在MySQL字段中查询LIKE字符串

Question

I have a field called 'areasCovered' in a MySQL database, which contains a string list of postcodes.

There are 2 rows that have similar data eg:

Row 1: 'B*,PO*,WA*'
Row 2: 'BB*, SO*, DE*'

Note - The strings are not in any particular order and could change depending on the user

Now, if I was to use a query like:

SELECT * FROM technicians WHERE areasCovered LIKE '%B*%'

I'd like it to return JUST Row 1. However, it's returning Row 2 aswell, because of the BB* in the string.

How could I prevent it from doing this?

Answer 1

The key to using like in this case is to include delimiters, so you can look for delimited values:

SELECT *
FROM technicians
WHERE concat(', ', areasCovered, ', ') LIKE '%, B*, %'

In MySQL, you can also use find_in_set() , but the space can cause you problems so you need to get rid of it:

SELECT *
FROM technicians
WHERE find_in_set('B', replace(areasCovered, ', ', ',') > 0

Finally, though, you should not be storing these types of lists as strings. You should be storing them in a separate table, a junction table, with one row per technician and per area covered. That makes these types of queries easier to express and they have better performance.

Answer 2

You are searching wild cards at the start as well as end.

You need only at end.

SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'

Reference:

Answer 3

Normally I hate REGEXP. But ho hum:

SELECT * FROM technicians 
WHERE concat(",",replace(areasCovered,", ",",")) regexp ',B{1}\\*';

To explain a bit:

Get rid of the pesky space:

select replace("B*,PO*,WA*",", ",",");

Bolt a comma on the front

select concat(",",replace("B*,PO*,WA*",", ",","));

Use a REGEX to match "comma B once followed by an asterix":

select concat(",",replace("B*,PO*,WA*",", ",",")) regexp ',B{1}\\*';

Answer 4

I could not check it on my machine, but it's should work:

SELECT * FROM technicians WHERE areasCovered <> replace(areaCovered,',B*','whatever')

In case the 'B*' does not exist, the areasCovered will be equal to replace(areaCovered,',B*','whatever'), and it will reject that row. In case the 'B*' exists, the areCovered will NOT be eqaul to replace(areaCovered,',B*','whatever'), and it will accept that row.

Answer 5

You can Do it the way Programming Student suggested

SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'

Or you can also use limit on query

SELECT * FROM technicians WHERE areasCovered LIKE '%B*%' LIMIT 1

Answer 6

%B*%在每一侧都包含％，这使其可以返回文本中任意位置上值包含B*所有行，但是您的要求是查找所有包含以B*开头的值的行，因此以下查询应工作。

SELECT * FROM technicians WHERE areasCovered LIKE 'B*%'