问题描述:

I would like to check if the website can connect to mySQL. If not, I would like to display an error saying that the user should try to access the page again in a few minutes...

I really do not know how to do this ;)

Any help would be greatly appreciated!

string mysql_error ([ resource $link_identifier ] )

But how do I use this?

This just gives me the error, but I want the message to display with any error.

Thanks

网友答案:

Try this:

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

$mysqlConnection = mysql_connect($server, $username, $password);
if (!$mysqlConnection)
{
  echo "Please try later.";
}
else
{
mysql_select_db($database, $mysqlConnection);
}
?>
网友答案:

very basic:

<?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');
?>

result:

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'user'@'localhost' (using password: YES) in /home/user/public_html/zdel1.php on line 6 try again in some minutes, please

as per Wrikken's recommendation below, check out a complete error handler for more complex, efficient and elegant solutions: http://www.php.net/manual/en/function.set-error-handler.php

网友答案:

Please check this:

$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
)"; 
$value="INSERT INTO employee (firstname,lastname,email) VALUES ('john', 'steve', '[email protected]')";
if(!mysqli_query($connection, $table)){echo "Error creating table: " . $connection->error;}
if(!mysqli_query($connection, $value)){echo "Error inserting values: " . $connection->error;}*/
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