doubi8383 2013-10-02 11:59
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使用transaction插入mysql数据库中的两个不同的表

i am working on a project where i have to insert records in two diffetent table at a time so i am using transactions but it donot insert the record when i execute the query from .php file but if i echo query and copy its output from browser and paste it inside phpmyadmin SQL window it works fine

$insert_device_order = "START TRANSACTION;
                                INSERT INTO request 
                                VALUES ('', '', '$spoc_id', '$spoc_username', '', '', 'Device', 'Submitted','$spoc_id' );
                                INSERT INTO devices 
                                VALUES ( '', LAST_INSERT_ID( ) , '$device_msisdn', '$device_payment','$device_offer', '$device_device' );  
                            COMMIT;";

    $result = mysql_query($insert_device_order);

    if($result){header("Location: addDevices2.php");}else{ echo "cannot insert";}

i am unable to find the issue as i checked my config file is also included at the top.

i use this method to to over come this issue but it also do not worked

mysql_query("START TRANSACTION");
    mysql_query("INSERT INTO request VALUES ('', '', '$spoc_id', '$spoc_username', '', '', 'Device', 'Submitted','$spoc_id' )");
    mysql_query("INSERT INTO devices VALUES ( '', LAST_INSERT_ID( ) , '$device_msisdn', '$device_payment','$device_offer', '$device_device' )");
    mysql_query('COMMIT');
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  • doutan1875 2013-10-02 12:02
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    If you read the fine manual:

    Warning

    This extension is deprecated as of PHP 5.5.0, and will be removed in the future.

    ...

    mysql_query() sends a unique query (multiple queries are not supported) to the currently active database on the server that's associated with the specified link_identifier.

    Parameters

    query

    An SQL query

    The query string should not end with a semicolon.

    Emphasis mine. You're sending four queries (and ending them all with a semicolon, to boot).

    Call mysql_query multiple times. Or better yet, stop using mysql_*.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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