Question

我在数据库中保存多行，所以我在jquery中的ajax调用中传递值。

$("button#submitbutton").click(function() {
 var data = [];
 var name, email, message;
 $("table tbody tr").each(function(index) {
name = $(this).find('.name').text();
email = $(this).find('.email').text();
message = $(this).find('.message').text();
//---->Form validation goes here
data.push({
  name: name,
  email: email,
  message: message
});
 });
 submitFormData(data);
});    
function submitFormData(formData) {
var url= '/userrecords';
alert(url);
 $.ajax({
        type: 'POST',
        data: formData,
        cache: false,
        processData: false,
        contentType: false,
        beforeSend: beforeSendHandler,
        url: url,
        success: function(result){
        if(result.success == true) {
          $('.alert-success').show();
          $('.alert-danger').hide();
          $("#successmsg").html(result.msg);
          setTimeout(function() {
            $(".alert-success").alert('close');
          }, 10000);
        } else {
          $('.alert-danger').show();
          $('.alert-success').hide();
          $("#error").html(result.msg);
          setTimeout(function() {
            $(".alert-danger").alert('close');
          }, 10000);
        }
        }
});
}

在Spring MVC中我写了控制器类

@RequestMapping(value = "/userrecords")
public @ResponseBody Response saveList(Model model)
throws ParseException, SQLIntegrityConstraintViolationException {
//read all values here
}

在控制器类中如何访问formData值。我对此并不了解。

Answer 1

    client side:

    data: {model:formData.serialize()},
    cache: false,
    processData: false,
    contentType: false

    server side:1.> your api method should have the 
    same model structure 
    2.> then deserialize your model 
    3.> name of the object also be same like : model here

    @RequestMapping(value = "/userrecords")
    public @ResponseBody Response saveList(Model model)
    throws ParseException, SQLIntegrityConstraintViolationException {
    //read all values here
    }

Answer 2

需要纠正以下几点

使用POST } @RequestMapping(value = "/userrecords,method=RequestMethod.POST")装饰方法
在解析为ajax请求.serialize
确保您的Model班级具有相同结构的姓名，电子邮件和消息

Answer 3

创建dto模型（lombok +你可以使用@NotNull @Size（min = 1）代替@NotEmpty，如果你不想要hibernate依赖，你可以为有效的电子邮件实现自己的注释和验证器）

@Data
@AllArgsConstructor
@NoArgsConstructor
public class DataModel {

    @NotEmpty
    private String name;

    @Email
    private String email;

    @NotEmpty
    private String message;
}

在您的控制器中生成/使用指定请求/响应正文的类型，并且有效注释会在反序列化模型中触发字段验证。

@RequestMapping(value = "/userrecords", method = RequestMethod.POST, 
                consumes = { MediaType.APPLICATION_JSON_VALUE }, produces = { MediaType.APPLICATION_JSON_VALUE } )
@ResponseBody
public Response saveList(@Valid @RequestBody DataModel data, BindingResult result, Model data)
       throws ParseException, SQLIntegrityConstraintViolationException {
      //read all values here
}

如何在Controller类中访问formData值

3 个答案: