2015-11-27 51 views
1

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); 
+2

混合mysql和mysqli !! – Saty

回答

3

你必須更改查詢:

$sqlCD = mysql_query("UPDATE nmc_cd SET CDTitle=$pCDTitle, CDYear=$pCDYear, CDPrice=$pCDPrice"); 
$sqlCD2 = mysql_query("UPDATE nmc_category SET catDesc=$pCDCategory"); 
$sqlCD3 = mysql_query("UPDATE nmc_publisher SET pubName=$pCDPubName"); 

原因: 如果雙qoutes「」的字符串有一個像$測試一個PHP變量的值將自動在字符串中,但如果您使用單個qoute'',則會打印其變量名稱。 例如:

$test = 5; 
echo "The number is $test"; //The number is 5 prints 
echo 'The number is $test'; //The number is $test prints 
1

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;  
 

 
$sqlCD = mysqli_query($conn,"UPDATE nmc_cd SET CDTitle='$pCDTitle', CDYear='$pCDYear', CDPrice='$pCDPrice'") or die(mysqli_error($conn)); 
 
$sqlCD2 = mysqli_query($conn,"UPDATE nmc_category SET catDesc='$pCDCategory'") or die(mysqli_error($conn)); 
 
$sqlCD3 = mysqli_query($conn,"UPDATE nmc_publisher SET pubName='$pCDPubName'") or die(mysqli_error($conn)); 
 
mysqli_close($conn);

嘗試。

+0

謝謝,但實現它取代整個數據庫..我認爲它缺乏行ID,並指示它只更新選定的行... –