mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-09-12 02:41:42 +08:00
存量题库数据更新
This commit is contained in:
@@ -1,4 +1,4 @@
|
||||
<p>部门表 <code>Department</code>:</p>
|
||||
<p>表 <code>Department</code>:</p>
|
||||
|
||||
<pre>
|
||||
+---------------+---------+
|
||||
@@ -8,19 +8,26 @@
|
||||
| revenue | int |
|
||||
| month | varchar |
|
||||
+---------------+---------+
|
||||
(id, month) 是表的联合主键。
|
||||
在 SQL 中,(id, month) 是表的联合主键。
|
||||
这个表格有关于每个部门每月收入的信息。
|
||||
月份(month)可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。
|
||||
月份(month)可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p>编写一个 SQL 查询来重新格式化表,使得新的表中有一个部门 id 列和一些对应 <strong>每个月 </strong>的收入(revenue)列。</p>
|
||||
<p>重新格式化表格,使得 <strong>每个月 </strong>都有一个部门 id 列和一个收入列。</p>
|
||||
|
||||
<p>查询结果格式如下面的示例所示:</p>
|
||||
<p>以 <strong>任意顺序</strong> 返回结果表。</p>
|
||||
|
||||
<p>结果格式如以下示例所示。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
Department 表:
|
||||
<b>输入:</b>
|
||||
Department table:
|
||||
+------+---------+-------+
|
||||
| id | revenue | month |
|
||||
+------+---------+-------+
|
||||
@@ -30,8 +37,7 @@ Department 表:
|
||||
| 1 | 7000 | Feb |
|
||||
| 1 | 6000 | Mar |
|
||||
+------+---------+-------+
|
||||
|
||||
查询得到的结果表:
|
||||
<b>输出:</b>
|
||||
+------+-------------+-------------+-------------+-----+-------------+
|
||||
| id | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
|
||||
+------+-------------+-------------+-------------+-----+-------------+
|
||||
@@ -39,6 +45,5 @@ Department 表:
|
||||
| 2 | 9000 | null | null | ... | null |
|
||||
| 3 | null | 10000 | null | ... | null |
|
||||
+------+-------------+-------------+-------------+-----+-------------+
|
||||
|
||||
注意,结果表有 13 列 (1个部门 id 列 + 12个月份的收入列)。
|
||||
</pre>
|
||||
<b>解释:</b>四月到十二月的收入为空。
|
||||
请注意,结果表共有 13 列(1 列用于部门 ID,其余 12 列用于各个月份)。</pre>
|
||||
|
Reference in New Issue
Block a user