MySQLi如果条件警报消息未显示输出 [英] MySQLi If condition alert msg not showing the output

查看:58
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) ){
 ..
}

这篇关于MySQLi如果条件警报消息未显示输出的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆