每個(gè)表(表1和表2)具有其自己的DATETIME字段.

我正在嘗試捕獲兩個(gè)表的ID,并按其DATETIME字段對其進(jìn)行排序.

例：

Table 1 Table 2

------------ -------------

id | datetime1 id | table1id | datetime2

------------------------ -----------------------

1 | 2014-09-21 20:31:26 1 | 2 | 2014-09-21 20:31:29

2 | 2014-09-21 20:31:27 2 | 3 | 2014-09-21 20:31:30

3 | 2014-09-21 20:31:28

Table 3

------------

id | user

------------------------

2 | phil

3 | nathalie

我的輸出未通過以下查詢正確排序：

SELECT *

FROM (

SELECT

1 AS selection,

table1.id, table1.datetime1,

table2.datetime2

table3.user

FROM Table1

LEFT OUTER JOIN table2

ON table1.id = table2.table1id

LEFT OUTER JOIN table3

ON table1.id = table3.id

UNION ALL

SELECT

2 AS selection,

table1.id, table1.datetime1,

table2.datetime2

table3.user

FROM Table1

INNER JOIN table2

ON table1.id = table2.table1id

INNER JOIN table3

ON table1.id = table3.id

) AS query

ORDER BY table1.datetime1 DESC, table2.datetime2 DESC

所需數(shù)據(jù)：

從表2 id：2,1,

從表1 id：3,2,1

因此：2、1、3、2、1

編輯

對于可能要處理冗長而復(fù)雜的MySQL請求的人們,請?jiān)赑hpmyAdmin中嘗試一下！它會(huì)告訴您錯(cuò)誤！

編輯

解決方法:

您真正需要做的是更仔細(xì)地考慮架構(gòu).考慮將日期時(shí)間列命名為相同的名稱,然后運(yùn)行如下查詢-http://sqlfiddle.com/#!2/a3b4c/7/0

SELECT selection, id, datetimefoo, user FROM (

SELECT

1 AS selection,

table1.id, table1.datetimefoo,

table3.user

FROM table1

LEFT OUTER JOIN table2

ON table1.id = table2.table1id

LEFT OUTER JOIN table3

ON table1.id = table3.id

UNION

SELECT

2 AS selection,

table1.id, table1.datetimefoo,

table3.user

FROM table1

INNER JOIN table2

ON table1.id = table2.table1id

INNER JOIN table3

ON table1.id = table3.id

) AS T2

ORDER BY datetimefoo DESC

在SQL提琴中,這會(huì)產(chǎn)生與您要查找的結(jié)果更接近的結(jié)果.我仍然不確定為什么您需要在第二個(gè)查詢上使用INNER JOINS-在這里您不需要做什么.

SELECT * FROM (

SELECT

1 AS selection,

table1.id, table1.datetimefoo AS sort_date, -- alias on first table's date

table2.datetimebar,

table3.user

FROM table1

LEFT OUTER JOIN table2

ON table1.id = table2.table1id

LEFT OUTER JOIN table3

ON table1.id = table3.id

UNION

SELECT

2 AS selection,

table1.id, table1.datetimefoo,

table2.datetimebar AS sort_date, -- alias on second table's date

table3.user

FROM table1

INNER JOIN table2

ON table1.id = table2.table1id

INNER JOIN table3

ON table1.id = table3.id

) AS T2

ORDER BY sort_date DESC

標(biāo)簽：php,sql,mysql

來源： https://codeday.me/bug/20191012/1898765.html

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