入门选定项目，没有选择

Question

我想找出选择什么项目，什么没被选中成为0

这是列表框（或下拉菜单）是有用的，什么是强调它的意思是它已经加入（LEFT JOIN)

我不能让SQL工作：

SELECT StoreID, ft.id, name, count(1) as Selected 
FROM `store_food_type` AS sft 
    LEFT JOIN food_type AS ft 
     ON ft.id = sft.food_type_id 
WHERE StoreID = 17633 
GROUP BY ft.id 

UNION 

SELECT 0, ft.id, name, 0 as Selected FROM food_type AS ft 

编辑：它并没有产生一个错误，它显示的结果，但name是重复的，我不想

它看起来是这样的：

+------------+------+----------+----------+ 
| StoreID | id | name  | Selected | 
+------------+------+----------+----------+ 
|  17633 | 1 | Indian |  1 | 
|  17633 | 8 | American |  1 | 
|   0 | 1 | Indian |  0 | 
|   0 | 2 | English |  0 | 
|   0 | 25 | Kurdish |  0 | 
|   0 | 5 | Chinese |  0 | 
|   0 | 7 | Turkish |  0 | 
|   0 | 8 | American |  0 | 
+------------+------+----------+----------+ 

我想要的结果出现这样的：因为在你查询UNION的

+------------+------+----------+----------+ 
| StoreID | id | name  | Selected | 
+------------+------+----------+----------+ 
|  17633 | 1 | Indian |  1 | 
|  17633 | 8 | American |  1 | 
|   0 | 25 | Kurdish |  0 | 
|   0 | 5 | Chinese |  0 | 
|   0 | 7 | Turkish |  0 | 
|   0 | 8 | American |  0 | 
+------------+------+----------+----------+ 

Answer 1

你行得到复制。事实上，你的整个查询是反向的 - 你试图获得所有食物类型的列表，并且如果它们被列在给定商店中，则有指示符。

试试这个：

SELECT ft.id, ft.name, COALESCE(sft.storeId, 0) as storeId, 
     CASE WHEN sft.storeId IS NOT NULL THEN '1' 
      ELSE '0' END as selected 
FROM food_type a ft 
LEFT JOIN store_food_type as sft 
ON sft.food_type_id = ft.id 
AND sft.storeId = 17633 

虽然你不严格需要selected列，因为你可以简单地测试结果列storeId是否0。

Answer 2

也许最简单的方法是将集团的结果从被联合查询 - 就像这样：

SELECT MAX(StoreID) StoreID, id, MAX(name) name, MAX(Selected) Selected FROM 
(SELECT StoreID, ft.id, name, count(1) as Selected 
FROM `store_food_type` AS sft 
    LEFT JOIN food_type AS ft 
     ON ft.id = sft.food_type_id 
WHERE StoreID = 17633 
GROUP BY ft.id 
UNION 
SELECT 0, ft.id, name, 0 as Selected FROM food_type AS ft) sq 
GROUP BY id 

Answer 3

我不知道这个答案是比任何由马克·班尼斯特提供的UNION选择更好的，但它是另一种方式来做到这一点。

SELECT sft.storeid, ft.id, ft.name, count(sft.storeid) selected 
FROM store_food_type sft 
     FULL JOIN food_type ft ON ft.id = sft.food_type_id 
WHERE sft.storeid = 17633 or sft.storeid is null 
GROUP BY ft.id, ft.name, sft.storeid 

如果一个食品只能在任何给定的商店中选择一次，那么“selected”字段和GROUP BY是多余的。