嵌套SQL的问题

时间:2011-05-16 19:24:48

标签: php mysql sql nested

我在MySQL中嵌套选择有问题。

这个有效,但我需要两个。

示例:

SELECT `a`.`title` , `a`.`askprice` , `a`.`picture` 
, `a`.`description` , `a`.`userid` , `a`.`id`
FROM (
`mm_ads_fields_values` AS afv
)
LEFT JOIN `mm_ads` AS a ON `a`.`id` = `afv`.`aid`
WHERE `afv`.`value` = '38'
AND `a`.`category` = '227'
AND `a`.`status` =1
AND a.id
IN (
    SELECT a.id
    FROM mm_ads AS a
    LEFT JOIN mm_ads_fields_values AS afv ON afv.aid = a.id
    WHERE afv.value = '2913'
)
ORDER BY `id` DESC
LIMIT 20 

这个有效。但我需要像这样的SQL语句:

SELECT `a`.`title` , `a`.`askprice` , `a`.`picture` 
, `a`.`description` , `a`.`userid` , `a`.`id`
FROM (
`mm_ads_fields_values` AS afv
)
LEFT JOIN `mm_ads` AS a ON `a`.`id` = `afv`.`aid`
WHERE `afv`.`value` = '38'
AND `a`.`category` = '227'
AND `a`.`status` =1
AND a.id
IN (
   SELECT a.id
   FROM mm_ads AS a
   LEFT JOIN mm_ads_fields_values AS afv ON afv.aid = a.id
   WHERE afv.value = '2913'
)
AND a.id
IN (
    SELECT a.id
    FROM mm_ads AS a
    LEFT JOIN mm_ads_fields_values AS afv ON afv.aid = a.id
    WHERE afv.value = '51'
)
ORDER BY `id` DESC
LIMIT 20 

这一个,最后一个,将无法运作。它的装载和装载从来没有发生任何事情。

我做错了什么?

问候,马里奥

抱歉我的英语不好..

1 个答案:

答案 0 :(得分:0)

Glib回答:第二个IN语句不应该使用OR代替AND吗? (编辑:并使用正确的运算符优先级。)

...
AND 
( a.id
  IN (
     SELECT a.id
     FROM mm_ads AS a
     LEFT JOIN mm_ads_fields_values AS afv ON afv.aid = a.id
     WHERE afv.value = '2913'
  )
  OR a.id
  IN (
      SELECT a.id
      FROM mm_ads AS a
      LEFT JOIN mm_ads_fields_values AS afv ON afv.aid = a.id
      WHERE afv.value = '51'
  )
)
...