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我寫了下面的代碼,當它運行時,無論SQL查詢是什麼,我都會得到「錯誤:查詢是空的」。我知道生成的SQL很好,因爲如果我將它粘貼到SQL數據庫中,它將返回行。記錄集的PHP代碼已從其他工作正常的頁面粘貼,因此我無法查看錯誤的位置。PHP記錄集返回「錯誤:查詢是空的」
殼體4://檢查是否用戶先前已加載的清單
$mySQL = "SELECT * FROM tools_userChklists WHERE chklistID = '" . $_GET['chklistID'] . "'";
echo $mySQL;
$query_rsChecklists = $mySQL;
$rsChecklists = mysql_query($query_rsChecklists) or die(mysql_error());
$row_rsChecklists = mysql_fetch_assoc($rsChecklists);
$totalRows_rsChecklists = mysql_num_rows($rsChecklists);
if ($totalRows_rsChecklists <> 0){
//the user has already opened this checklist
$_SESSION['redirectorAction'] = "";
$_SESSION['redirectorAction'] = "Location: actions.php?action=5&chklistID=" . $_GET['chklistID'];
}else{
//the user has never opened this checklist
$_SESSION['redirectorAction'] = "";
$_SESSION['redirectorAction'] = "Location: actions.php?action=6&chklistID=" . $_GET['chklistID'];
}
break;
任何幫助理解。謝謝。
如前所述,我已經使用相同的格式代碼記錄幾年現在沒有問題。我會試一試,但我不太確定它會有所作爲。感謝您的輸入。 – MingMing 2011-12-29 23:23:48