mysql - 存储过程和 mysql_insert_id 问题

Question

我的问题是: 我创建了一个从 php 到 mysql 的存储过程。现在，在有人谈论这种做法之前，该程序运行良好。 问题是这样的，当我从 php 调用该过程并且我知道已经输入了一条记录时，mysql_insert_id() 返回 0...有什么想法吗？

哦，是的，我的 id 字段是 AUTO_INCREMENT 和 PRIMARY KEY

附注 如果这里有一种简单的方法来格式化我的代码，请告诉我。我把 pre 放在了 html 里，但是好像效果不是很好。

提前联系。

$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'; 
    }       

}

Answer 1

用谷歌十秒找到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.