PHP mysqli 函数 return 值失败

Question

在此代码块中，它以前设计为使用 session_id。我正在尝试从使用 session_id 转换为使用从数据库检索到的用户 ID。我不确定我做错了什么，但函数没有返回变量。如有任何建议，我们将不胜感激。

 protected function get_user_id() {
        //previous code used the session id
        //@session_start();
        //return session_id();

        // New code to use User ID instead of session_id
        // Connecting to the database
        include ("../../../admin/includes/connect.php");

        // Let's get the user ID from the database for use with the widget
        $user_id_query = "SELECT nonadmin_user_id FROM `nonadmin_user_login` WHERE email = '$_SESSION[email]'";
        $run_query = mysqli_query($conn, $user_id_query);
        while($row=mysqli_fetch_array($run_query)){

        // Create variable for the user's id
        $nonadmin_user_id = $row['nonadmin_user_id']; }
        return $nonadmin_user_id;
    }
    // This function needs to use the variable $nonadmin_user_id
    protected function get_user_path() {
        if ($this->options['user_dirs']) {
            return $this->get_user_id().'/';
        }
        return '';
    }

Answer 1

"Fred you're the man! It was the session. I removed the comment out from in front of the session start and now it works perfect. What baffles me on this is I was under the impression that if you start a session in a file and then include other files the included files did not require the session to be started."

需要启动会话，以便在您的查询中识别并成功传递会话数组。

此外，session_start(); 需要驻留在所有使用会话的文件中。

http://php.net/manual/en/function.session-start.php

将 error reporting 添加到您的文件的顶部，这将有助于查找错误。

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

旁注：错误报告只应在试运行中进行，绝不能在生产中进行。

PHP mysqli 函数 return 值失败

PHP mysqli function return value failing

php

mysqli