SELECT column_name(s)
FROM table1
INNER JOIN table2
ON table1.column_name=table2.column_name;
或:
SELECT column_name(s)
FROM table1
JOIN table2
ON table1.column_name=table2.column_name;
注释:INNER JOIN 与 JOIN 是相同的。
假设有两个表:Students 和 Enrollments。
Students 表:
| StudentID | Name | Age |
|---|---|---|
| 1 | Alice | 22 |
| 2 | Bob | 23 |
| 3 | Charlie | 24 |
Enrollments 表:
| EnrollmentID | StudentID | Course |
|---|---|---|
| 101 | 1 | Math |
| 102 | 2 | Science |
| 103 | 4 | History |
SELECT Students.Name, Enrollments.Course
FROM Students
INNER JOIN Enrollments
ON Students.StudentID = Enrollments.StudentID;
得到结果如下:
| Name | Course |
|---|---|
| Alice | Math |
| Bob | Science |
解释:INNER JOIN 返回的是 Students 和 Enrollments 表中 StudentID 匹配的数据。没有匹配上的记录(如 Charlie 和 EnrollmentID 为 103 的数据)将被排除。
在本教程中,我们将使用 RUNOOB 样本数据库。
下面是选自 "Websites" 表的数据:
+----+--------------+---------------------------+-------+---------+
+----+--------------+---------------------------+-------+---------+
<table>
<tr>
<th>1</th>
<th>Google</th>
<th>https://www.google.cm/</th>
<th>1</th>
<th>USA</th>
</tr>
<tr>
<td>3</td>
<td>菜鸟教程</td>
<td>http://www.runoob.com/</td>
<td>4689</td>
<td>CN</td>
</tr>
<tr>
<td>4</td>
<td>微博</td>
<td>http://weibo.com/</td>
<td>20</td>
<td>CN</td>
</tr>
<tr>
<td>5</td>
<td>Facebook</td>
<td>https://www.facebook.com/</td>
<td>3</td>
<td>USA</td>
</tr>
<tr>
<td>7</td>
<td>stackoverflow</td>
<td>http://stackoverflow.com/</td>
<td>0</td>
<td>IND</td>
</tr>
</table>
+----+---------------+---------------------------+-------+---------+
下面是 "access_log" 网站访问记录表的数据:
mysql> SELECT * FROM access_log;
+-----+---------+-------+------------+
+-----+---------+-------+------------+
<table>
<tr>
<th>1</th>
<th>1</th>
<th>45</th>
<th>2016-05-10</th>
</tr>
<tr>
<td>3</td>
<td>1</td>
<td>230</td>
<td>2016-05-14</td>
</tr>
<tr>
<td>4</td>
<td>2</td>
<td>10</td>
<td>2016-05-14</td>
</tr>
<tr>
<td>5</td>
<td>5</td>
<td>205</td>
<td>2016-05-14</td>
</tr>
<tr>
<td>6</td>
<td>4</td>
<td>13</td>
<td>2016-05-15</td>
</tr>
<tr>
<td>7</td>
<td>3</td>
<td>220</td>
<td>2016-05-15</td>
</tr>
<tr>
<td>8</td>
<td>5</td>
<td>545</td>
<td>2016-05-16</td>
</tr>
<tr>
<td>9</td>
<td>3</td>
<td>201</td>
<td>2016-05-17</td>
</tr>
</table>
+-----+---------+-------+------------+
9 rows in set (0.00 sec)
下面的 SQL 语句将返回所有网站的访问记录:
SELECT Websites.name, access_log.count, access_log.date
FROM Websites
INNER JOIN access_log
ON Websites.id=access_log.site_id
ORDER BY access_log.count;
执行以上 SQL 输出结果如下:

注释:INNER JOIN 关键字在表中存在至少一个匹配时返回行。如果 "Websites" 表中的行在 "access_log" 中没有匹配,则不会列出这些行。
来源:https://www.runoob.com/sql/sql-join-inner.html