假设我有以下表格:
ID | title
---+----------
1 | sandwich
2 | spaghetti
ID | food_reference | type | location | bought
----+----------------+------+----------+----------
100 | 1 | ham | storeA | 11-1-2013
101 | 1 | jam | storeB | 11-1-2013
102 | 2 | tuna | storeB | 11-6-2013
我想要一个选择查询,它将找到所有匹配的行,所需的输出:
{id:1,title:sandwich,ingridients:[
{id:100,type:ham,location:storeA,bought:11-1-2013},
{id:101,type:jam,location:storeB,bought:11-1-2013}],
id:2,title:spaghetti,ingridients:[
{id:102,type:tuna,location:storeB,bought:11-6-2013}]}
到目前为止,我有这样的事情:
SELECT F.id, F.title, group_concat(I.d_id ,I.type,I.location,I.bought SEPARATOR ',') as ingridients,
FROM food F, ingridients I
WHERE F.id=I.food_reference
GROUP BY F.id
问题是,所有这些值被连接(显然)导致此输出:
{id:1,title:sandwich,ingridients:100hamstoreA11-1-2013,101jamstoreB11-1-2013},
{id:2,title:spaghetti,ingridients:102tunastoreB11-6-2013}
注意:此输出是使用对象
上的函数json_encode创建的那么你有什么建议怎么做吗?
答案 0 :(得分:1)
为什么你不能使用内连接,只需尝试
select * from food inner join Ingredients on food.id=Ingridients.food_reference;
然后尝试
SELECT foo.id, foo.title,
group_concat(ing.d_id ,ing.type,ing.location,ing.bought SEPARATOR ',')
as ingredients,FROM food foo, ingridients ing
WHERE foo.id=ing.food_reference;