php - MYSQL 错误在第 1 行 - 但 phpmyadmin 说没有错误

Question

这是我的 SQL 查询

INSERT INTO `TABLE` SET `cdOwnerId` = '0', `cdArtist` = 'Amarillo', `cdTitle` = 'Tony Christie', `cdOrder` = '1', `cdType` = 'album', `cdNotes` = 'MAKE THIS WORK LATER', `cdId` = '0006', `cdObtainable` = '8'; 

INSERT INTO `TABLE` SET `cdOwnerId` = '0', `cdArtist` = 'My Way', `cdTitle` = 'Frank Sinatra', `cdOrder` = '2', `cdType` = 'album', `cdNotes` = 'MAKE THIS WORK LATER', `cdId` = '0001', `cdObtainable` = '8'; 

INSERT INTO `TABLE` SET `cdOwnerId` = '0', `cdArtist` = 'Piano concerto 21', `cdTitle` = 'Mozart', `cdOrder` = '3', `cdType` = 'album', `cdNotes` = 'MAKE THIS WORK LATER', `cdId` = '0003', `cdObtainable` = '8';

我得到的结果是......

您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在 'INSERT INTO TABLE SET cdId = '0', cdArtist = 'My Way', ` 附近使用的正确语法m' 在第 1 行

但是通过 phpMyAdmin，它需要它，但是通过 PHP，不...

更新

仍然没有..

 INSERT INTO `TABLE` (`cdOwnerId`, `cdArtist`,`cdTitle`, `cdOrder`, `cdType`, `cdNotes`, `cdId`, `cdObtainable`) VALUES ('2', 'Time to say goodbye', 'Katherine Jenkins', '1', 'album', 'MAKE THIS WORK LATER', '0005', '8'),('2', 'Piano concerto 21 andante', 'Mozart', '2', 'album', 'MAKE THIS WORK LATER', '0003', '8'),('2', 'Wind Beneath My Wings', 'Bette Midler', '3', 'album', 'MAKE THIS WORK LATER', '0002', '8');

您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在 '('2', 'Time to say goodbye', 'Katherine Jenkins', '1', 'album', 'MAKE THIS WORK ' 行附近使用的正确语法1

表结构...

`musicCdsId` int(8) NOT NULL,
`cdOwnerId` int(20) NOT NULL,
`cdArtist` varchar(250) NOT NULL,
`cdTitle` varchar(250) NOT NULL,
`cdOrder` tinyint(2) NOT NULL,
`cdType` varchar(10) NOT NULL,
`cdNotes` varchar(250) NOT NULL,
`cdId` varchar(11) NOT NULL,
`cdObtainable` tinyint(1) NOT NULL,
`cdCost` int(5) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

循环播放多张 CD

if(isset($_POST['musicDetails']) && is_array($_POST['musicDetails'])){
        $orderCount = 1;
        foreach($_POST['musicDetails'] as $detail){
            if(!empty($detail['title']) || !empty($detail['artist'])){
                if(!empty($cdDetails)){$cdDetails .= ",";}
                $cdDetails .= "(".dbsafe(mysql_insert_id()).", ".dbsafe($detail['artist']).", ".dbsafe($detail['title']).", ".dbsafe($orderCount).", ".dbsafe($detail['order']).", ".dbsafe("MAKE THIS WORK LATER").", ".dbsafe($detail['mycdid']).", ".dbsafe(8).")";
                $orderCount++;
            }
        } 
    }   

Answer 1

如果使用“INSERT INTO table-name SET”，则没有问题。但更常用的是“UPDATE SET”和“INSERT INTO table-name (column-names ...) VALUES (...)”。这里，SET 子句明确指示列名称。

这里的问题可能有所不同(因为我不确定 cdOwnerId/cdId 的数据类型是 int 还是其他)。在某些情况下，“整数”数据类型不接受字符方式的值插入(cdOwnerId = '0'、cdId = '0006')。

您可以尝试将它们替换为 (cdOwnerId = 0 , cdId = 6) 并检查

http://mysqlrockstar.blogspot.in/