→ aptx1596:解決了... 12/16 17:22
推 JoeHorn:aupdate1() 裡面的 SQL 會有問題，建議用一次 query 解決 12/17 03:17
→ JoeHorn:UPDATE `$qwe` SET `$asd`=`$asd`+1 12/17 03:19
最近在做一個網站，需要處理大量的資料庫，步驟都相同！
讀出-> +1 ->放回去
用函式來寫卻一直不成功耶！
關鍵程式碼如下：
<?
Function aupdate1($asd,$qwe)
{
$sql = "select $asd from `$qwe`";
$result=mysql_db_query( $mysql_database, $sql,$conn );
$row=mysql_fetch_assoc($result);
$IM_C1=$row[$asd]+1;
$sql = "UPDATE `$qwe` SET $asd ='$IM_C1'";
$result=mysql_db_query( $mysql_database, $sql,$conn );
}
echo"第一題";echo"<BR>";
if($_POST['C1'] == 1)
{
echo "校園安全";
$asd = "C1";
$qwe = "subject_1";
aupdate1($asd,$qwe);
echo"<BR>";
}
?>
她一直顯示：
Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link
resource in C:\AppServ\www\award_act\x_3_1_panel.php on line 7
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result
resource in C:\AppServ\www\award_act\x_3_1_panel.php on line 8
Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link
resource in C:\AppServ\www\award_act\x_3_1_panel.php on line 11
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