我有一个PHP代码,当用户删除记录时会发出警报,但是在提醒时刷新页面,然后后台页面变成空白,这看起来很烦人。如果不刷新页面,我该怎么做?下面是我的示例代码deleteuser.php。
<?
include 'mysqlcon.php'; // my connection credentials to php
$postudel = $_POST['enterudel'];
if($postudel!="")
{
$sqldel = "DELETE FROM User WHERE UserName = '$postudel'";
$getdel = mysql_query($sqldel,$con) or die(mysql_error());
echo "<script language='javascript'>";
echo "alert('User $postudel successfully deleted.');";
echo "</script>";
}
?>
<html>
<head>
<title>Delete User</title>
<SCRIPT language=JavaScript>
function deluser()
{
var r=confirm("Are you sure you want to do this?");
if (r==true)
{
var objsel = document.getElementById("duser");
var udel = objsel.options[objsel.selectedIndex].id;
document.formdel.enterudel.value = udel;
document.formdel.submit();
}
else
return false;
}
</script>
</head>
<body>
<table width="50%" align="left">
<tr><td width="50%" colspan="2">
<big>Delete User:</big>
</td></tr>
<tr><td width="25%">
Username:
<SELECT id="duser">
<option>Choose One</option>
<?
$sqldu=mysql_query("SELECT UserName from User");
while($row=mysql_fetch_array($sqldu))
{
echo "<OPTION ID=".$row['UserName'];
echo ">".$row['UserName']."</OPTION>";
}
?>
</SELECT>
</td>
<td width="25%">
<input type="button" value="Delete" id="delete" onClick="deluser()">
</td></tr>
</table>
<form name="formdel" action="../deleteuser.php" method="post">
<input type="hidden" name="enterudel" value="<?=$postudel?>">
</form>
</body>
</html>
答案 0 :(得分:2)
拆分
if($postudel!="")
{
$sqldel = "DELETE FROM User WHERE UserName = '$postudel'";
$getdel = mysql_query($sqldel,$con) or die(mysql_error());
echo "<script language='javascript'>";
echo "alert('User $postudel successfully deleted.');";
echo "</script>";
}
分为两部分:
if($postudel!="")
{
$sqldel = "DELETE FROM User WHERE UserName = '$postudel'";
$getdel = mysql_query($sqldel,$con) or die(mysql_error());
}
现在在哪里,
<?
if($postudel!="")
{
echo "<script language='javascript'>";
echo "alert('User $postudel successfully deleted.');";
echo "</script>";
}
?>
在</body>
之前。然后,您将在警报之前删除并绘制页面,避免空白页