我試圖更新mysql多個表,雖然沒有錯誤,但更改沒有反映在mysql數據庫中。如果我需要使用JOIN將查詢更新放入一個字符串中,請讓我。更新mysql中的多表
難道我也仍然需要
mysqli_query($conn, $sqlCD) or die(mysqli_error($conn)); in the below ?
$pCDTitle = filter_has_var(INPUT_GET, 'CDTitle') ? $_GET['CDTitle']: null; // store all parameter in variable
$pCDPubName = filter_has_var(INPUT_GET, 'CDPub') ? $_GET['CDPub']: null;
$pCDYear = filter_has_var(INPUT_GET, 'CDYear') ? $_GET['CDYear']: null;
$pCDCategory = filter_has_var(INPUT_GET, 'CDCat') ? $_GET['CDCat']: null;
$pCDPrice = filter_has_var(INPUT_GET, 'CDPrice') ? $_GET['CDPrice']: null;
$pCDID = filter_has_var(INPUT_GET, 'CDID') ? $_GET['CDID']: null;
$pCDPubID = filter_has_var(INPUT_GET, 'pubID') ? $_GET['pubID']: null;
$sqlCD = mysql_query("UPDATE nmc_cd SET CDTitle='$pCDTitle', CDYear='$pCDYear', CDPrice='$pCDPrice'");
$sqlCD2 = mysql_query("UPDATE nmc_category SET catDesc='$pCDCategory'");
$sqlCD3 = mysql_query("UPDATE nmc_publisher SET pubName='$pCDPubName'");
//mysqli_query($conn, $sqlCD) or die(mysqli_error($conn));
//mysqli_query($conn, $sqlCD2) or die(mysqli_error($conn));
//mysqli_query($conn, $sqlCD3) or die(mysqli_error($conn));
mysqli_close($conn);
混合mysql和mysqli !! – Saty