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我写了一个函数来更新记录。表单显示所有正确的数据,但在编辑数据和提交表单后,isset函数不起作用,记录未更新。需要一点帮助。谢谢。Isset检查表单提交不起作用
/*功能来编辑员工*/
function emp_edit()
}
global $link;
$employeeid = $_GET['id'];
$query = "SELECT * FROM employees where employeeid='$employeeid'";
$result = mysqli_query($link, $query);
$row = mysqli_fetch_assoc($result);
$fname = $row['fname'];
$lname = $row['lname'];
$email = $row['email'];
$zip = $row['zip'];
print "<h3>Edit Employee Record</h3>";
print "<form method='post' action='emp_maintenance.php'>
<br><input type='hidden' name='employeeid' value='$employeeid'>
<br><input type='text' name='fname' value='$fname'> First Name<br>
<br><input type='text' name='lname' value='$lname'> Last Name<br>
<br><input type='text' name='email' value='$email'> Email<br>
<br><input type='text' name='zip' value='$zip'> Zip Code<br>
<br><input type='submit' name='edit3' value='Update'><br>
</form>";
if(isset($_POST['edit3']))
{
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$zip = $_POST['zip'];
$query = "UPDATE employees SET fname='$fname', lname='$lname', email = '$email', zip = '$zip' WHERE employeeid = '$employeeid'";
mysqli_query($link, $query);
}
}
'function emp_edit() }' - 这不太可能是“真正的”脚本.... – VolkerK
函数必须在关闭之前打开。在你的脚本中你并没有打开函数,而是声明了$ link;当您尝试运行mysql查询时可能为null –