php - 计算第二个表中的外键数量并将其与第一个表中的行一起显示的有效方法 - PHP - MySQL

Question

我有两个表 tablea 和 tableb 如下；

表格

+--------+-------+-------+-------+------+
|   fa   |   fb  |  fc   |  fd   |  fe  |
+--------+-------+-------+-------+------+
|  col1  |  f11  |  f12  |  f13  |  x1  |
+--------+-------+-------+-------+------+
|  col2  |  f21  |  f22  |  f23  |  x2  |
+--------+-------+-------+-------+------+
|  col3  |  f31  |  f32  |  f33  |  x3  |
+--------+-------+-------+-------+------+
|  col4  |  f41  |  f42  |  f43  |  x4  |
+--------+-------+-------+-------+------+

表格

+--------+-------+-------+-------+------+
|  tbba  |  tbbb |  tbbc |  tbbd | tbbe |
+--------+-------+-------+-------+------+
|  cola  |  fa1  |  fa2  |  fa3  |  x1  |
+--------+-------+-------+-------+------+
|  colb  |  fb1  |  fb2  |  fb3  |  x1  |
+--------+-------+-------+-------+------+
|  colc  |  fc1  |  fc2  |  fc3  |  x1  |
+--------+-------+-------+-------+------+
|  cold  |  fd1  |  fd2  |  fd3  |  x2  |
+--------+-------+-------+-------+------+
|  cole  |  fe1  |  fe2  |  fe3  |  x2  |
+--------+-------+-------+-------+------+
|  colf  |  ff1  |  ff2  |  ff3  |  x3  |
+--------+-------+-------+-------+------+
|  colg  |  fg1  |  fg2  |  fg3  |  x3  |
+--------+-------+-------+-------+------+
|  colh  |  fh1  |  fh2  |  fh3  |  x3  |
+--------+-------+-------+-------+------+
|  coli  |  fi1  |  fi2  |  fi3  |  x3  |
+--------+-------+-------+-------+------+
|  colj  |  fj1  |  fj2  |  fj3  |  x4  |
+--------+-------+-------+-------+------+

我想生成一个表格；

+--------+-------+-----+
|  col1  |  f11  |  3  |
+--------+-------+-----+
|  col2  |  f21  |  2  |
+--------+-------+-----+
|  col3  |  f31  |  4  |
+--------+-------+-----+
|  col4  |  f41  |  1  |
+--------+-------+-----+

该数字表示tablea 的fe 字段中的条目在tableb 中出现的次数。目前，我正在从 tablea 中获取所有行，并且在每个循环中，我使用另一个查询来获取 tableb 中的行数，使用 tablea 的 fe 字段。因此，我有 1 个主查询和 4 个子查询。有什么有效的方法，比如加入什么的吗？

谢谢

Answer 1

试试这个:

SELECT a.fa , a.fb , COUNT( b.tbbe) 
FROM  tablea a
LEFT JOIN tableb b ON a.fe = b.tbbe
GROUP BY a.fa