Question

我的表的查询结果需要按以下标准排序：

1）排名值为100或更高（从最高到最低排序）

2）List_Date - 过去90天内最新到最旧的。

3）排名小于100的值（从最高到最低排序）

我尝试了以下内容，但意识到它不会按照MySQL文档工作，“因为UNION默认会产生一组无序的行。”

(SELECT Ranking, List_Date 
FROM example_table
WHERE Ranking > 100
ORDER BY Ranking DESC)

UNION

(SELECT Ranking, List_Date 
FROM example_table
WHERE List_Date BETWEEN CURDATE() - INTERVAL 90 DAY AND CURDATE()
ORDER BY List_Date DESC)

UNION

(SELECT Ranking, List_Date 
FROM example_table
WHERE Ranking < 100
ORDER BY Ranking DESC)

我想弄清楚如何保持三个陈述的顺序。可能的？

Answer 1

您似乎不需要使用UNION，因为您可以使用OR将条件组合在一起。试试这个：

SELECT Ranking, List_Date 
FROM example_table
WHERE Ranking <> 100 
OR List_Date BETWEEN CURDATE() - INTERVAL 90 DAY AND CURDATE()
ORDER BY Ranking, List_Date DESC

Answer 2

我假设您确实需要temp/ temp/txt/，因为正如其他答案所述，您的示例可以在没有它的情况下解决，通常会影响性能。您可以将它全部包装在新的UNION语句中。

SELECT

Answer 3

嗯，虽然我收到的答案没有直接解决我的问题，但他们指出了我正确的方向。这是我的解决方案。它有效，但可能不优雅。

SELECT * FROM (
(SELECT (@cnt := @cnt + 1) AS rowNumber, Ranking, List_Date 
FROM example_table
CROSS JOIN (SELECT @cnt := 0) AS dummy
WHERE Ranking > 100
ORDER BY Ranking DESC)

UNION

(SELECT (@cnt := @cnt + 1) AS rowNumber, Ranking, List_Date 
FROM example_table
CROSS JOIN (SELECT @cnt := (SELECT COUNT(*) FROM `example_table `)) AS dummy
WHERE List_Date BETWEEN CURDATE() - INTERVAL 90 DAY AND CURDATE()
ORDER BY List_Date DESC)

UNION

(SELECT (@cnt := @cnt + 1) AS rowNumber, Ranking, List_Date 
FROM example_table
CROSS JOIN (SELECT @cnt := ((SELECT COUNT(*) FROM `example_table `) * 2)) AS dummy
WHERE Ranking < 100
ORDER BY Ranking DESC)
) as T
ORDER BY rowNumber DESC

将三个MySQL查询结果合并到一个正确排序的结果中

3 个答案: