Question

我正在尝试从我的数据库中获取具有垂直管道分隔值的数字的记录。

我有这样的查询：

SELECT * FROM level_3 

WHERE 
    date_end = -6 

AND 
    ccode = 'GB'

AND ( 
    responsible LIKE '%5%' OR 
    accountable LIKE '%5%' OR 
    consulted LIKE '%5%' OR 
    informed LIKE '%5%'
    )

以下是结果：

date_end    ccode    responsible    accountable    consulted    informed
-6          GB       |6|            |6|            |8|9|15|     |8|
-6          GB       |6|5|          |6|            |8|          |16|

但 LIKE％5％匹配15但我只希望它与确切的数字匹配。我怎么能这样做呢？

Answer 1

尝试此查询：

SELECT * FROM level_3 

WHERE 
    date_end = -6 

AND 
    ccode = 'GB'

AND ( 
    responsible LIKE '%|5|%' OR 
    accountable LIKE '%|5|%' OR 
    consulted LIKE '%|5|%' OR 
    informed LIKE '%|5|%'
    )

Answer 2

LIKE有一个正则表达式挂件。

SELECT * FROM level_3 
WHERE date_end = -6 
AND ccode = 'GB'
AND ( 
    responsible REGEXP '\|[0-9]+\|' OR 
    ...
)

（备注用于其他用途：使用'^ ... $'进行从开始^到结束$的总匹配。）

Answer 3

您想要的是

Where concat('|',responsible,'|') like '%|5|%'

MySQL LIKE仅匹配整数

3 个答案: