獲取SQL中兒童的記錄總數

我有兩個表格：shop_item（父母）和shop_products（孩子）。獲取SQL中兒童的記錄總數

在表shop_products有列shop_item_id，它表明，他的父母是什麼。有些父母有很多孩子。在表shop_products中也有專欄shop_number，這是唯一的。我的目標是找到所有有孩子的父母，例如123（例如）。我可以通過它achive：

SELECT shop_products.shop_number, shop_products.shop_item_id, shop_item.* 
FROM shop_products, 
    shop_item 
WHERE shop_products.shop_number LIKE '123%' 
    AND shop_item.id = shop_products.shop_item_id;

這一個工作，但我想還可以得到這些孩子的數量（因爲作爲結果，我得到父母的），但我不知道這是否是甚至有可能。

編輯：

實施例表shop_item：

+----+-------+-----+ 
| id | name | ... | 
+----+-------+-----+ 
| 1 | test1 | ... | 
| 2 | test2 | ... | 
| 3 | test3 | ... | 
| 4 | test4 | ... | 
+----+-------+-----+

實施例表shop_products：

+----+--------------+-------------+-----+ 
| id | shop_item_id | shop_number | ... | 
+----+--------------+-------------+-----+ 
| 1 | 1   | 12345  | ... | 
| 2 | 2   | 1234567  | ... | 
| 3 | 1   | 14486  | ... | 
| 4 | 3   | 32333  | ... | 
| 5 | 1   | 12399  | ... | 
| 6 | 4   | 12325  | ... | 
| 7 | 2   | 25511  | ... | 
| 8 | 1   | 42387  | ... | 
+----+--------------+-------------+-----+

期待的結果：

+----+----------+-----+ 
| id | children | ... | 
+----+----------+-----+ 
| 1 | 2  | ... | 
| 2 | 1  | ... | 
| 4 | 1  | ... | 
+----+----------+-----+

來源

2016-08-09 debute

添加一些示例表和預期結果（只是足以描述問題）。 – jarlh

@jarlh見編輯。 – debute

首先，學會使用正確的，明確的join語法。簡單的規則：從不在FROM子句中使用逗號。總是使用JOIN。

然後，你的問題的答案是GROUP BY。所以，找父母

SELECT s.shop_number, COUNT(*) 
FROM shop_products p JOIN 
    shop_item i 
    ON p.shop_number LIKE '123%' AND 
     i.id = p.shop_item_id 
GROUP BY s.shop_number;

如果你想與列的原始詳細數據。接着，添加使用相關子查詢：

SELECT s.shop_number, si.*, 
     (SELECT COUNT(*) FROM shop_item si2 WHERE si2.id = p.shop_item_id) as cnt 
FROM shop_products p JOIN 
    shop_item i 
    ON p.shop_number LIKE '123%' AND 
     i.id = p.shop_item_id;

來源

2016-08-09 10:41:54

你可以試試這個，如果你需要得到唯一的共同項目（即僅擁有父母的孩子，有孩子家長）：

select shop_products.shop_number, shop_products.shop_item_id, 
     shop_item.*, Count(shop_products.shop_item_id) over (partition by shop_products.shop_item_id) as ChildrenCount 
from shop_item 
inner join shop_products 
on shop_products.shop_item_id = shop_item.shop_item_id 
where shop_products.shop_number LIKE '123%'

否則，如果你想獲得數據，而不管共同使用左外連接：

select shop_products.shop_number, shop_products.shop_item_id, 
    shop_item.*, Count(shop_products.shop_item_id) over (partition by shop_products.shop_item_id) as ChildrenCount 
from shop_item 
left outer join shop_products 
on shop_products.shop_item_id = shop_item.shop_item_id 
where shop_products.shop_number LIKE '123%'

來源

2016-08-09 10:58:48 Zeina

獲取SQL中兒童的記錄總數

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