我看过其他关于此问题的帖子，但仍然没有找到合适的答案。
我提交的表单有三个参数而不是两个。
这是我的 CustomAuthenticationProvider:
@Component
public class CustomAuthenticationProvider implements AuthenticationProvider {
private static final Logger logger = LoggerFactory.getLogger(CustomAuthenticationProvider.class);
@Override
public Authentication authenticate(Authentication auth) throws AuthenticationException {
String userName = auth.getName().trim();
String password = auth.getCredentials().toString().trim();
String companyName ;
if (userName.equals("admin") && password.equals("123456")) {
List<GrantedAuthority> grantedAuths = new ArrayList<>();
grantedAuths.add(new SimpleGrantedAuthority("ROLE_ADMIN"));
Authentication upat = new UsernamePasswordAuthenticationToken(userName, password, grantedAuths);
logger.info("{}:{}",userName,grantedAuths);
return upat;
} else {
return null;
}
}
@Override
public boolean supports(Class<?> auth) {
return (UsernamePasswordAuthenticationToken.class.isAssignableFrom(auth));
}
我想从登录表单获取额外参数来验证 CustomAuthenticationProvider 中的 companyName
如何从登录表单获取参数？
最佳答案
经过研究和对另一篇文章进行重大修改后，我通过注入(inject) AuthenticationDetailsSource 找到了解决方案和WebAuthenticationDetails 。所以我在这里分享解决方案:
首先为自定义过滤器创建新类来获取参数:
class ExtraParam extends WebAuthenticationDetails {
private static final long serialVersionUID = 1L;
private final String company;
public ExtraParam(HttpServletRequest request) {
super(request);
this.company = request.getParameter("company");
}
public String getCompanyName() {
return company;
}
}
然后创建用于注入(inject)的源类:
class ExtraParamSource implements AuthenticationDetailsSource<HttpServletRequest, ExtraParam> {
public ExtraParam buildDetails (HttpServletRequest context) {
return new ExtraParam(context);
}
}
然后，修改您的 spring security xml 以识别类和过滤器:
添加authentication-details-source-ref="ExtraParam"在你的<form-login像这样:
<form-login ... authentication-details-source-ref="ExtraParamSource"... />
不要忘记 bean :
<beans:bean id="ExtraParamSource" class="com.xxx.xxx.xxx.ExtraParamSource"/>
然后完成!就这样。要获取参数，只需使用下面的示例:
@Component
public class CustomAuthenticationProvider implements AuthenticationProvider{
@Override
public Authentication authenticate(Authentication auth) throws AuthenticationException {
String userName = auth.getName().trim();
String password = auth.getCredentials().toString().trim();
String companyName = ((ExtraParam)auth.getDetails()).getCompanyName();
....
引用文献:
1) How to pass an additional parameter with spring security login page
关于java - 如何从登录表单调用额外参数到 CustomAuthenticationProvider，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35215065/