Question

我有一个问题，我写了一个存储过程，将数据插入到一堆表中，最后通过选择它返回和id。当我在PHPMyAdmin中执行存储过程时，插入数据并返回ID。

当使用MySQLi从PHP执行过程时，数据被插入到表中，但是当我检查结果以检索新id时，结果似乎是空白的。任何帮助在这里将不胜感激。我不确定该过程在返回包含具有新ID的行的结果之前是否返回空结果。 phpMyAdmin执行结果中的空行对我来说有点麻烦？

例程sp_createEvent

的执行结果

的EventID 30

存储过程：

DELIMITER $$
CREATE PROCEDURE `sp_createEvent` (IN varEventName varchar(200), IN varUserID INT, IN varPrintingRequested INT, IN varEventCode BIGINT, IN varActiveFrom DATETIME, IN varActiveTo DATETIME, IN varLimitGuestPhotos BIT)
    LANGUAGE SQL
    DETERMINISTIC
    SQL SECURITY DEFINER
    COMMENT 'A procedure'
    BEGIN
    DECLARE albumname VARCHAR(50);
    DECLARE albumslug VARCHAR(50);
    DECLARE usertopalbum INT;
    DECLARE eventalbumid INT;
    DECLARE fullphotosgalleryid INT;
    DECLARE photostripgalleryid INT;
    DECLARE eventid INT;

    DECLARE CONTINUE HANDLER FOR NOT FOUND BEGIN END;

    START TRANSACTION; 

    SELECT 
        CONCAT(u.`display_name`, " - ",varEventName), 
        REPLACE(LOWER(CONCAT(RTRIM(u.`display_name`), "-",varEventName)), " ", "-"), 
        uasc.`album_id`
    INTO
        albumname,
        albumslug,
        usertopalbum
    FROM `wp_users` u
    LEFT JOIN `wp_pocketbooth_useralbumshortcode` uasc on uasc.`wp_user_id` = u.ID
    WHERE u.`ID` = varUserID and
    uasc.`album_id` is not null;

    INSERT INTO `wp_bwg_album`(`id`, `name`, `slug`, `description`, `preview_image`, `random_preview_image`, `order`, `author`, `published`) 
    VALUES (NULL,albumname,albumslug,'smartbooth.co.za Album','','','1','1','1');

    SET eventalbumid = LAST_INSERT_ID();

    INSERT INTO `wp_bwg_gallery`(`id`, `name`, `slug`, `description`, `page_link`, `preview_image`, `random_preview_image`, `order`, `author`, `published`, `gallery_type`, `gallery_source`, `autogallery_image_number`, `update_flag`) 
    VALUES (NULL, CONCAT(albumname, ' (Full Photos)'), CONCAT(albumslug, '-full-photos'),'smartbooth.co.za Gallery','','','','','1','1',1,'','12','');

    SET fullphotosgalleryid = LAST_INSERT_ID();

    INSERT INTO `wp_bwg_gallery`(`id`, `name`, `slug`, `description`, `page_link`, `preview_image`, `random_preview_image`, `order`, `author`, `published`, `gallery_type`, `gallery_source`, `autogallery_image_number`, `update_flag`) 
    VALUES (NULL, CONCAT(albumname, ' (Photo Strips)'), CONCAT(albumslug, '-photo-strips'),'smartbooth.co.za Gallery','','','','','1','1',1,'','12','');

    SET photostripgalleryid = LAST_INSERT_ID();

    INSERT INTO `wp_bwg_album_gallery`(`id`, `album_id`, `is_album`, `alb_gal_id`, `order`) 
    VALUES (NULL,eventalbumid,'0',fullphotosgalleryid,'1');

    INSERT INTO `wp_bwg_album_gallery`(`id`, `album_id`, `is_album`, `alb_gal_id`, `order`) 
    VALUES (NULL,eventalbumid,'0',photostripgalleryid,'1');

    INSERT INTO `wp_pocketbooth_events` (`eventid`, `eventname`, `printingrequested`, `printimages`, `eventcode`, `toplevelalbumid`,`activefrom`,`activeto`, `limitguestphotototime`) 
    VALUES (NULL, varEventName, varPrintingRequested, 0, varEventCode, eventalbumid, varActiveFrom, varActiveTo, varLimitGuestPhotos);

    SET eventid = LAST_INSERT_ID();

    INSERT INTO `wp_pocketbooth_eventsubscriptions` (`subscriptionid`, `userid`, `eventid`, `isowner`) 
    VALUES (NULL, varUserID, eventid, 1);

    commit;

    select eventid;

    END $$
    DELIMITER ;

PHP代码：

<?php
include 'connection.php';

$userid = $_POST['userid'];
$eventname = $_POST['eventname'];
$printingrequested = $_POST['printingrequested'];
$eventcode = $_POST['eventcode'];
$activefrom = $_POST['activefrom'];
$activeto = $_POST['activeto'];
$limitphotos = $_POST['limitphotos'];

$sql = "CALL `sp_createEvent` ('$eventname' , '$userid' , '$printingrequested' , '$eventcode' , '$activefrom' ,'$activeto', '$limitphotos');";

$res = $connection->query($sql);

if (!$res) 
{
    echo "Error: " . $sql . "<br>" . mysqli_error($connection);
}
else 
{
    $data = array();
    while($row = mysql_fetch_array($res)) {
        $data[] = $row['eventid'];
    }

    echo json_encode($data);
}



@mysqli_close($conn);

?>

Answer 1

而不是SET eventid = LAST_INSERT_ID（）; 写

select LAST_INSERT_ID() into eventid;

MySQL过程将空结果返回给PHP

1 个答案: