Question

问题示例：

source | target
apple  |   dog
dog    |   cat
door   |   cat
dog    |   apple
cat    |   dog

结果：

apple dog 2
dog   cat 2
door  cat 1

以下是我的问题，例如：

我正在尝试计算来自源和目标的apple和dog事件。该数是2，即; apple dog和dog cat。

以同样的方式; dog cat和cat dog，它们会出现2次。

如何使用mysql？

执行此操作

数据非常大，所以这只是一个简单的例子。

Answer 1

正如我所看到的，您的问题是：计算您的值，使其与列中的顺序无关。因此，对<'foo', 'bar'>应计为<'bar', 'foo'>。为此您可以使用：

SELECT 
  *, 
  COUNT(*) 
FROM 
  test 
GROUP BY 
  LEAST(source, target),
  GREATEST(source, target)

请注意：

将非组列与组功能混合仅适用于MySQL。它是extension，因此服务器可以自由选择任何一行。

Answer 2

假设Source和Target加入了ID，我会这样做：

SELECT
    FirstValue,
    SecondValue,
    COUNT(*) As MyCount
FROM
(SELECT
    SourceTable.Value FirstValue,
    TargetTable.Value SecondValue
FROM
    SourceTable 
    INNER JOIN TargetTable ON SourceTable.IDValue = TargetTable.IDValue
UNION ALL
SELECT
    TargetTable.Value FirstValue,
    SourceTable.Value SecondValue
FROM
    TargetTable 
    INNER JOIN SourceTable ON TargetTable.IDValue = SourceTable.IDValue) 
GROUP BY
    FirstValue,
    SecondValue

再次阅读问题我不确定这些是否是同一个表中的两列。如果是，那么查询可以简化为：

SELECT
    FirstValue,
    SecondValue,
    COUNT(*) As MyCount
FROM
(SELECT
    SourceColumn FirstValue,
    TargetColumn SecondValue
FROM
    MyTable
UNION ALL
SELECT
    TargetColumn FirstValue,
    SourceColumn SecondValue
FROM
    MyTable) 
GROUP BY
    FirstValue,
    SecondValue

Answer 3

 DROP TABLE IF EXISTS my_table;

 CREATE TABLE my_table
 (source VARCHAR(12) NOT NULL
 ,target VARCHAR(12) NOT NULL
 ,PRIMARY KEY(source,target)
 );

 INSERT INTO my_table VALUES
 ('apple','dog'),
 ('dog','cat'),
 ('door','cat'),
 ('dog','apple'),
 ('cat','dog');

 SELECT * FROM my_table;
 +--------+--------+
 | source | target |
 +--------+--------+
 | apple  | dog    |
 | cat    | dog    |
 | dog    | apple  |
 | dog    | cat    |
 | door   | cat    |
 +--------+--------+


 SELECT GREATEST(source,target),LEAST(source,target),COUNT(*) FROM my_table GROUP BY GREATEST(source,target),LEAST(source,target);
 +-------------------------+----------------------+----------+
 | GREATEST(source,target) | LEAST(source,target) | COUNT(*) |
 +-------------------------+----------------------+----------+
 | dog                     | apple                |        2 |
 | dog                     | cat                  |        2 |
 | door                    | cat                  |        1 |
 +-------------------------+----------------------+----------+

mysql - 从两列中查找出现次数

3 个答案: