我有來自ajax調用的PHP變量。但是,它們在通過if語句和while循環時未定義,我該如何解決此問題。我無法得到任何工作。代碼的PHP變量在while循環中不可用
$userID = $_POST['id'];
$clubID = $_POST['clubID'];
$type = $_POST['type'];
$delAtt = $_POST['delAtt'];
休息
if($delAtt == 1)
{
$result = mysql_query('SELECT * FROM events WHERE clubID = "'.$clubID.'"');
while ($row == mysql_fetch_assoc($result))
{
$eventID = $row['eventID'];
echo $eventID;
mysql_query('DELETE FROM eventmember WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
mysql_query('DELETE FROM eventmember2 WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
mysql_query('DELETE FROM eventmember3 WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
mysql_query('DELETE FROM attendance WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
mysql_query('DELETE FROM attendance2 WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
mysql_query('DELETE FROM attendance3 WHERE userID = "'.$userID.'" AND eventID = "'.$eventID.'"');
}
}
所有的變量if語句,但不能同時內部循環
沒有什麼錯,你做一個ajax POST或GET?另外,檢查拼寫,區分大小寫。 – Niels
你確定你正在進入if {}塊嗎?也許delAtt是2或「1」 –
我將它設置爲1僅用於測試,此外,當我回顯$ clubID時,它在if {}語句內工作,但不在while循環內 – nshah