Question

我尝试在登录后从页面上的数据库中显示用户，但我不断收到错误消息：

致命错误：调用未定义的函数fetch_assoc（）

这是代码：

    <?php
session_start(); ?>

<link rel="stylesheet" href="form.css">
<div class="body content">

    <div class="welcome">
        <div class="alert alert-success"><?= $_SESSION['message'] ?></div>
        <img src="<?= $_SESSION['avatar'] ?>"><br />
        Welcome <span class="user"><?= $_SESSION['username'] ?></span>

        <?php
        $mysqli = new mysqli("localhost","user","password","database");
        //Select queries return a resultset
        $sql = "SELECT username, avatar FROM users";
        $result = $mysqli->query($sql); //$result = mysqli_result object
        //var_dump($result);
        ?>

        <div id='registered'>
        <span>All registered users:</span>

        <?php
        //while($row = $result->fetch_assoc()){ //returns associative array of fetched row
         while($row = fetch_assoc($result)){ //returns associative array of fetched row
            //echo '<pre>';
            //print_r($row);
            //echo '</pre>';
            echo "<div class='userlist'><span>$row[username]</span>";
            echo "<img src='$row[avatar]'></div>";
        }
        ?>  
        </div>
    </div>
</div>

Answer 1

这就是你如何做到这一点，你正在混合面向对象和程序：

while($row = $result->fetch_assoc()){ //returns associative array of fetched row
            //echo '<pre>';
            //print_r($row);
            //echo '</pre>';
            echo "<div class='userlist'><span>$row[username]</span>";
            echo "<img src='$row[avatar]'></div>";
        }

Answer 2

以下是与数据库进行过程连接的正确脚本：

<?php    
    $con = mysqli_connect("localhost","root","","db1");
    $sql = "SELECT username, avatar FROM users";
    $result = mysqli_query($con,$sql);
    $row = mysqli_fetch_assoc($result);
    echo $row["username"] . " " . $row["avatar"] . "<br>";
    mysqli_close($con);
?>

致命错误：调用未定义的函数fetch_assoc（）给出了什么？

2 个答案: