Question

我已经在angular中实现了一个演示客户端，它将用户名和密码作为POST发送到REST Api，然后读取它并执行一些auth逻辑。问题是在控制器中始终捕获为null（没有用户没有通过）。我尝试在Chrome中的Advanced Rest Client中发送相同的数据，一切正常。

这是我的代码段：

Angular客户：

function LoginCtrl($scope, $rootScope, $location, $http, LoginService) {

$scope.submit = function() {
    LoginService.authenticate($.param({username: $scope.inputUsername, password: $scope.inputPassword}), function(user) {
        $rootScope.user = user;
        $http.defaults.headers.common['Authentication'] = user.token;
        $location.path("/");
    });
};

};

Controller java（First Try）：

public @ResponseBody UserTransfer authenticate(HttpServletRequest request) throws IOException {
    //tried this one first but NULL this two params.
    String username = request.getParam("username");
    String password = request.getParam("password");               


    //somer logic
    return new UserTransfer(userDetails.getUsername(), roles, TokenUtils.createToken(userDetails));

Controller java（Second Try）：

public @ResponseBody UserTransfer authenticate(@RequestParam("username") String username, @RequestParam("password") String password) throws IOException {

    //tried secondly this one but in this case the server responded with ERROR 400 - Bad Request              


    //somer logic
    return new UserTransfer(userDetails.getUsername(), roles, TokenUtils.createToken(userDetails));

将这些数据从Angular传递给我的控制器的正确方法是什么？我看不出错误......

感谢您的帮助!!!

Answer 1

有同样的问题用泽西来处理那个

的web.xml

<servlet>
<servlet-name>RestService</servlet-name>
<servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
<init-param>
    <param-name>com.sun.jersey.config.property.packages</param-name>
    <param-value>com.pearson.taxonomy.cm.web.controller</param-value>
</init-param>
<init-param>
    <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
    <param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>RestService</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

UserResource.java

import javax.ws.rs.FormParam;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import java.util.HashMap;
import java.util.Map;

@Component
@Path("/user")
public class UserResource {

    @Autowired
    private UserDetailsService userService;

    @Autowired
    @Qualifier("authenticationManager")
    private AuthenticationManager authManager;


    @Path("authenticate")
    @POST
    @Produces(MediaType.APPLICATION_JSON)
    public UserView authenticate(@FormParam("username") String username, @FormParam("password") String password) {

        UsernamePasswordAuthenticationToken authenticationToken = new UsernamePasswordAuthenticationToken(username, password);
        Authentication authentication = this.authManager.authenticate(authenticationToken);
        SecurityContextHolder.getContext().setAuthentication(authentication);

        Map<String, Boolean> roles = new HashMap<String, Boolean>();

        UserDetails userDetails = this.userService.loadUserByUsername(username);

        for (GrantedAuthority authority : userDetails.getAuthorities()) {
            roles.put(authority.toString(), Boolean.TRUE);
        }

        return new UserView(userDetails.getUsername(), roles, TokenUtils.createToken(userDetails));
    }

}

如何从angularJS客户端捕获用户和密码到spring security

1 个答案: