Question

我遇到以下查询的sql问题，它返回错误...

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''

我无法理解为什么这会引发错误。当我删除部分连接时，它正确运行。

此查询的作业是根据给定的位置（纬度/经度）返回属性列表，其中包含来自单独offers表的任何匹配数据以及存储的城镇和县的url slug因此，在单独的表towns和counties中。如果一个房产没有特别优惠，它仍然应该返回数据。

查询是......

    SELECT COUNT(agencyid) AS `count_offers`, prop.*, offers.*, twn.slug AS `slug_town`, cnty.slug AS `slug_county`, ROUND(((3959 * acos(cos(radians(50.1854670)) * cos(radians(prop.latitude)) * cos(radians(prop.longitude) - radians(-5.4209100)) + sin(radians(50.1854670)) * sin( radians(prop.latitude)))) * 2),0)/2 AS `distance` 
FROM `properties` AS `prop`

JOIN `towns` AS `twn` ON twn.name=prop.town
JOIN `counties` AS `cnty` ON cnty.name=twn.county
LEFT JOIN `offers` ON offers.agencyid = prop.mID
AND 
offers.dt_expire>1377985886 
AND 
(prop.sleeps >= offers.sleeps_min AND prop.sleeps <= offers.sleeps_max)

AND 
(
  prop.slug_code = offers.the_property 
  OR 
  (
    (offers.the_property='' OR offers.the_property=NULL) 
    AND 
    (
      (
        prop.county = offers.the_county 
        AND
        prop.town=offers.the_town
      ) 
      OR 
      (
        prop.county = offers.the_county
        AND
        twn.name=offers.the_town
        AND
        prop.place = offers.the_place
      ) 
      OR
      prop.county = offers.the_county 
      OR
      prop.region = offers.the_region 
      OR
      prop.country = offers.the_country
    )
  )

感谢您的帮助。

Answer 1

在MySQL中，如果您需要与NULL进行比较，则语法为Colunname IS NULL

SELECT COUNT(agencyid) AS `count_offers`, prop.*, offers.*, twn.slug AS `slug_town`, cnty.slug AS `slug_county`, ROUND(((3959 * acos(cos(radians(50.1854670)) * cos(radians(prop.latitude)) * cos(radians(prop.longitude) - radians(-5.4209100)) + sin(radians(50.1854670)) * sin( radians(prop.latitude)))) * 2),0)/2 AS `distance` 
FROM `properties` AS `prop`

JOIN `towns` AS `twn` ON twn.name=prop.town
JOIN `counties` AS `cnty` ON cnty.name=twn.county
LEFT JOIN `offers` ON offers.agencyid = prop.mID
AND 
offers.dt_expire>1377985886 
AND 
(prop.sleeps >= offers.sleeps_min AND prop.sleeps <= offers.sleeps_max)

AND 
(
  prop.slug_code = offers.the_property 
  OR 
  (
    (offers.the_property='' OR offers.the_property IS NULL) 
    AND 
    (
      (
        prop.county = offers.the_county 
        AND
        prop.town=offers.the_town
      ) 
      OR 
      (
        prop.county = offers.the_county
        AND
        twn.name=offers.the_town
        AND
        prop.place = offers.the_place
      ) 
      OR
      prop.county = offers.the_county 
      OR
      prop.region = offers.the_region 
      OR
      prop.country = offers.the_country
    )
  )
)

Answer 2

缺少一个右括号（可能在最后一行）。

Answer 3

如果没有一些样本数据或表结构可以回答这样的问题是非常棘手的，但我建议你这就是你想要的。

如果我没记错，MySql不会将''评估为NULL，但如果我错了，请从上面的评论中取出条件。

SELECT 
  COUNT(agencyid) AS `count_offers`, 
  prop.*, 
  offers.*, 
  twn.slug AS `slug_town`, 
  cnty.slug AS `slug_county`, 
  ROUND(((3959 * acos(cos(radians(50.1854670)) * cos(radians(prop.latitude)) * cos(radians(prop.longitude) - radians(-5.4209100)) + sin(radians(50.1854670)) * sin( radians(prop.latitude)))) * 2),0)/2 AS `distance` 
FROM 
  `properties` AS `prop`
  JOIN `towns` AS `twn` ON 
    twn.name=prop.town
  JOIN `counties` AS `cnty` ON 
    cnty.name=twn.county
  LEFT JOIN `offers` ON 
    offers.agencyid = prop.mID AND 
    offers.dt_expire>1377985886 AND 
    prop.sleeps >= offers.sleeps_min AND 
    prop.sleeps <= offers.sleeps_max AND
    (
      prop.slug_code = offers.the_property OR 
      (
        (IfNull(offers.the_property, '') = '') AND 
        (
          (
            prop.county = offers.the_county AND
            prop.town=offers.the_town
          ) 
          OR 
          (
            prop.county = offers.the_county AND
            twn.name=offers.the_town AND
            prop.place = offers.the_place
          ) 
          OR
          prop.county = offers.the_county OR
          prop.region = offers.the_region OR
          prop.country = offers.the_country
        )
      )
    )

MySQL使用LEFT JOIN从多个表中进行选择

3 个答案: