wso2 身份服务器 5.7.0 如何为 Role1 编写脚本？

Question

我想用这个 link:

https://docs.wso2.com/display/IS570/Adaptive+Authentication

我希望 Role1 无法登录到我的应用程序并编写此代码：它不起作用，Role1 可以登录到我的应用程序。怎么写这个脚本？

  // Role-Based from Template...

// This script will step up authentication for any user belonging
// to one of the given roles
// If the user has any of the below roles, authentication will be stepped up
var rolesToStepUp = ['Role1'];

function onLoginRequest(context) {
    executeStep(1, {
        onSuccess: function (context) {
            // Extracting authenticated subject from the first step
            var user = context.currentKnownSubject;
            // Checking if the user is assigned to one of the given roles
            var hasRole = hasAnyOfTheRoles(user, rolesToStepUp);
            if (hasRole) {
                Log.info(user.username + ' Has one of Roles: ' + rolesToStepUp.toString());
              return false;
            }
        }
    });
}

// End of Role-Based.......

Answer 1

如果您没有收到日志 Log.info(user.username + ' Has one of Roles: ' + rolesToStepUp.toString());，请检查您的情况。如果您的角色是内部角色，请按如下方式更改 rolesToStepUp 变量并尝试。

var rolesToStepUp = ['Internal/Role1'];

查看了解更多详情
为了让用户登录失败，只返回false是不行的。使用 sendError 效用函数 (https://docs.wso2.com/display/IS570/Adaptive+Authentication+JS+API+Reference#AdaptiveAuthenticationJSAPIReference-sendError(url,parameters))。检查基于用户年龄的脚本是否使用 sendError (https://docs.wso2.com/display/IS570/Configuring+User-Age-Based+Adaptive+Authentication)

wso2 身份服务器 5.7.0 如何为 Role1 编写脚本？

wso2 identity server 5.7.0 How write Script for Role1?

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