Alawn/Interview-Notebook
1
0
Fork 0
mirror of https://github.com/CyC2018/Interview-Notebook.git synced 2026-10-21 00:03:41 +00:00
Code Issues Releases Wiki Activity

auto commit

Browse Source
This commit is contained in:
CyC2018
2019-12-11 00:26:43 +08:00
parent 3b3fac2878
commit 612d266233
2 changed files with 32 additions and 14 deletions
Download Patch File Download Diff File Expand all files Collapse all files
docs/notes/Leetcode-Database 题解.md
+16 -7
View File
@@ -230,7 +230,7 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
对 class 列进行分组之后，再使用 count 汇总函数统计每个分组的记录个数，之后使用 HAVING 进行筛选。HAVING 针对分组进行筛选，而 WHERE 针对每个记录（行）进行筛选。
```sql
SELECT
@@ -293,7 +293,7 @@ https://leetcode.com/problems/duplicate-emails/description/
## Solution
对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
对 Email 进行分组，如果并使用 COUNT 进行计数统计，结果大于等于 2 的表示 Email 重复。
```sql
SELECT
@@ -354,7 +354,7 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
连接：
连接查询：
```sql
DELETE p1
@@ -373,7 +373,14 @@ DELETE
FROM
Person
WHERE
id NOT IN ( SELECT id FROM ( SELECT min( id ) AS id FROM Person GROUP BY email ) AS m );
id NOT IN (
SELECT id
FROM (
SELECT min( id ) AS id
FROM Person
GROUP BY email
) AS m
);
```
应该注意的是上述解法额外嵌套了一个 SELECT 语句，如果不这么做，会出现错误：You can't specify target table 'Person' for update in FROM clause。以下演示了这种错误解法。
@@ -383,7 +390,11 @@ DELETE
FROM
Person
WHERE
id NOT IN ( SELECT min( id ) AS id FROM Person GROUP BY email );
id NOT IN (
SELECT min( id ) AS id
FROM Person
GROUP BY email
);
```
参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
@@ -392,8 +403,6 @@ WHERE
与 182 相同。
# 175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/
notes/Leetcode-Database 题解.md
+16 -7
View File
@@ -230,7 +230,7 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
对 class 列进行分组之后，再使用 count 汇总函数统计每个分组的记录个数，之后使用 HAVING 进行筛选。HAVING 针对分组进行筛选，而 WHERE 针对每个记录（行）进行筛选。
```sql
SELECT
@@ -293,7 +293,7 @@ https://leetcode.com/problems/duplicate-emails/description/
## Solution
对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
对 Email 进行分组，如果并使用 COUNT 进行计数统计，结果大于等于 2 的表示 Email 重复。
```sql
SELECT
@@ -354,7 +354,7 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
连接：
连接查询：
```sql
DELETE p1
@@ -373,7 +373,14 @@ DELETE
FROM
Person
WHERE
id NOT IN ( SELECT id FROM ( SELECT min( id ) AS id FROM Person GROUP BY email ) AS m );
id NOT IN (
SELECT id
FROM (
SELECT min( id ) AS id
FROM Person
GROUP BY email
) AS m
);
```
应该注意的是上述解法额外嵌套了一个 SELECT 语句，如果不这么做，会出现错误：You can't specify target table 'Person' for update in FROM clause。以下演示了这种错误解法。
@@ -383,7 +390,11 @@ DELETE
FROM
Person
WHERE
id NOT IN ( SELECT min( id ) AS id FROM Person GROUP BY email );
id NOT IN (
SELECT min( id ) AS id
FROM Person
GROUP BY email
);
```
参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
@@ -392,8 +403,6 @@ WHERE
与 182 相同。
# 175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/