jQuery 将 JSON 数据传递给 Spring MVC 控制器
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passing JSON data to a Spring MVC controller
提问by user2702205
I need to send a JSON string to Spring MVC controller.But I do not have any form bindings to it , I just need to send a plain JSON data to Controller class.I am making jQuery AJAX call to the Controller method like the below code.
我需要向 Spring MVC 控制器发送一个 JSON 字符串。但是我没有任何形式绑定到它，我只需要向控制器类发送一个普通的 JSON 数据。我正在对控制器方法进行 jQuery AJAX 调用，如下面的代码.
$.ajax ({
url: "./save",
type: "POST",
data: JSON.stringify(array),
dataType: "json",
contentType: "application/json; charset=utf-8",
success: function(){
alert("success ");
}
});
But how do I retrieve it in the Controller method?(Note: It is just plain JSON data and not a form submission).
但是如何在 Controller 方法中检索它？（注意：它只是普通的 JSON 数据而不是表单提交）。
回答by Vineeth Bhaskaran
Add the following dependencies
添加以下依赖项
<dependency>
<groupId>org.codehaus.Hymanson</groupId>
<artifactId>Hymanson-mapper-asl</artifactId>
<version>1.9.7</version>
</dependency>
<dependency>
<groupId>org.codehaus.Hymanson</groupId>
<artifactId>Hymanson-core-asl</artifactId>
<version>1.9.7</version>
</dependency>
Modify request as follows
修改请求如下
$.ajax({
url:urlName,
type:"POST",
contentType: "application/json; charset=utf-8",
data: jsonString, //Stringified Json Object
async: false, //Cross-domain requests and dataType: "jsonp" requests do not support synchronous operation
cache: false, //This will force requested pages not to be cached by the browser
processData:false, //To avoid making query String instead of JSON
success: function(resposeJsonObject){
// Success Message Handler
}
});
Controller side
控制器端
@RequestMapping(value = urlPattern , method = RequestMethod.POST)
public @ResponseBody Person save(@RequestBody Person jsonString) {
Person person=personService.savedata(jsonString);
return person;
}
@RequestBody- Covert Json object to java @ResponseBody- convert Java object to json
@RequestBody- 将 Json 对象@ResponseBody转换为 Java - 将 Java 对象转换为 json
回答by surya handoko
Html
$('#save').click(function(event) { var jenis = $('#jenis').val(); var model = $('#model').val(); var harga = $('#harga').val(); var json = { "jenis" : jenis, "model" : model, "harga": harga}; $.ajax({ url: 'phone/save', data: JSON.stringify(json), type: "POST", beforeSend: function(xhr) { xhr.setRequestHeader("Accept", "application/json"); xhr.setRequestHeader("Content-Type", "application/json"); }, success: function(data){ alert(data); } }); event.preventDefault(); });Controller
@Controller @RequestMapping(value="/phone") public class phoneController { phoneDao pd=new phoneDao(); @RequestMapping(value="/save",method=RequestMethod.POST) public @ResponseBody int save(@RequestBody Smartphones phone) { return pd.save(phone); }Dao
public Integer save(Smartphones i) { int id = 0; Session session=HibernateUtil.getSessionFactory().openSession(); Transaction trans=session.beginTransaction(); try { session.save(i); id=i.getId(); trans.commit(); } catch(HibernateException he){} return id; }
html
$('#save').click(function(event) { var jenis = $('#jenis').val(); var model = $('#model').val(); var harga = $('#harga').val(); var json = { "jenis" : jenis, "model" : model, "harga": harga}; $.ajax({ url: 'phone/save', data: JSON.stringify(json), type: "POST", beforeSend: function(xhr) { xhr.setRequestHeader("Accept", "application/json"); xhr.setRequestHeader("Content-Type", "application/json"); }, success: function(data){ alert(data); } }); event.preventDefault(); });控制器
@Controller @RequestMapping(value="/phone") public class phoneController { phoneDao pd=new phoneDao(); @RequestMapping(value="/save",method=RequestMethod.POST) public @ResponseBody int save(@RequestBody Smartphones phone) { return pd.save(phone); }道
public Integer save(Smartphones i) { int id = 0; Session session=HibernateUtil.getSessionFactory().openSession(); Transaction trans=session.beginTransaction(); try { session.save(i); id=i.getId(); trans.commit(); } catch(HibernateException he){} return id; }
回答by Leonardo Silva
You can stringify the JSON Object with JSON.stringify(jsonObject) and receive it on controller as String.
您可以使用 JSON.stringify(jsonObject) 对 JSON 对象进行字符串化，并在控制器上将其作为字符串接收。
In the Controller, you can use the javax.jsonto convert and manipulate this.
在控制器中，您可以使用javax.json来转换和操作它。
Download and add the .jar to the project libs and import the JsonObject.
下载 .jar 并将其添加到项目库中并导入 JsonObject。
To create an json object, you can use
要创建一个 json 对象，您可以使用
JsonObjectBuilder job = Json.createObjectBuilder();
job.add("header1", foo1);
job.add("header2", foo2);
JsonObject json = job.build();
To read it from String, you can use
要从字符串中读取它，您可以使用
JsonReader jr = Json.createReader(new StringReader(jsonString));
JsonObject json = jsonReader.readObject();
jsonReader.close();
