我有三个表：RewriteEngine on
#don't do anything if index.php is called directly
RewriteRule ^index\.php - [L]
#if other php file called(user typed it in), redirect to extensionless uri
RewriteCond %{THE_REQUEST} ^GET\ /(.+)\.php
RewriteRule ^ /%1? [R=302,L]
#if extensionless uri maps to a real php file, forward to index.php
RewriteCond %{DOCUMENT_ROOT}/$1\.php -f [OR]
#or if not a real file or directory forward to index.php
RewriteCond %{SCRIPT_FILENAME} !-d
RewriteCond %{SCRIPT_FILENAME} !-f
RewriteRule ^(.+)/?$ /index.php [L,NC]
，OnInit: function(){
...
this.oPlanningCalendar = this.byId("PC1")
var oBinding = oPlanningCalendar.getBinding("rows");
oBinding.attachDataReceived(this.fnDataReceived, this);
和OnInit: function(){
...
this.oPlanningCalendar = this.byId("PC1");
this.oPlanningCalendar.updateRows = function(sReason) {
this.oPlanningCalendar.updateAggregation("rows");
var oBinding = this.oPlanningCalendar.getBinding("rows");
if (oBinding) {
jQuery.sap.log.info("max rows = " + oBinding.getLength() || 0);
}
}.bind(this);
。
表格如下所示：
albums表：
songs images表：
albums id | title | image_id
5 | First Album | 1
6 | Another Album | 2
表：
songs我正在尝试显示带有标题和相应图像的歌曲列表。
如果歌曲没有图像（如果id | album_id | title | image_id
32 | 5 | My Song | 3
33 | 5 | Another One | 4
34 | 5 | First Song | 0
35 | 6 | My Song #2 | 0
36 | 6 | Fancy Title | 0
37 | 6 | My Love Song | 5
列为images），那么我只想使用相册图片。
到目前为止我唯一的代码就是这个，我被困住了：
id | path
1 | path/to/image1.jpg
2 | path/to/image2.jpg
3 | path/to/image3.jpg
4 | path/to/image4.jpg
5 | path/to/image5.jpg
如果它是一个可以处理大型表的高效查询，那就太好了。
答案 0 :(得分：3)
这样的事情应该有效：
SELECT s.title, COALESCE(si.path, ai.path)
FROM albums AS a
INNER JOIN songs s ON a.id = s.album_id
LEFT JOIN images ai ON i.id = a.image_id
LEFT JOIN images si ON i.id = s.image_id
请注意，您必须将images表加入albums表和songs表。 ai和si充当这些联接的单独命名空间，以便COALESCE知道如何查找可供选择的选项。
（请注意INNER JOIN是值得商榷的，具体取决于您对数据的期望。）
答案 1 :(得分：1)
您也可以这样使用UNION：
SELECT songs.title, images.path
FROM songs
LEFT JOIN images ON (images.id = songs.image_id)
WHERE songs.image_id != 0
UNION
SELECT songs.title, images.path
FROM songs
LEFT JOIN albums ON (songs.album_id = albums.id)
LEFT JOIN images ON (images.id = albums.image_id)
WHERE albums.image_id != 0 AND songs.image_id = 0
基本上，您首先选择带有图像的所有歌曲，然后如果歌曲本身没有，则将所有歌曲附加到专辑图像。
答案 2 :(得分：1)
这是一个使用多个LEFT JOIN的解决方案，使用您提供的数据进行演示。
最终查询：
SELECT
s.title,
COALESCE(i1.path, i2.path)
FROM songs s
LEFT JOIN albums a ON s.album_id = a.id
LEFT JOIN images i1 ON s.image_id = i1.id
LEFT JOIN images i2 ON a.image_id = i2.id
ORDER BY s.id;
以下是完整演示。
SQL：
-- Data for demo
create table albums(id int, title char(100), image_id int);
insert into albums values
(5 , 'First Album' , 1),
(6 , 'Another Album' , 2);
create table songs(id int, album_id int, title char(100), image_id int);
insert into songs values
(32 , 5 , 'My Song' , 3),
(33 , 5 , 'Another One' , 4),
(34 , 5 , 'First Song' , 0),
(35 , 6 , 'My Song #2' , 0),
(36 , 6 , 'Fancy Title' , 0),
(37 , 6 , 'My Love Song' , 5);
create table images(id int, path char(200));
insert into images values
(1 , 'path/to/image1.jpg'),
(2 , 'path/to/image2.jpg'),
(3 , 'path/to/image3.jpg'),
(4 , 'path/to/image4.jpg'),
(5 , 'path/to/image5.jpg');
SELECT * FROM albums;
SELECT * FROM songs;
SELECT * FROM images;
-- SQL needed
SELECT
s.title,
COALESCE(i1.path, i2.path)
FROM songs s
LEFT JOIN albums a ON s.album_id = a.id
LEFT JOIN images i1 ON s.image_id = i1.id
LEFT JOIN images i2 ON a.image_id = i2.id
ORDER BY s.id;
输出：
mysql> SELECT
-> s.title,
-> COALESCE(i1.path, i2.path)
-> FROM songs s
-> LEFT JOIN albums a ON s.album_id = a.id
-> LEFT JOIN images i1 ON s.image_id = i1.id
-> LEFT JOIN images i2 ON a.image_id = i2.id
-> ORDER BY s.id;
+--------------+----------------------------+
| title | COALESCE(i1.path, i2.path) |
+--------------+----------------------------+
| My Song | path/to/image3.jpg |
| Another One | path/to/image4.jpg |
| First Song | path/to/image1.jpg |
| My Song #2 | path/to/image2.jpg |
| Fancy Title | path/to/image2.jpg |
| My Love Song | path/to/image5.jpg |
+--------------+----------------------------+
6 rows in set (0.00 sec)