【发布时间】:2012-12-10 12:07:02
【问题描述】:
我正在尝试将子查询生成的两个虚拟字段操作到一个新字段中,但 MySQL 告诉我“GP”是一个未知列,但它已经被声明了。请看一下我的查询:
SELECT *,
(SELECT COUNT(id_gol) FROM tb_gol as gol
INNER JOIN tb_jogo as jg ON(gol.fk_id_jogo = jg.id_jogo)
WHERE gol.ic_excluido != '*' AND gol.fk_id_equipe = e.id_equipe AND jg.fk_id_campeonato = g.fk_id_campeonato) as 'GP',
(SELECT COUNT(id_gol) FROM tb_gol as gol
INNER JOIN tb_jogo as jg ON(gol.fk_id_jogo = jg.id_jogo)
WHERE gol.ic_excluido != '*' AND gol.fk_id_equipe != e.id_equipe AND (jg.fk_id_equipe1 = e.id_equipe OR jg.fk_id_equipe2 = e.id_equipe) AND jg.fk_id_campeonato = g.fk_id_campeonato) as 'GC',
(SELECT COUNT(id_wo) as WOs FROM tb_wo as w INNER JOIN tb_jogo as j ON (w.fk_id_jogo = j.id_jogo) WHERE w.fk_id_equipe = e.id_equipe AND j.fk_id_campeonato = g.fk_id_campeonato) as 'WO',
(GP+(GC*-1)) as 'SALDO'
FROM tb_equipe as e
INNER JOIN tb_gruposEquipes as ge ON (e.id_equipe = ge.fk_id_equipe)
INNER JOIN tb_grupos as g ON (g.id_grupo = ge.fk_id_grupo)
WHERE g.fk_id_campeonato = 23
ORDER BY WO ASC
如您所见,“SALDO”将是“GP-GC”的结果。但是 MySQL 不识别这些列
我该如何解决这个问题?
解决方案
非常感谢你们的帮助,伙计们,但我找到了问题所在。 “SALDO”不能通过减去GP的GC来创建,因为“GP”和“GC”在运行时不存在。
因此,当您需要在运行时操作虚拟字段时,您将不得不重复生成虚拟字段的代码。
那么,如果你有这个选择:
SELECT *, (field1+1) as 'GP', (field2+1) as 'GC', (GP+GC) as 'SALDO' FROM (...)
你需要用这个替换它:
SELECT *, (field1+1) as 'GP', (field2+1) as 'GC', ((field1+1)+(field2+1)) as 'SALDO' FROM (...)
【问题讨论】:
-
可以在查询选择列表中使用别名来为列指定不同的名称。您可以在
GROUP BY、ORDER BY或HAVING子句中使用别名来引用该列。查看手册 dev.mysql.com/doc/refman/5.0/en/problems-with-alias.html 。您是在某种脚本中使用此查询还是只是 sql? -
能否添加
tb_equipe、tb_gruposEquipes和tb_grupos的表结构。这是一个漂亮的小网络应用程序,可帮助进行组调试:sqlfiddle.com
标签: mysql subquery virtual field