Question

人们，有人可以告诉我如何在联接中实现以下查询，与以下查询相比，该查询具有更高的性能

SELECT *, count(*) c FROM question_ans_word where answer_id in
  (select answer_id from question_ans_word where keywords_id in
     (SELECT id FROM uniqueword where word like "minimum")
  )
group by answer_id;

Answer 1

以下是如何使用联接执行此操作。我是否能说出它是否会更好：

SELECT * FROM 
job
WHERE 
1= (CASE WHEN jobbusiness=1 AND (jobstartdate >= jobenddate AND NOW() >= jobstartdate) THEN 1
   CASE WHEN jobbusiness=0 AND jobfixedstartdate=0 AND jobfixedenddate=0 AND date_diff(NOW(),jobcreatedtimestamp)>60 THEN 1
   CASE WHEN jobbusiness=0 AND jobfixedstartdate=1 AND jobfixedenddate=0 AND jobstartdate>= NOW() THEN 1
   CASE WHEN jobbusiness=0 AND jobfixedstartdate=1 AND jobfixedenddate=1 AND jobenddate>=NOW()    THEN 1
   ELSE 0
   END)

好运。

Answer 2

根据我的理解，这看起来像这样...

From question_ans_word qaw
JOIN uniqueword uw on qaw.keywords_id = uw.id and word like "minimum"

Answer 3

尝试一下，

select *, count(*) c from question_ans_word qw INNER JOIN uniqueword uw ON uw.id = qw.keywords_id WHERE uw.word like "minimum" group by answer_id

如何在mysql中实现join

3 个答案: