angularjs向spring发布json
400错误请求 从JSJSON到spring mvc 我使用angular js创建登录页面模块 它从angular到spring转换为json格式 如何在运行400错误请求时接收错误代码 myjsp代码:angularjs向spring发布json,json,angularjs,spring,Json,Angularjs,Spring,400错误请求 从JSJSON到spring mvc 我使用angular js创建登录页面模块 它从angular到spring转换为json格式 如何在运行400错误请求时接收错误代码 myjsp代码: <div id="showProjects" ng-controller="crescentGetProjects"> <table> <tr><td>User Name</td><td><
<div id="showProjects" ng-controller="crescentGetProjects">
<table>
<tr><td>User Name</td><td><input type="text" id="pname" name="pname" ng-model="project.UserName"></td></tr>
<tr><td>Password</td><td><input type="text" id="aname" name="aname" ng-model="project.Password"></td></tr>
<tr><td><input type="hidden" ng-model="project.projectId" /></td>
<td><input type="button" value="Save" ng-click="saveProjectDetails(project)"></td></tr>
</table>
</div>
控制器代码
@RequestMapping(value = "/register", method = RequestMethod.POST,consumes="application/json")
public @ResponseBody String registerUser(@RequestBody LoginEntity userModel,HttpServletRequest request) {
String resp=null;
try{
System.out.println("$$$$$$$$$$ "+userModel.getUserName());
crescentManager.LoginCheck(userModel);
resp="success";
}catch(Exception e){
resp="fail";
System.out.println(e);
}
return resp;
}
我的实体类是
@Entity
@Table(name="LOGININFO")
public class LoginEntity {
@Id
@Column(name="LOGINID")
@GeneratedValue
private Integer LoginId;
@Column(name="USERNAME")
private String UserName;
@Column(name="PASSWORD")
private String Password;
public Integer getLoginId() {
return LoginId;
}
public void setLoginId(Integer loginId) {
LoginId = loginId;
}
public String getUserName() {
return UserName;
}
public void setUserName(String userName) {
UserName = userName;
System.out.println("@@@@@@@@@@@@@@@username is settled in entity "+userName);
}
public String getPassword() {
return Password;
}
public void setPassword(String password) {
Password = password;
}
}
埃罗味精
400
错误的请求您可以发布Firebug或开发工具捕获的HTTP请求吗?另外,如果您提供SpringTry提供的错误日志输出,并遵循camel case变量名的约定,那就太好了。JsonParser需要getUserName()的用户名变量名,而您的代码提供的是用户名。因此,这个400错误的请求错误。@RequestMapping(value=“research/api/register”像这样更改此错误将显示我更改了请求映射值WARN org.springframework.web.servlet.PageNotFound-在名为“login model”的DispatcherServlet中找不到URI为[/LoginTest/rest/research/api/register]的HTTP请求的映射
@Entity
@Table(name="LOGININFO")
public class LoginEntity {
@Id
@Column(name="LOGINID")
@GeneratedValue
private Integer LoginId;
@Column(name="USERNAME")
private String UserName;
@Column(name="PASSWORD")
private String Password;
public Integer getLoginId() {
return LoginId;
}
public void setLoginId(Integer loginId) {
LoginId = loginId;
}
public String getUserName() {
return UserName;
}
public void setUserName(String userName) {
UserName = userName;
System.out.println("@@@@@@@@@@@@@@@username is settled in entity "+userName);
}
public String getPassword() {
return Password;
}
public void setPassword(String password) {
Password = password;
}
}