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SELECT LAST_INSERT_ID() 在直接 MySQL 查詢中工作，但不是間歇性 PHP QUERY

[英]SELECT LAST_INSERT_ID() works in Direct MySQL Query but not intermittently PHP QUERY

 原文  2011-06-08 13:37:53  2   5    php/ mysql

問題描述

（當我寫這篇文章時，它適用於 MSSQL 數據庫沒有問題，現在我正在使用 SELECT LAST_INSERT_ID() 而不是 @SCOPE_IDENTITY 為 MySQL 寫同樣的東西）我有一段代碼進入公司然后該公司的客戶詳細信息。 第一個 SELECT LAST_INSERT_ID() 適用於將 company_id 從公司表插入到客戶表中，但隨后會引發錯誤以檢索 client_id。

代碼如下： //// 設置初始 SQL

$SQLINSERTCOMPANY = "insert into companies(company_name,is_organisation,company_regno,company_url,is_active,entry_date)
            VALUES ('$company_name',$is_organisation,'$company_regno','$company_url',1,'$entry_date');";

$SQLINSERTCOMPANYDETAILS = "insert into clients (client_name,client_tel,client_mobile,client_email,client_usern,client_pwdx,mailing_list,savedsession,salt,company_id)
            VALUES ('$client_name','$client_tel','$client_mobile','$client_email','$client_email','$psw',$sign_up,'$PHPSESSIONID','$salt'";

$SQLUPDATECOMPANY = "update clients 
        set company_name = '$company_name',
        is_organisation = $is_organisation,
        company_regno = '$company_regno',
        company_url = '$company_url',
        is_active = 1,
        entry_date = '$entry_date' ";

$SQLUPDATECOMPANYDETAILS = "update clients 
            set client_name = '$client_name',
            client_tel = '$client_tel',
            client_mobile = '$client_mobile',
            client_email = '$client_email',
            client_usern = '$client_email',
            client_pwdx = '$psw',
            mailing_list = $sign_up,
            savedsession = '$PHPSESSIONID' ";

////get the cart for later use if the cart has items
$cart = $_SESSION['cart'];
if($cart){
    $strSQL = "select *
        from category_items 
        where category_item_id IN ($cart) order by category_items.item_code;";          
    $query_get_value = mysql_query($strSQL) or die ('query failed ' . mysql_error());
}   

///// add a new client here
if($_POST['action'] == "add"){  

    ////// Insert new company or just update the details
    if($no_of_client_entry == 0) {

        ////// update clients
        $strSQL1 = $SQLINSERTCOMPANY;

        ////// Insert Client Details
        $strSQL2 = $SQLINSERTCOMPANYDETAILS . ",(SELECT LAST_INSERT_ID())); SELECT LAST_INSERT_ID() as NEWID;";

        $sent_to = $client_email;
        $headers = "MIME-Version: 1.0" . "\r\n";
        $headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
        $headers .= "FROM: TheImageLounge<no-reply>";
        $subject = "Welcome to The Image Lounge";
        $message = "<html><head><title>Error Report</title>";
        $message .= "<style type='text/css'>
                        body{margin: 0 auto;text-align:left;}
                        h1, h2, h3 { margin:10px;padding:10px;background-color:#ccc;}                   
                     </style>";
        $message .= "</head><body>";

        $message .= "<p>Welcome to Website... TBC/p>";
        $message .= "<p>Your Details Are:</p>";
        $message .= "<p>Your username: $client_email<br />";
        $message .= "<p>Your password: " . $_SESSION['client_psw'] . "</p>";
        $message .= "<p>We hope that you will enjoy using our website.</p>";
        $message .= "<p>Thank You and best regards,<br />";
        $message .= "<p>The Website.</p>";
        $message .= "</body></html>";

        $sentOK = mail($sent_to,$subject,$message,$headers);    

    } else { ///// just update the client details

        $client_id = $clients_rows['client_id'];

        ////// update clients
        $strSQL1 = $SQLUPDATECOMPANY . " where client_id = $client_id;";

        ////// Insert Client Details
        $strSQL2 = $SQLUPDATECOMPANYDETAILS . " where client_id = $client_id; SELECT client_id as NEWID from clients where client_id = $client_id;";            

    } ///// End if
    ///echo $strSQL2;
    ///// Run transactions

    echo $strSQL1 . "<br />";
    echo $strSQL2;

    $query_insert_client = mysql_query($strSQL1) or die('' . mysql_error());
    $query_insert_clients = mysql_query($strSQL2) or die('' . mysql_error());   

    $last_client = mysql_fetch_assoc($query_insert_clients);
    $client_id =  $last_client['NEWID'];
    mysql_free_result($query_insert_clients);

現在當我 output SQL 看起來像這樣：

insert into companies(company_name,is_organisation,company_regno,company_url,is_active,entry_date) VALUES ('Individual',0,'','',1,'2011-06-08 14:16:33');

insert into clients (client_name,client_tel,client_mobile,client_email,client_usern,client_pwdx,mailing_list,savedsession,salt,company_id) VALUES ('Andis','0777','','andrew@com','andrew@com','6be86ab8355979352cdd28bbe7026be71a196936fc1ab0c315629f68c0d5f9162acd20b201a27d0ced0cfd53a8e880a1e8bf3714833ed9c4cbe6d0fe3bf15ca0',0,'8730dfec5ce01b315e1a31007185c486','SALTYDOG',(SELECT LAST_INSERT_ID())); SELECT LAST_INSERT_ID() as NEWID;

但后來我得到這個錯誤

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '; SELECT LAST_INSERT_ID() as NEWID' at line 2

如果我將輸出的 SQL 復制並粘貼到 MySQL 中，它工作正常。 所以我不確定為什么在第二個查詢中失敗。 這在使用 MSSQL 數據庫時工作正常。

任何幫助將不勝感激。

謝謝

和我

5 個解決方案

解決方案1
 2  2011-06-08 13:42:33

您正嘗試在一個 go 中發送多個查詢。 來自文檔：

mysql_query() 發送唯一查詢（不支持多個查詢）

您可能還想考慮使用 mysql_insert_id function 而不是發出手動查詢： http://www.php.net/manual/en/function.mysql-insert-id.ZE1BFD762321E409CEE4ACB8401

解決方案2
 1  2011-06-08 13:45:57

error:

 1. u did not close the insert tag

just do simple change
$addorder = "INSERT INTO table_name () 
                        VALUES  ()"; 
$id = "SELECT LAST_INSERT_ID() as new_id"; 
mysql_query($addorder) or die("not added in to the orders db:<br />".mysql_error()."<br />".$addorder."<hr />"); 
echo($id);

解決方案3
 1 已采納  2011-06-08 13:47:39

您不能使用mysql_query()同時運行 2 個查詢，但是您可以使用 PDO 執行此操作。

為了實現您的目標，您可以完全刪除SELECT LAST_INSERT_ID() as NEWID並使用mysql_insert_id()

解決方案4
 0  2011-06-08 13:43:27

SELECT LAST_INSERT_ID() is a query that runs in mysql, but php only allows you to run one query at a time, thus running the SELECT LAST_INSERT_ID() within your query does not work.

我的建議是將變量分配給 SELECT LAST_INSERT_ID() 的值，然后在查詢中插入變量。

解決方案5
 0  2011-06-08 13:44:55

It's not the SELECT statement that's wrong, it's the fact that you're doing two statements in one go, which is not allowed by the MySQL library of PHP. 只需執行兩個不同的查詢即可解決問題。

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