MySQLi如果条件警报消息未显示输出 [英] MySQLi If condition alert msg not showing the output
本文介绍了MySQLi如果条件警报消息未显示输出的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
下面给出的代码工作完美. db表也插入值. 但是回显消息未显示在输出中
Below Given code working perfect. Db table also values are inserted. But Echo Message not showing in output
<?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_DATABASE', 'admin');
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
if(isset($_POST["register"]))
{
$id = $_POST["id"];
$name = $_POST["user_name"];
$password = $_POST["password"];
$query = mysqli_query($db, "INSERT INTO user (id, user_name, password)VALUES ('$id', '$name', '$password')");
$result = mysqli_query($query);
if($result)
{
echo "Thank You! you are now registered.";
}
}
?>
推荐答案
您两次以错误的方式调用了mysqli_query
,您只需要调用一次即可.
You called mysqli_query
2 times and second time in wrong way, you just need to call it once instead
// Change
$query = mysqli_query($db, "INSERT INTO user (id, user_name, password)VALUES ('$id', '$name', '$password')");
$result = mysqli_query($query);
// To
$result = mysqli_query($db, "INSERT INTO user (id, user_name, password) VALUES ('$id', '$name', '$password')");
// And change your condition to
if( mysqli_affected_rows($db) ){
..
}
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