一種有效的方法來計算第二張表中的外鍵數量並與第一張表中的行一起顯示-PHP-MySQL

Question

我有兩個表tablea和tableb如下；

表A

+--------+-------+-------+-------+------+
|   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B

+--------+-------+-------+-------+------+
|  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  |
+--------+-------+-------+-------+------+

我想生成一個表

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

數字表示tablea fe字段中的條目出現在tableb 。 當前，我正在從tablea獲取所有行，並且在每個循環中，我使用tablea的fe字段使用另一個查詢來獲取tableb的行數。 因此，我有1個主查詢和4個子查詢。 有什么有效的方法，例如加入或什么？

謝謝

Answer 1

嘗試這個：

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