嘿,我正在尝试找出我的sql查询是否失败。我想要它,所以如果它确实失败了,使用下面的代码重定向到表单页面:
$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!');
}它所说的一切就是这样:
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发布于 2011-08-13 20:26:57
试试这个
$checkconnection = @mysql_connect('localhost', $dbuser, $dbpass)它将隐藏默认错误并触发您自己的错误
发布于 2011-08-13 20:32:02
mysql_connect返回值为false仅表示失败。如果返回FALSE,则or die()表达式将退出php脚本。这就是为什么你不会输出任何数据的原因。
删除or die()命令,并显示if( !$checkconnection )子句中的实际错误。可以使用mysql_error()检索报告的错误。
发布于 2011-08-13 20:27:07
它只是显示警告,因为您的or die()没有输出任何内容(空参数列表)。试着这样做:
<?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!');
}
?>https://stackoverflow.com/questions/7050373
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