Question

<?php
<form name "fav_button" action="" method="post">
<input type="submit" name="fav_button"  value="Add to Favorite" />
</form>

if(isset($_POST['fav_button']) ){
        $conn=mysqli_connect("localhost","root","alaa","registration_database");
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);

            }
          $query= ("INSERT INTO favs (uId, itemId) VALUES ('1','2'); ");
        }
 ?>

我正在尝试将数据添加到表格中，点击按钮“fav_button”页面刷新并且没有错误显示，但是表格没有更新，任何人都能解释我做错了什么吗？ /> 我正在运行我在phpmyadmin中使用的查询，它运行正常。

Answer 1

您错过了$conn->query($query);并且运行查询非常重要..此行实际执行您的查询

    <form name="fav_button" action="" method="post">
    <input type="submit" name="fav_button"  value="Add to Favorite" />
    </form>

    if(isset($_POST['fav_button']) ){
            $conn=mysqli_connect("localhost","root","alaa","registration_database");
            if ($conn->connect_error) {
                die("Connection failed: " . $conn->connect_error);

                }
              $query= ("INSERT INTO favs (uId, itemId) VALUES ('1','2'); ");
              $conn->query($query);//this is the line which actually executes your query
            }
     ?>

我正在尝试通过php将数据插入数据库

1 个答案: