Вопрос

My problem is this: I Created a store procedure from php to mysql. Now before someone says something about that practice, The procedure works fine. The problem is this, when I call the procedure from php and I KNOW that a record has been entered, mysql_insert_id(), returns 0... Any ideas why?

Oh, and yes, my id field is AUTO_INCREMENT and PRIMARY KEY

p.s If there's an easy way to format my code in here, pls tell me. I put pre in html, but it doesn't seem to work very well.

Ty in advance.

$PROCEDURE_INSERT_UPDATE_BOARD = 
"
CREATE PROCEDURE InsertUpdateBoard(IN `pNewBoardUUID` CHAR(36),
                                   IN `pOldBoardUUID` CHAR(36),
                                   IN `pSimName` VARCHAR(40), 
                                   IN `pOwnerName` VARCHAR(64),
                                   IN `pOwnerUUID` CHAR(36),
                                   IN `pLandmark` VARCHAR(80),
                                   IN `pVersion` VARCHAR(10),
                                   IN `pManagerUUID` CHAR(36),
                                   IN `pBoardURL`    CHAR(80),
                                   IN `pPassword` CHAR(8))
    BEGIN   
        CALL InsertUpdateSims(`pSimName`); 
        CALL InsertUpdateAvatars(`pOwnerName`, `pOwnerUUID`);

        INSERT INTO boards(`boardUUID`, `simId`, `ownerId`, `landmark`,
                                   `version`, `managerId`, `boardURL`, `password`)
                            VALUES(`pOldBoardUUID`,                                      
                            (SELECT rS.id FROM 2starsglobal.sims AS rS
                                    WHERE rS.name=`pSimName`),
                            (SELECT rA.id FROM 2starsglobal.avatars AS rA
                                    WHERE rA.UUID = `pOwnerUUID`),
                            `pLandmark`, `pVersion`,
                            (SELECT rA2.id FROM 2starsglobal.avatars AS rA2 
                                    WHERE rA2.UUID = `pManagerUUID`),
                            `pBoardURL`, 
                            `pPassword`)
        ON DUPLICATE KEY UPDATE `boardUUID`=`pNewBoardUUID`, 
                    `landmark`=`pLandmark`, `version`=`pVersion`,
                    `boardURL`=`pBoardURL`; 
    END
";

And the Php code

function InsertUpdateBoard($boardNewUUID, $boardOldUUID, $simName, 
                           $ownerName, $ownerUUID, $boardLandmark,
                           $versionNumber, $managerUUID, $boardURL)
{ 
    $password = generatePassword(8);

    $query = "CALL InsertUpdateBoard('$boardNewUUID', '$boardOldUUID', '$simName', 
                                         '$ownerName', '$ownerUUID', '$boardLandmark', 
                                         '$versionNumber', '$managerUUID', '$boardURL',
                                         '$password')";

    mysql_query($query) or die("ERROR:QUERY_FAILED " . mysql_error());  

    if(mysql_affected_rows() > 0)
    {
        if(mysql_insert_id() > 0)
        {    
            echo "SUCCESS,$password";
        }
        else 
        {
            echo "SUCCESS"; 
        }
    }
    else
    {
        echo 'FAILED:BOARD_REGISTRATION_UPDATE'; 
    }       

}
Это было полезно?

Решение

Ten seconds with google found this page

Note that if you Call a MySQL stored procedure to insert a new record and then reference $db->insert_id; you will get 0 back, not the last inserted ID.

It is therefore necessary to add a line to your MySQL Stored Procedure such as

select last_insert_id() as intRecordKey;

after the insert so that the query will return the new key value.

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