mysql - SQL:如何过滤以仅返回结果集中的行，其中字段中的某些数据出现在未过滤结果集中的多行中？

Question

如果我有这些数据:

code1 code2
  1    10       <-- Desired (1 appears more than once)
  1    11       <-- Desired (1 appears more than once)
  2    20
  3    30       <-- Desired (3 appears more than once)
  3    31       <-- Desired (3 appears more than once)
  4    40
  5    50

...我想编写一个单个 SQL 查询，其结果如下:

code1 code2
  1    10       <-- This result appears because 1 appears more than once above
  1    11       <-- This result appears because 1 appears more than once above
  3    30       <-- This result appears because 3 appears more than once above
  3    31       <-- This result appears because 3 appears more than once above

(即，返回 code1 列中的任何数据多次出现的所有行的单个 SQL 查询)...

我可以写什么SQL？可能吗？

这是我目前所拥有的，但不起作用:

// WARNING!
// INVALID SQL

SELECT 
  code1,
  code2 
FROM
  mytable 
GROUP BY code1,
  code2 
HAVING COUNT(code1) > 1    <-- This line is invalid

// WARNING!
// INVALID SQL

与其继续与之抗争……我想我会在 StackOverflow 上提问。谢谢!

(允许使用任何特定于 MySQL 的命令，如果它们可能有帮助的话。)

Answer 1

你可以这样试试:

SELECT code1, code2
FROM myTable
WHERE code1 IN 
    (SELECT code1 FROM myTable GROUP BY code1 HAVING COUNT(code1) > 1)

或者像这样使用 INNER JOIN :

SELECT t.code1, code2
FROM myTable t
  INNER JOIN
    (SELECT code1 FROM myTable GROUP BY code1 HAVING COUNT(code1) > 1)
     s on s.code1 = t.code1

SQLFiddle Demo