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Commit 3133ba5

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feat: add sql solutions to lc problems: No.1083,1084 (doocs#1198)
* No.1083.Sales Analysis II * No.1084.Sales Analysis III
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-78
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6 files changed

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‎solution/1000-1099/1083.Sales Analysis II/README.md‎

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@@ -71,6 +71,10 @@ id 为 1 的买家购买了一部 S8,但是却没有购买 iPhone,而 id 为
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:JOIN + GROUP BY + HAVING**
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我们先将 `Sales` 表和 `Product` 表连接起来,然后根据 `buyer_id` 分组,最后用 `HAVING` 子句筛选出购买了 S8 却没有购买 iPhone 的买家。
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<!-- tabs:start -->
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### **SQL**
@@ -79,23 +83,10 @@ id 为 1 的买家购买了一部 S8,但是却没有购买 iPhone,而 id 为
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# Write your MySQL query statement below
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SELECT buyer_id
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FROM
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(
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SELECT
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buyer_id,
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CASE
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WHEN p.product_name = 'S8' THEN 1
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ELSE 0
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END AS s8,
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CASE
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WHEN p.product_name = 'iPhone' THEN 1
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ELSE 0
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END AS iPhone
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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) AS t
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GROUP BY buyer_id
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HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1083.Sales Analysis II/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -87,23 +87,10 @@ Sales table:
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# Write your MySQL query statement below
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SELECT buyer_id
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FROM
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(
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SELECT
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buyer_id,
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CASE
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WHEN p.product_name = 'S8' THEN 1
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ELSE 0
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END AS s8,
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CASE
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WHEN p.product_name = 'iPhone' THEN 1
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ELSE 0
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END AS iPhone
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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) AS t
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GROUP BY buyer_id
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HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;
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```
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<!-- tabs:end -->
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Original file line numberDiff line numberDiff line change
@@ -1,20 +1,7 @@
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# Write your MySQL query statement below
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SELECT buyer_id
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FROM
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(
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SELECT
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buyer_id,
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CASE
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WHEN p.product_name = 'S8' THEN 1
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ELSE 0
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END AS s8,
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CASE
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WHEN p.product_name = 'iPhone' THEN 1
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ELSE 0
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END AS iPhone
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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) AS t
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GROUP BY buyer_id
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HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;

‎solution/1000-1099/1084.Sales Analysis III/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -85,22 +85,22 @@ id 3的产品在2019年春季之后销售。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:JOIN + GROUP BY + HAVING**
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我们可以通过 `JOIN` 将 `Sales` 表和 `Product` 表连接起来,然后通过 `GROUP BY` 和 `HAVING` 来筛选出符合条件的产品。
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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p.product_id,
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p.product_name
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SELECT product_id, product_name
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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GROUP BY p.product_id
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HAVING
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SUM(s.sale_date < '2019年01月01日') = 0
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AND SUM(s.sale_date > '2019年03月31日') = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING count(1) = sum(sale_date BETWEEN '2019年01月01日' AND '2019年03月31日');
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```
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<!-- tabs:end -->

‎solution/1000-1099/1084.Sales Analysis III/README_EN.md‎

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@@ -87,16 +87,12 @@ We return only product 1 as it is the product that was only sold in the spring o
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```sql
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# Write your MySQL query statement below
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SELECT
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p.product_id,
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p.product_name
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SELECT product_id, product_name
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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GROUP BY p.product_id
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HAVING
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SUM(s.sale_date < '2019年01月01日') = 0
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AND SUM(s.sale_date > '2019年03月31日') = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING count(1) = sum(sale_date BETWEEN '2019年01月01日' AND '2019年03月31日');
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```
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<!-- tabs:end -->
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@@ -1,11 +1,7 @@
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# Write your MySQL query statement below
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SELECT
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p.product_id,
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p.product_name
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SELECT product_id, product_name
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FROM
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Product AS p
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JOIN Sales AS s ON p.product_id = s.product_id
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GROUP BY p.product_id
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HAVING
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SUM(s.sale_date < '2019年01月01日') = 0
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AND SUM(s.sale_date > '2019年03月31日') = 0;
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING count(1) = sum(sale_date BETWEEN '2019年01月01日' AND '2019年03月31日');

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