1

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')

我还希望将附件表中的结果数据包含在结果中，问题是表中的任何数据行都没有content附件，如果帖子有附件，它会保存在行附件的内容表中，带有 1 , 如果没有附件，则为 0。

现在的问题是我的查询只显示有附件的数据，我猜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：将数据合并在一起以显示已上传的帖子，也可以显示没有上传的帖子，还显示来自已上传的附件表的数据。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) ""
  }
4

2 回答 2

2

看起来你只需要LEFT OUTER JOIN像这样的附件表：

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'

将OUTER JOIN返回所有结果，无论是否存在a.content_id与 匹配的c.id。附件表中的那些选定字段将简单地显示为NULL。

于 2013-04-09T00:17:26.250 回答
1

尝试使用左外连接：

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')

有一个很好的关于 joins 的 Wikipedia 页面。

于 2013-04-09T00:17:36.510 回答