【发布时间】:2017-06-20 12:39:43
【问题描述】:
我有简单的登录控制器。在其中,我有一个名为 signIn() 的操作方法。我正在提交我的凭据以通过 ajax 调用登录网站。
这是我的 ajax 调用函数-
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);
“人”数据在哪里 -
{"emailId":"aaa@gmail.com","password":"12345"}
而登录控制器的signIn()方法代码是-
@PostMapping(value = "/login",consumes = MediaType.APPLICATION_JSON_VALUE)
public String signIn(@RequestBody Person person) {
if(person.getEmailId().equals("aaa@gmail.com") && person.getPassword().equals("12345")){
return "Success";
}
else{
return "Invalid";
}
}
我创建了简单的 Person 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,它给了我 415 Unsupported Media Type 错误。你能帮我为什么会发生这个错误???提前谢谢..
【问题讨论】:
-
@Surely 我做了这些更改 jQuery.ajax({ 'type': type, 'url': url, 'headers': header ? header : {}, 'contentType': contentType, '数据':JSON.stringify(内容),'成功':成功,'错误':错误});
-
stackoverflow.com/questions/7181534/…。整个信息是什么?您有两个应用程序/XXX 值。你试过从命令行 curl 吗?
-
谢谢。我已经解决了这个问题。我已经指定了
,但我忘了提到将使用哪个 HttpMessageConverter。我需要对 Jackson 库(jackson-databind)的依赖,以便我可以将 HttpRequestBody 转换为 Person 对象,现在,我可以支持 MappingJackson2HttpMessageConverter
标签: java ajax spring spring-mvc