php - 如何从连接表结果中具有相同名称的两个值中获取单个值？

Question

我运行以下查询并获得结果:

现在，如果我运行以下命令，那么我得到的 user_id 为 0。

<?php 
        session_start();
        require_once('connectvars.php');
        $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

        $user_id = $_SESSION['user_id'];

        $result = mysqli_query($dbc,"SELECT * FROM `user_question`join question WHERE user_question.qid = question.qid and user_question.user_id = $user_id")
        or die("Error in fetching the value.");

        while ( $row = mysqli_fetch_array($result))
        {
            echo 'user id : ';
            echo $row['user_id'];
            echo "<br>";
            echo 'qid: ';
            echo $row['qid'];
            echo "<br>";
            echo 'answer_key: ';
            echo $row['answer_key'];
                    echo "<br>";

        }


     ?>

如何获取 user_id : 4 ？

Answer 1

您可以按如下方式进行:

$result = mysqli_query($dbc,"SELECT user_question.user_id,user_question.qid,user_question.answer_key FROM `user_question`join question WHERE user_question.qid = question.qid and user_question.user_id = $user_id")

现在 echo $row['user_id']...