {
    if (result.hasErrors()) {
        return "redirect:/main/user/setter/settingpage";
    }
    else
    if(module != null){

         try {

             MultipartFile filea = module.getFileData();

             InputStream inputStream = null;
             OutputStream outputStream = null;
             if (filea.getSize() > 0) {
             inputStream = filea.getInputStream();
             outputStream = new FileOutputStream("C:\\Test\\"
             + filea.getOriginalFilename());
             System.out.println("Uploaded File Name");
             System.out.println(filea.getOriginalFilename());
                             int readBytes = 0;
             byte[] buffer = new byte[8192];
             while ((readBytes = inputStream.read(buffer, 0, 8192)) != 

                             -1) {
             outputStream.write(buffer, 0, readBytes);
             }
             outputStream.close();
             inputStream.close();
             session.setAttribute("success", "File Uploaded");
             session.setAttribute("uploadFile", "C:\\Test\\"
             + filea.getOriginalFilename());
             }
             } catch (Exception e) {
             e.printStackTrace();
             }

}
// Delegate to service
moduleService.add(userId, module);

return "redirect:/main/user/setter/settingpage";
    }

形式本身

 <c:url var="processUrl" value="/main/user/setter/addpage?userId=2" />
 <form:form modelAttribute="module" method="POST" action="${processUrl}" name="module" 

 enctype="multipart/form-data">

 <form:label path="fileName">Document Title:</form:label>
 <form:input path="fileName" name="name"/><br/>

 <form:label path="documentPath">Coursework Sample:</form:label>
 <form:input path="documentPath" type="file" name="file" id="file" size="43.9"/><br/>

 <form:label path="liveDate">Live Date:</form:label>
 <form:input path="liveDate"/><br/>

  <input type="submit" name="save" value="Save" id="save"/>
  <input type="submit" name="send" value="Send" id="send"/>

 </form:form>

我正在尝试保存表单,我将使用单独的方法发送它.

Answer 1

您的方法声明如下:

public String saveProcess(@ModelAttribute("module") Module module, 
@RequestParam("userId") Integer userId, 
@RequestParam("name") String name,
@RequestParam("file") MultipartFile file, 
BindingResult result, HttpSession session) ...

假设你正试图以应用BindingResult为Module实例,您需要更改的方法定义,以便BindingResult在之后谈到Module实例.

public String saveProcess(@ModelAttribute("module") Module module,  
BindingResult result,
@RequestParam("userId") Integer userId, 
@RequestParam("name") String name,
@RequestParam("file") MultipartFile file, 
HttpSession session) ...

有关详细信息,请参阅参考

Errors或BindingResult参数必须遵循立即绑定的模型对象,因为方法签名可能包含多个模型对象,而Spring将为每个模型对象创建单独的BindingResult实例[...]

看到这里.