我创建了一个PHP,可以在另一个页面上从原始表单成功收集数据。我知道这是因为我运行了自己的调试技术,它显示所有变量都成功通过。
我的问题出现在下面的脚本中。我试图让它更新表格中的多个字段,但当$firstop
和$secondop
及其值出现时,它只会更新$firstop
选项。怎么解决这个问题?
<?php
$fname = trim($_POST['upname']);
$sname = trim($_POST['upsurename']);
$firstop = trim($_POST['firstup']);
$update1 = trim($_POST['update1']);
$secondop = trim($_POST['secondup']);
$update2 = trim($_POST['update2']);
$var = trim($_POST['submit']);
$con=mysqli_connect("localhost","root","","fypannadale");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($firstop)) {
switch ($firstop) {
case "FirstName":
mysqli_query($con,"UPDATE members SET FirstName='$update1'
WHERE FirstName='$fname' AND LastName='$sname'");
break;
case "LastName":
mysqli_query($con,"UPDATE members SET LastName='$update1'
WHERE FirstName='$fname' AND LastName='$sname'");
break;
}
}
if (isset($secondop)) {
switch ($secondop) {
case "FirstName1":
mysqli_query($con,"UPDATE members SET FirstName='$update2' WHERE FirstName='$fname' AND LastName='$sname'");
break;
case "LastName1":
mysqli_query($con,"UPDATE members SET LastName='$update2' WHERE FirstName='$fname' AND LastName='$sname'");
break;
}
}?>