Question

SELECT COUNT(news_id) AS count 
FROM fusion_news 
WHERE (news_subject LIKE '%новым%' || news_news LIKE '%новым%')

有没有办法可以使用别名或设置一些标志来查看哪个列匹配？目前我只是在完成news_news或news_subject匹配的查询之后，在PHP中进行手动检查，但是如果可能的话，我可以通过单独的SQL以某种方式标记这一点吗？

Answer 1

你可以使用UNION ..考虑：

Select count(1) as Count, "in News Subject" as Source
  from fusion_news
 where news_subject LIKE '%новым%'
UNION
Select count(1) as Count, "in News Body" as Source
  from fusion_news
 where news_news LIKE '%новым%'

或者，如果你想要计算细分选择（选择count（1）作为Count 来自fusion_news 其中news_subject LIKE'％новым％'）作为SubjectCount，选择（选择count（1）作为Count 来自fusion_news news_news LIKE'％новым％'）as Content Count

Answer 2

您可以将其作为布尔值查询。当被视为数字时，Mysql会将true转换为1并将false转换为0，因此您只需sum where中的布尔表达式子句：

SELECT SUM(news_subject LIKE '%новым%') AS news_subject_num_matches, 
       SUM(news_news LIKE '%новым%') AS news_news_num_matches
FROM   fusion_news 
WHERE  (news_subject LIKE '%новым%' || news_news LIKE '%новым%')

从多个LIKE子句中查找哪个列与LIKE匹配？

2 个答案: