MySQL FIND_IN_SET 带数组字符串

Question

I have a field in a table I am querying that looks similar to this:

Name          Phone          Category_IDS          Category_Labels
Sample        1111111111     ["1"]                 ["foo", "bar"]

I am trying to use the FIND_IN_SET function to find all rows that contain one of the values listed in the comma separated list. Something like this returns nothing:

SELECT * FROM sampletable WHERE FIND_IN_SET('1', category_ids) <> 0

It does work if I do this:

SELECT * FROM factual_usplaces WHERE FIND_IN_SET('["1"]', category_ids) <> 0

But of course that limits to searches to rows where the category_ids or labels only contains a single value in the comma separated list. So ["1"] would be found but ["1", "2"] would not.

Is there a way to remove the brackets and quotations from the string on the fly in the query?

Answer 1

If data is stored exactly how you showed it then you can use REPLACE() to strip double quotes and brackets before feeding category_ids to FIND_IN_SET() .

SELECT * 
  FROM Table1 
 WHERE FIND_IN_SET(1, REPLACE(
                        REPLACE(
                          REPLACE(category_ids, '"', ''), 
                        '[', ''), 
                      ']','')) > 0

Here is SQLFiddle

---

Now if you will use it a lot then you may consider to create a user defined function to simplify your code

CREATE FUNCTION UNQOUTE_LIST(_list VARCHAR(512)) 
RETURNS VARCHAR(512)
RETURN 
REPLACE(
  REPLACE(
    REPLACE(_list, '"', ''), 
  '[', ''), 
']','');

And use it

SELECT * 
  FROM Table1 
 WHERE FIND_IN_SET(1, UNQOUTE_LIST(category_ids)) > 0

Here is SQLFiddle

Answer 2

try below sql query with like operator

SELECT * FROM factual_usplaces WHERE category_ids LIKE '%1,2%'

Hope This help you

Answer 3

SELECT * 
  FROM Table1 
 WHERE FIND_IN_SET("bar", REPLACE(REPLACE(REPLACE(`Category_Labels`, '\"', ''), '[', ''), ']',''))

This is working for me