PHP MySQL 读取数据网址内容以超链接方式输出
2020-11-15学无止境南夏30°c
A+ A-<?php //数据库信息 $servername = "localhost"; $username = "username"; $password = "password"; $dbname = "myDB"; // 创建连接 $conn = new mysqli($servername, $username, $password, $dbname); //以下实例中我们从 myDB 数据库的 MyGuests 表读取了 link 和 type 列的数据并显示在页面上: $sql = "SELECT link,type FROM MyGuests"; $result = $conn->query($sql); $ar = array(); if ($result->num_rows > 0) { while($row = $result->fetch_assoc()) { echo '<a href=" '.$row["link"].'" target="_blank">'.$row["type"].'</a>'."<hr>"; } } $conn->close(); ?> ?>
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