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php - 验证登录用户后返回 URL

问题描述

在验证用户通过移动应用程序正确登录并向他们提供 URL 后，我需要返回一个 URL，该 URL 在我的 PHP 函数中打开移动应用程序上的支持聊天窗口。下面是我的代码示例。

<?php

if((isset($_POST['key']) && $_POST['key'] != "") && isset($_POST['userName']) && $_POST['userName'] != "")
    {
        $acc = new userAccount();

        //if($acc->clientloginIntoAccount($_POST['userName'],$_POST['passWord'],$_POST['key']) == 1)
        $Response = array();

        if($acc->ClientLoginEncryptionUpgrade($_POST['userName'], $_POST['passWord'], $_POST['key'], $Response))
        {
            $_SESSION['clientloggedin'] = $Response['details']['session_data'];
            $_SESSION['clientloggedin']['cost_centre'] = $Response['details']['company'];
            $_SESSION['clientloggedin']['time'] = $_POST['time'];
            $_SESSION['primary_color_1'] = $Response['details']['primary_color_1'];
            $_SESSION['primary_color_2']= $Response['details']['primary_color_2'];
            $_SESSION['footer_colour']= $Response['details']['footer_colour'];
            $_SESSION['email_color'] = $Response['details']['email_color'];
            $_SESSION['client_logo'] = $Response['details']['client_logo'];
            $_SESSION['link_color'] = $Response['details']['link_color'];
            $_SESSION['menu_items'] = $Response['details']['menu_items'];
            $_SESSION['menu_items_icons'] = $Response['details']['menu_items_icons'];
            $employee = getTableValue("personnel","id","employee_number='{$_POST['key']}'");
            $_SESSION['loggedin']['personnelID'] = $employee;

            echo "<script>window.location.href='home';</script>";
            return "<script>javascript:void($zopim.livechat.window.openPopout())</script>";
        }
        else
        {
            echo "<div class = 'bg-danger'> Incorrect username / password. Please try again. </div>";

?>

标签： php

解决方案

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