SQL语法错误，语法无效

Question

我有以下查询：

SELECT s0_.id, s0_.caption, s0_.lowresimageurl, s0_.medresimageurl, s0_.highresimageurl, s0_.price
FROM app_instagram_shop_picture s0_
LEFT JOIN app_instagram_shop s1_ force index (idx_locked_deletedAt)
LEFT JOIN app_instagram_picture_category s2_
LEFT JOIN app_instagram_first_level_category s3_
ON s0_.shop_id = s1_.id 
AND s1_.deletedAt IS NULL
AND s0_.id = s2_.picture_id
AND s2_.first_level_category_id = s3_.id
WHERE (s0_.isShown = 1
AND s1_.id != 32179
AND s1_.expirydate IS NULL
AND s1_.deletedAt IS NULL
AND s1_.isLocked = 0
AND s1_.owner_id IS NULL 
AND s0_.id != 2598561
AND s0_.deletedAt IS NULL
AND s0_.isStyleInspiration = 0 
AND s0_.isLocked = 0 
AND s3_.id = 11 
) 
AND (s0_.deletedAt IS NULL) 
ORDER BY s0_.updated DESC 
LIMIT 16

由于某种原因，这导致我以下错误：

SQLSTATE[42000]: Syntax error or access violation: 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 'WHERE (s0_.isShown = 1
AND s1_.id != 32179
AND s1_.expirydate IS N' at line 10

但是，我似乎无法弄清楚问题出在哪里。

Answer 1

请记住，每个SELECT语句仅具有一个WHERE子句，并在检查的一开始就使用它，此后仅使用AND（我被很多次的我遗忘了，因为忘记了这一点，并在后面的链中放置了额外的WHERE子句）

Answer 2

    SELECT s0_.id, s0_.caption, s0_.lowresimageurl, s0_.medresimageurl, s0_.highresimageurl,                          s0_.price
FROM app_instagram_shop_picture s0_
LEFT JOIN app_instagram_shop s1_ force index (idx_locked_deletedAt)
LEFT JOIN app_instagram_picture_category s2_
LEFT JOIN app_instagram_first_level_category s3_
ON s0_.shop_id = s1_.id 
AND s1_.deletedAt IS NULL
AND s0_.id = s2_.picture_id
AND s2_.first_level_category_id = s3_.id
WHERE s0_.isShown = 1
AND s1_.id != 32179
AND s1_.expirydate IS NULL
AND s1_.deletedAt IS NULL
AND s1_.isLocked = 0
AND s1_.owner_id IS NULL 
AND s0_.id != 2598561
AND s0_.deletedAt IS NULL
AND s0_.isStyleInspiration = 0 
AND s0_.isLocked = 0 
AND s3_.id = 11 

AND s0_.deletedAt IS NULL
ORDER BY s0_.updated DESC 
LIMIT 16

Answer 3

使用JOIN ，您需要为每个联接表指定ON或USING ，以指示联接关系。

您有3个LEFT JOINs但只有一个ON条件。 您可能打算这样做：

FROM app_instagram_shop_picture s0_
  LEFT JOIN app_instagram_shop s1_ force index (idx_locked_deletedAt)
  ON s0_.shop_id = s1_.id 
  AND s1_.deletedAt IS NULL
  LEFT JOIN app_instagram_picture_category s2_
  ON s0_.id = s2_.picture_id
  LEFT JOIN app_instagram_first_level_category s3_
  ON s2_.first_level_category_id = s3_.id

（在第一个ON s0_.shop_id = s1_.id指定其他谓词不适用于其他ON s0_.shop_id = s1_.id表）

Answer 4

支架我的原因错误，请卸下支架并使用

SELECT s0_.id, s0_.caption, s0_.lowresimageurl, s0_.medresimageurl, s0_.highresimageurl, s0_.price
FROM app_instagram_shop_picture s0_
LEFT JOIN app_instagram_shop s1_ force index (idx_locked_deletedAt)
LEFT JOIN app_instagram_picture_category s2_
LEFT JOIN app_instagram_first_level_category s3_
ON s0_.shop_id = s1_.id 
AND s1_.deletedAt IS NULL
AND s0_.id = s2_.picture_id
AND s2_.first_level_category_id = s3_.id
WHERE s0_.isShown = 1
AND s1_.id != 32179
AND s1_.expirydate IS NULL
AND s1_.deletedAt IS NULL
AND s1_.isLocked = 0
AND s1_.owner_id IS NULL 
AND s0_.id != 2598561
AND s0_.deletedAt IS NULL
AND s0_.isStyleInspiration = 0 
AND s0_.isLocked = 0 
AND s3_.id = 11 

AND (s0_.deletedAt IS NULL) 
ORDER BY s0_.updated DESC 
LIMIT 16