我有简单的登录控制器。在它里面,我有一个叫做signIn()的动作方法。我正在提交我的凭证以通过ajax呼叫登录到网站。Spring MVC中不受支持的媒体类型
这是我的AJAX调用程序功能 -
var AJAXCaller = function() {
};
AJAXCaller.prototype.call = function (type, uri, header, contentType, content, success, error) {
var config = new SSKSConfig();
var url = config.getLocation() + uri;
return jQuery.ajax({
'type': type,
'url': url,
'headers': header,
'data': JSON.stringify(content),
'success': success,
'error': error
});
};
var caller= new AJAXCaller();
caller.call('POST', '/login', { 'Content-Type': 'application/json' }, 'application/x-www-form-urlencoded', person, fnSuccess, fnError);
,其中 '人' 的数据,如 -
{"emailId":"[email protected]","password":"12345"}
并登录控制器的登入()方法的代码是─
@PostMapping(value = "/login",consumes = MediaType.APPLICATION_JSON_VALUE)
public String signIn(@RequestBody Person person) {
if(person.getEmailId().equals("[email protected]") && person.getPassword().equals("12345")){
return "Success";
}
else{
return "Invalid";
}
}
我已创建简单的人POJO类 -
public class Person implements PersonSupport, AddressSupport {
private String code;
private String name;
private String emailId;
private String password;
private String contactNo;
private final Address address = new Address();
public Person() {
}
@Override
public void setCode(String code) {
this.code = code;
}
@Override
public String getCode() {
return this.code;
}
@Override
public void setName(String name) {
this.name = name;
}
@Override
public String getName() {
return name;
}
@Override
public void setContactNo(String contactNo) {
this.contactNo = contactNo;
}
@Override
public String getContactNo() {
return contactNo;
}
@Override
public void setEmailId(String emailId) {
this.emailId = emailId;
}
@Override
public String getPassword() {
return this.password;
}
@Override
public void setPassword(String password) {
this.password = password;
}
@Override
public String getEmailId() {
return emailId;
}
@Override
public String getRoadNo() {
return address.getRoadNo();
}
@Override
public String getRoadName() {
return address.getRoadName();
}
@Override
public String getCity() {
return address.getCity();
}
@Override
public String getPinCode() {
return address.getPinCode();
}
@Override
public StateList getStateList() {
return address.getStateList();
}
@Override
public void setRoadNo(String roadNo) {
address.setRoadNo(roadNo);
}
@Override
public void setRoadName(String roadName) {
address.setRoadName(roadName);
}
@Override
public void setCity(String city) {
address.setCity(city);
}
@Override
public void setPinCode(String pinCode) {
address.setPinCode(pinCode);
}
@Override
public void setStateList(StateList stateList) {
address.setStateList(stateList);
}
}
现在,当我尝试登录时,它给我415不支持的媒体类型错误。你能帮我解释为什么这个错误发生了吗?在此先感谢..
@Surely我已经作出这些变化jQuery.ajax({ '类型':类型, 'URL':URL, '报头'?:头标头:{}, '的contentType':contentType中, ' data':JSON.stringify(content), 'success':success, 'error':error }); –
https://stackoverflow.com/questions/7181534/http-post-using-json-in-java。什么是整个信息?你有两个应用程序/ XXX值。你有没有尝试从命令行卷曲? – efekctive
谢谢。我已经解决了这个问题。我已经指定了 ,但我忘了提及将使用哪个HttpMessageConverter。我需要一个依赖到Jackson库(jackson-databind),以便我可以将HttpRequestBody转换为Person对象,现在,我可以支持MappingJackson2HttpMessageConverter –