用聚合函数计算SQL

Question

I have a database with a table that holds all codes and a second table that holds codes that apply to data that can't be turned into primary key(hence a huge problem with it). Sometimes the codes are held in such a way "E1/D2/E3" in the comments section so multiple codes can appear in one comment

What I need is to get a list of codes of certain level and then use it with a LIKE statement to find a count of how many times each code exists in this particular database for example.

Count how many times each code exists in the codes column of the comments table. I've figured to get a list of Codes of this level and then I'm stuck with it. Please help

Code Table:

+----+-----------+-----+
|Code|Description|Level|
+----+-----------+-----+
|A1  |desc       |1    |
+----+-----------+-----+
|A2  |desc       |1    |
+----+-----------+-----+
|A3  |desc       |1    |
+----+-----------+-----+
|A4  |desc       |1    |
+----+-----------+-----+
|A1.1|desc       |2    |
+----+-----------+-----+
|A2.1|desc       |2    |
+----+-----------+-----+
|A3.1|desc       |2    |
+----+-----------+-----+
|A4.1|desc       |2    |
+----+-----------+-----+

Comments Table:

+-------------+----------+-------------+
|Comment      |Codes     |Location     |
+-------------+----------+-------------+
|Hello there  |A1/A1.1/A2|Somewhere    |
+-------------+----------+-------------+
|How are you  |A4/A7/B1.1|Long time ago|
+-------------+----------+-------------+
|My name is...|B1/B2/B3  |Somewhere    |
+-------------+----------+-------------+

Now what I need is to count the amount of times each code from level 1 occured in the comments section for each tuple.

Answer 1

If you mean how often each code appears in the codes column of the comments table:

select co.code, count(*)
from comments c join
     codes co
     on find_in_set(co.code, replace(c.codes, '/', ',')) > 0 and
        co.level = 1
group by co.code;

EDIT:

You can also express this using like :

select co.code, count(*)
from comments c join
     codes co
     on concat('/', c.codes, '/') like concat('%/', co.code, '/%')
        co.level = 1
group by co.code;

Note that appending of delimiters at the beginning and end to ensure full matches.