Question

所以这是我的查询

SELECT `c`.`location`, `c`.`id`, `c`.`user_id`, `c`.`date`, `c`.`attachment`, 
`ud`.`first_name`, `ud`.`last_name`, 
`a`.`file_name`, `a`.`folder_name`, `a`.`server_key`, `a`.`type` FROM `content`
AS `c` INNER JOIN `users_details` AS `ud` ON ud.user_id = c.user_id INNER JOIN
`attachments` AS `a` ON c.id = a.content_id WHERE (c.location = 'New York')

我希望还包含附件表中的结果数据，问题是表content中的任何数据行都没有附件，如果帖子有附件，则保存在内容中带有1的行附件中的表，如果没有附件，则为0。

现在问题是我的查询只显示有附件的数据，我想这个问题来自与附件表的连接。

那么我该如何做到这一点，所以我有以下输出：有点合并数据，以显示已上传的帖子，但也没有，并显示来自附件表的数据与上传的数据。有点像这样：

  [0] => array(11) {
    ["location"] => string(15) "New York"
    ["id"] => string(2) "25"
    ["user_id"] => string(1) "1"
    ["date"] => string(10) "1364348772"
    ["attachment"] => string(1) "1"
    ["first_name"] => string(8) "John"
    ["last_name"] => string(7) "Doe"
    ["file_name"] => string(36) "c2638acdac24dc2efe4e5971db5f4cc5.jpg"
    ["folder_name"] => string(13) "5133b99030ac4"
    ["server_key"] => string(1) "1"
    ["type"] => string(1) "1"
  }
  [1] => array(11) {
    ["location"] => string(15) "New York"
    ["id"] => string(2) "26"
    ["user_id"] => string(1) "1"
    ["date"] => string(10) "1364348812"
    ["attachment"] => string(1) "0"
    ["first_name"] => string(8) "John"
    ["last_name"] => string(7) "Doe"
    ["file_name"] => string(36) ""
    ["folder_name"] => string(13) ""
    ["server_key"] => string(1) ""
    ["type"] => string(1) ""
  }

Answer 1

看起来你只需LEFT OUTER JOIN这样的附件表：

SELECT
  `c`.`location`,
  `c`.`id`,
  `c`.`user_id`,
  `c`.`date`,
  `c`.`attachment`, 
  `ud`.`first_name`,
  `ud`.`last_name`, 
  `a`.`file_name`,
  `a`.`folder_name`,
  `a`.`server_key`,
  `a`.`type`
FROM `content` AS `c`
INNER JOIN `users_details` AS `ud`
  ON ud.user_id = c.user_id
LEFT OUTER JOIN `attachments` AS `a`
  ON c.id = a.content_id
WHERE c.location = 'New York'

OUTER JOIN将返回所有结果，无论是否有a.content_id与c.id匹配。附件表中的选定字段只显示为NULL。

Answer 2

尝试使用LEFT OUTER JOIN：

SELECT `c`.`location`, `c`.`id`, `c`.`user_id`, `c`.`date`, `c`.`attachment`, 
`ud`.`first_name`, `ud`.`last_name`, 
`a`.`file_name`, `a`.`folder_name`, `a`.`server_key`, `a`.`type` FROM `content`
AS `c` INNER JOIN `users_details` AS `ud` ON ud.user_id = c.user_id LEFT OUTER JOIN
`attachments` AS `a` ON c.id = a.content_id WHERE (c.location = 'New York')

有一个很好的Wikipedia page about joins。

加入查询仅根据一个匹配显示结果

2 个答案: