Question

我想检查网站是否可以连接到mySQL。如果没有，我想显示一个错误，说用户应该在几分钟内再次尝试访问该页面...

我真的不知道该怎么做;）

非常感谢任何帮助！

string mysql_error ([ resource $link_identifier ] )

但我该怎么用？

这只是给了我错误，但我希望显示消息时出现任何错误。

由于

Answer 1

试试这个：

<?php
$servername   = "localhost";
$database = "database";
$username = "user";
$password = "password";

// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
}
  echo "Connected successfully";
?>

Answer 2

非常基本：

<?php 
$username = 'user';
$password = 'password';
$server = 'localhost'; 
// Opens a connection to a MySQL server
$connection = mysql_connect ($server, $username, $password) or die('try again in some minutes, please');
//if you want to suppress the error message, substitute the connection line for:
//$connection = @mysql_connect($server, $username, $password) or die('try again in some minutes, please');
?>

结果：

警告：mysql_connect（）[function.mysql-connect]：第6行/home/user/public_html/zdel1.php中用户“user”@“localhost”（使用密码：YES）拒绝访问请稍后再试一次

根据Wrikken的建议，请查看完整的错误处理程序，以获得更复杂，更高效，更优雅的解决方案：http://www.php.net/manual/en/function.set-error-handler.php

Answer 3

请检查：

$servername='localhost';
$username='root';
$password='';
$databasename='MyDb';

$connection = mysqli_connect($servername,$username,$password);

if (!$connection) {
die("Connection failed: " . $conn->connect_error);
}

/*mysqli_query($connection, "DROP DATABASE if exists MyDb;");

if(!mysqli_query($connection, "CREATE DATABASE MyDb;")){
echo "Error creating database: " . $connection->error;
}

mysqli_query($connection, "use MyDb;");
mysqli_query($connection, "DROP TABLE if exists employee;");

$table="CREATE TABLE employee (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)"; 
$value="INSERT INTO employee (firstname,lastname,email) VALUES ('john', 'steve', 'johnsteve@yahoo.com')";
if(!mysqli_query($connection, $table)){echo "Error creating table: " . $connection->error;}
if(!mysqli_query($connection, $value)){echo "Error inserting values: " . $connection->error;}*/

检查数据库连接，否则显示消息

3 个答案: