固定!抱歉浪费任何人的时间:P
我试图在PHP 中使用mysql将两个列与一个列连接起来
$result = $link->query('SELECT Software.*, Genre.Naam AS Genre, Console.Naam AS ConsoleNaam, Draager.Naam AS MediumNaam, Studio.Naam AS StudioNaam, Publisher.Naam AS PublisherNaam FROM Software INNER JOIN Genre ON Genre.Genre_ID = Software.Genre_ID INNER JOIN Console ON Console.Console_ID = Software.Console_ID INNER JOIN Draager ON Draager.Draager_ID = Software.Medium_ID INNER JOIN Company AS Studio ON Software.Studio_ID = Studio.Company_ID INNER JOIN Company AS Publisher ON Software.Publisher_ID = Publisher.Company_ID ');
Table Company必须加入ID上的Table Software才能获得公司名称。
关于我必须对上面的脚本进行哪些更改才能达到目的的任何提示工作
只有一个答案:
JION在关联表时拼写为JOIN:D
$result = $link->query('SELECT Software.*,
Genre.Naam AS Genre,
Console.Naam AS ConsoleNaam,
Draager.Naam AS MediumNaam,
Studio.Naam AS StudioNaam,
Publisher.Naam AS PublisherNaam
FROM Software
INNER JOIN Genre ON Genre.Genre_ID = Software.Genre_ID
INNER JOIN Console ON Console.Console_ID = Software.Console_ID
INNER JOIN Draager ON Draager.Draager_ID = Software.Medium_ID
INNER JOIN Company AS Studio ON Software.Studio_ID = Studio.Company_ID
INNER JOIN Company AS Publisher ON Software.Publisher_ID = Publisher.Company_ID
');
您有一个打字错误。JOIN
拼错为JION
:
$result = $link->query('SELECT Software.*,
Genre.Naam AS Genre,
Console.Naam AS ConsoleNaam,
Draager.Naam AS MediumNaam,
Studio.Naam AS StudioNaam,
Publisher.Naam AS PublisherNaam
FROM Software
INNER JOIN Genre ON Genre.Genre_ID = Software.Genre_ID
INNER JOIN Console ON Console.Console_ID = Software.Console_ID
INNER JOIN Draager ON Draager.Draager_ID = Software.Medium_ID
/* Misspelled JOIN below *(/
INNER JOIN Company AS Studio ON Software.Studio_ID = Studio.Company_ID
INNER JOIN Company AS Publisher ON Software.Publisher_ID = Publisher.Company_ID
');