jQuery 如何从 Spring MVC 控制器返回一个对象以响应 AJAX 请求？
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How to return an object from a Spring MVC controller in response to AJAX request?
提问by Gorakh
I have to return a list of employees from a controller in response to jQuery AJAX request. How should I do for it?
我必须从控制器返回员工列表以响应 jQuery AJAX 请求。我该怎么办？
My controller:
我的控制器：
@RequestMapping("phcheck")
public ModelAndView pay(@RequestParam("empid") int empid, String fdate, String tdate) {
ModelAndView mav = new ModelAndView("phcheck");
List<Employee> employees = entityManager.createQuery(
"SELECT e FROM Employee e WHERE e.empId = " + empid, Employee.class)
.getResultList();
mav.addObject("employees", employees); // I need this list of employee in AJAX
return mav;
}
AJAX code in the related view:
相关视图中的 AJAX 代码：
$(document).ready(function () {
$("#empid").change(function () {
if ($("#fdate").val() != "" && $("#tdate").val() != "" && $("#empid").val() != "") {
jQuery.ajax({
url: "phcheck.htm?empid=" + $("#empid").val() +
"&&fdate=" + $("#fdate").val() +
"&&tdate=" + $("#tdate").val(),
success: function (data) {
alert(data + "success");
},
error: function (data) {
alert(data + "error");
}
});
} else {
alert("Please fill the from date and to date or select the employee id");
$("#empid .option").attr("selected", "selected");
}
});
});
Thanks in advance.
提前致谢。
回答by Abhishek Nayak
I need this list of employee in ajax
我需要这份 ajax 员工名单
In spring when you need object serialization, de-serialization and message conversion. in that case you need to annotate your controller handler method with @RequestBodyand @ResponseBody.
在 spring 需要对象序列化、反序列化和消息转换时。在这种情况下，您需要使用@RequestBody和注释控制器处理程序方法@ResponseBody。
Where:
在哪里：
- @ResponseBody: will inform spring that try to convert its return value and write it to the http response automatically.
- @RequestBody: will inform spring that try to convert the content of the incoming request body to your parameter object on the fly.
- @ResponseBody：将通知 spring 尝试转换其返回值并将其自动写入 http 响应。
- @RequestBody：将通知 spring 尝试将传入请求正文的内容动态转换为您的参数对象。
in your case you need JSON type, you have to add @ResponseBodyto your method signature or just above the method, and produces and consumes which are optional, like:
在您的情况下，您需要 JSON 类型，您必须添加@ResponseBody到您的方法签名或刚好在方法上方，并且生产和消费是可选的，例如：
@RequestMapping(value="phcheck", method=RequestMethod.GET
produces="application/json")
public @ResponseBody List<Employee> pay(@RequestParam("empid") int empid, String fdate, String tdate) {
//get your employee list here
return empList;
}
and in AJAX call use:
并在 AJAX 调用中使用：
contentType: 'application/json'attribute tells the type of data you're sending. anddataType: jsonattribute tells jquery what content type of response will receive.
contentType: 'application/json'属性告诉您发送的数据类型。和dataType: json属性告诉 jquery 将收到什么内容类型的响应。
in your case contentType: 'application/json'is not needed, default one i.e. 'application/x-www-form-urlencoded; charset=UTF-8'is enough.
在你的情况下contentType: 'application/json'不需要，默认一个 ie'application/x-www-form-urlencoded; charset=UTF-8'就足够了。
and you can receive list of employees in your AJAX success, to iterate over it do like:
并且您可以在 AJAX 成功中接收员工列表，对其进行迭代，例如：
success: function (data) {
$.each(data, function(index, currEmp) {
console.log(currEmp.name); //to print name of employee
});
},
Note:笔记：Hymanson mapperHyman逊映射器或任何其他映射器应该在构建路径上可用，以便进行 JSON 序列化和反序列化。
See Also:
也可以看看：
回答by geoand
@RequestMapping(value = "phcheck", produces = "application/json")
@ResponseBody
public ModelAndView pay(@RequestParam("empid") int empid, @RequestParam("fdate") String fdate, @RequestParam("tdate") String tdate) {
return entityManager.createQuery("select e from Employee e where e.empId="+empid, Employee.class).getResultList();
}
url:"phcheck.htm?empid="+$("#empid").val()+"&fdate="+$("#fdate").val()+"&tdate="+$("#tdate").val()
In Spring MVC you will have to have registered MappingHymanson2HttpMessageConverter like being done here
在 Spring MVC 中，您必须像在这里完成一样注册 MappingHymanson2HttpMessageConverter
回答by DavidA
The ModelAndView implies you plan to render a view, which you don't. To just return the object, use the @ResponseBody annotation:
ModelAndView 暗示您计划渲染视图，而您不打算渲染视图。要仅返回对象，请使用 @ResponseBody 注释：
@RequestMapping("phcheck")
public @ResponseBody List<Employee> pay(@RequestParam("empid") int empid, String fdate, String tdate) {
return entityManager.createQuery("select e from Employee e where e.empId="+empid, Employee.class).getResultList();
}
Also, you should be more careful about how you handle queries. Your query is insecure and would allow sql injection. For example, if someone called your method with empId = "'15' or '1'='1'", then it would return the entire list of employees.
此外，您应该更加小心处理查询的方式。您的查询不安全，将允许 sql 注入。例如，如果有人使用 empId = "'15' 或 '1'='1'" 调用您的方法，那么它将返回整个员工列表。
回答by cse
Make the method as @ResponseBody Type in the controller and in the ajax take the List from success function.
在控制器中将该方法设为 @ResponseBody Type，并在 ajax 中从成功函数中获取 List。
Put the Hymanson Mapper file in Pom.xml file if using Maven
如果使用 Maven，请将 Hymanson Mapper 文件放在 Pom.xml 文件中
