Question

我有两个表：questions和questions_lookup。如果放在网站上是一个很好的问题，用户会投票。

table: questions
    id
    question
    date_created
table: questions_lookup
    id
    question_id // (id linked to questions table)
    user_id // (the id of the user, I store this in a session variable called $me)
    show // (1 or 0, show or don't show)

我想有一个php页面，它可以从date_created排序的问题表中提取所有问题，然后显示用户是否已经回答。当我尝试进行任何连接时，我最终会出现重复的问题，因为它会提取其他用户的答案。

所以如果有10个问题。并且特定用户仅回答了3.我们仍然显示所有10个问题，但标记他们已经回答的问题。

所以我基本上想要显示如下内容：

Question 1
Question 2 (answered)
Question 3 (answered)
Question 4
Question 5
Question 6
Question 7 (answered)
Question 8
Question 9
Question 10

我试过了：

SELECT * FROM questions
RIGHT JOIN questions_lookup
ON (questions.id = questions_lookup.question_id)
WHERE questions_lookup.user_id = '$me'
ORDER BY questions.date_created DESC

Answer 1

这样的事情，假设questions_lookup中每个问题每个用户只能有一条记录。

select
  q.*,
  case when ql.question_id is null then
    'no'
  else
    'yes'
  end as user_has_answered
from
  questions q
  left join questions_lookup ql 
    on ql.question_id = q.id
    and ql.user_id = 5 /* User id of current user */

诀窍是查询所有questions和left join questions_lookup。通过将user_id添加到连接条件，您将遗漏其他用户的记录，同时仍返回没有当前用户记录的问题。如果将ql.user_id = 5移动到where子句，查询将不再有效，因为它会有效地将左连接转换为内连接。

[编辑]

我看到你添加了你的查询。那里有两个错误。右连接应该是左连接，因为你总是希望在左边有一个记录（问题），在右边有一个可选记录（查找）。此外，条件不应该在where子句中。

Answer 2

怎么样：

SELECT questions.*, max(questions_lookup.show) AS show 
FROM questions 
LEFT JOIN questions_lookup ON questions_lookup.question_id=questions.id 
WHERE (questions_lookup.user_id='youruserid' OR questions_lookup.user_id IS NULL) 
GROUP BY questions.id 
ORDER BY questions.date_created ASC

然后在您的搜索结果中，show=1表示用户已回答。

Answer 3

SELECT q.*, 
       l.user_id,
       l.show,
       IF(l.question_id IS NULL,'','answered') as answered
  FROM questions q LEFT JOIN
       questions_lookup l ON q.id = l.question_id AND
                             l.user_id = 5 <-- user id goes here
 ORDER BY q.date_created DESC

您可以使用计算出的answered列，具体取决于您进一步处理所需的输出：

IF(l.question_id IS NULL,'','answered') as answered <-- 'answered' if answered, empty string if not (like in your example)
IFNULL(l.question_id,0) as answered <-- question_id (if autogenerated unsigned  int will be > 0) if answered, 0-if not

或GolezTrol建议

CASE WHEN ql.question_id IS NULL THEN 'no' ELSE 'yes' END as answered <-- yes if answered and no if not

MySQL连接查询，无法搞清楚

3 个答案: