Question

嘿，我正在试图找出我的sql查询是否失败。我想要它，如果它使用下面的代码失败重定向到表单页面：

$checkconnection = mysql_connect('localhost', $dbuser, $dbpass)
                                                or die();


if(!$checkconnection)
{
$_SESSION['errormsg'] = "<div style='padding-left: 50px;color:#FF0000'>Cannot connect to specfied database!</div>";
header("Location: install.php");
}else{
echo('Connection Successful!');
}

使用它只是说：

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'nzcraftn_admin'@'localhost' (using password: YES) in /home/nzcraftn/public_html/phishnet/install/install_submit.php on line 17

Answer 1

试试这个

$checkconnection = @mysql_connect('localhost', $dbuser, $dbpass)

它会隐藏默认错误并触发您自己的

Answer 2

mysql_connect的返回值为false仅表示失败。如果它返回FALSE，or die()表达式将退出php脚本。这就是为什么你不输出它的任何输出的原因。

删除or die()命令，并在if( !$checkconnection )子句中显示实际错误。可以使用mysql_error()检索报告的错误。

Answer 3

它只显示警告，因为or die()没有输出任何内容（空参数列表）。试试这个：

<?php
    //Start the session
    session_start(); 

    //Do the conntection
    $checkconnection = @mysql_connect('localhost', $dbuser, $dbpass);

    //Check if it's valid
    if(!$checkconnection) {

        //Add it up to the session, and redirect
        $_SESSION['errormsg'] = "<div style='padding-left: 50px;color:#FF0000'>Cannot connect to specfied database!</div>";
        session_write_close();
        header("Location: install.php");
        exit();

    } else{

        //Yay
        echo('Connection Successful!');

    }
?>

Answer 4

创世记的答案只是压制了警告，但仍然有效

如果你想要'干净'，你可以尝试/捕捉错误：

（直接来自php.net/mysql_connect上的评论：

// Assign variables
        global $db_connection, $db_server, $db_database, $db_username, $db_password;
        $db_server = $server;
        $db_database = $database;
        $db_username = $username;
        $db_password = $password;

        // Attempt connection
        try
        {
            // Create connection to MYSQL database
            // Fourth true parameter will allow for multiple connections to be made
            $db_connection = mysql_connect ($server, $username, $password, true);
            mysql_select_db ($database);
            if (!$db_connection)
            {
                throw new Exception('MySQL Connection Database Error: ' . mysql_error());
            }
            else
            {
                $CONNECTED = true;
            }
        }
        catch (Exception $e)
        {
            echo $e->getMessage();
        }

检查mysql_connect（）是否失败？

4 个答案: