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我知道這可能有點簡單,但我看不出我的代碼有任何問題,但它不起作用。當點擊提交按鈕時,頁面會一直刷新。我認爲這不是輸入if(isset($_POST['sub']))
的代碼。這裏是我的代碼:提交按鈕不斷刷新頁面
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Add Client</title>
</head>
<body>
<?php
require("dbconnect.php");
?>
<form method="post">
<center>
<h1>Add Client</h1>
<table>
<tr><td>Client Name</td><td><input name="cname" type="text" /></td></tr>
<tr><td>Client Category</td><td><input name="cat" type="text" /></td></tr>
<tr><td>Client Address</td><td><input name"add" type="text" /></td></tr>
<tr><td>Client City</td><td><input name="city" type="text" /></td></tr>
<tr><td>Contact Person</td><td><input name="cper" type="text" /></td></tr>
<tr><td>Contact Details</td><td><input name="cdet" type="text" /></td></tr>
<tr><td>Sales Territory</td><td>
<?php
$a=mysql_query("select `sales-territory-no`, `sales-territory-name` from `sales-territory`") or die(mysql_error());
echo"<select name='terr'>";
while($b=mysql_fetch_array($a))
{
echo "<option value='".$b['sales-territory-no']."'>".$b['sales-territory-name']."</option>";
}
echo "</select>";
?>
</td></tr>
<tr><td>Sales Person ID</td><td>
<?php
$c=mysql_query("select * from `sales`") or die(mysql_error());
echo"<select name='sales'>";
while($d=mysql_fetch_array($c))
{
echo "<option value='".$d['sales-no']."'>".$d['firstname'].' '.$d['lastname']."</option>";
}
echo "</select>";
?>
</td></tr>
</table>
<br />
<input type="submit" name"sub" value="Submit" />
</center>
</form>
<?php
if(isset($_POST['sub']))
{
echo "<script type=\"text/javascript\">";
echo "alert(\"Hello!\");";
echo "</script>";
}
?>
</body>
</html>
我可以」我相信我沒有看到。 -___- – xjshiya
有時你只需要第二雙眼睛;) – Chitowns24