Question

我有3张桌子：

place (placeId,placeName,...)
event (eventId, eventName,...)
picture (pictureId,associateId,picturePath,type [1 place 2 event])

这就是我想要的：

当图片类型为1张图片时.associateId = place.placeId返回路径

当图片类型为2 picture.associateId = event.eventId返回路径

1个地方 - ＆gt;很多活动

1个地方 - ＆gt;一张照片

1个事件 - ＆gt;一张照片

可能是这样的（请帮忙）：

SELECT Place.placeName, Picture.picturePath, Event.eventId, ... FROM Place INNER JOIN Event ON Place.placeId = Event.eventPlace INNER JOIN Picture ON (IF Picture.type=1 return Picture.picturePath ELSE Picture.type=2 return Picture.picturePath)

这个咨询给我一个json来显示日期的事件：

SELECT Place.placeName, Place.placePopularity, Picture.picturePath, Event.eventId, Event.eventCount
FROM Place INNER JOIN Picture ON Place.placeId = Picture.associateId 
    INNER JOIN Event ON Place.placeId = Event.eventPlace
WHERE Picture.type = 2 AND Event.eventDate = '$date'
ORDER BY Event.eventCount DESC

Answer 1

使用此

运行if else

    if () { // picture type 1
        // this is the query
        $query = "SELECT pictureTable.picture_path FROM placeTable INNER JOIN picture_table ON placeTable.placeId = picture_table.associateId";
        } else { // picture type 2 means for event

        $query = "SELECT pictureTable.picture_path FROM eventTable INNER JOIN  picture_table ON eventTable.eventId = picture_table.associateId";
       // now you can get the result
       }

希望这有帮助。

MySQL选择IF INNER JOIN

1 个答案: