2013-06-30 76 views
0

在JavaScript代码中的php回来了,因为php statememnt而不是$ _GET值是否有需要更改的配置中的某些内容?这工作在一个大学的Apache服务器,但似乎并没有在这种情况下,PHP和JavaScript在其他网页上工作就好了。删除工作也很好。wamp服务器没有在javascript语句中运行PHP代码

<SCRIPT LANGUAGE="JavaScript"> 
    setTimeout(function(){ 
    alert("<?echo $_GET['deletename']?> Deleted, you will now return to the Artist Search...") 
    window.location = "http://www.aaaaa.tk/CI/index.php/about?deletedartist=<?echo $_GET['deletename']?>"; 
    }, 100); 

</script> 
<link rel="stylesheet" type="text/css" href="main.css"> 
</head> 
<?php 

$artistid = $_GET['deleteid']; 

$username="X"; 
$password="X"; 
$database="X"; 
mysql_connect(localhost,$username,$password) or die("Could not connect : " . mysql_error()); 
@mysql_select_db($database) or die("Unable to select database"); 
/* Performing SQL query */ 
$query = "DELETE FROM fbartist WHERE id =$artistid"; 
$query2 = "ALTER TABLE fbartist AUTO_INCREMENT=1;"; 
$result = mysql_query($query) or die("Delete Query failed :".$query."" . mysql_error()); 
$result2 = mysql_query($query2) or die("Auto Increment Query failed : " . mysql_error()); 

mysql_close(); 
?> 

感谢您给予的任何帮助。

+0

“alert”语句后缺少分号吗? –

回答

2

可尝试

alert("<?php echo .... 

检查php.inishort_open_tag = on,见here

+0

是的,它是在开始标记中缺少的php,我认为我已经完成了所有这些,我的坏。谢谢你的帮助。 – Purgatory