MySQL 查询在逗号分隔的字符串中查找值

Question

I have a field COLORS (varchar(50)) in a my table SHIRTS that contains a comma delimited string such as 1,2,5,12,15, . Each number representing the available colors.

When running the query select * from shirts where colors like '%1%' to get all the red shirts (color=1), I also get the shirts whose color is grey (=12) and orange (=15).

How should I rewrite the query so that is selects ONLY the color 1 and not all colors containing the number 1?

Answer 1

The classic way would be to add commas to the left and right:

select * from shirts where CONCAT(',', colors, ',') like '%,1,%'

But find_in_set also works:

select * from shirts where find_in_set('1',colors) <> 0

Answer 2

在这种情况下， FIND_IN_SET是你的朋友

select * from shirts where FIND_IN_SET(1,colors) 

Answer 3

Take a look at the FIND_IN_SET function for MySQL.

SELECT * 
    FROM shirts 
    WHERE FIND_IN_SET('1',colors) > 0

Answer 4

This will work for sure, and I actually tried it out:

lwdba@localhost (DB test) :: DROP TABLE IF EXISTS shirts;
Query OK, 0 rows affected (0.08 sec)

lwdba@localhost (DB test) :: CREATE TABLE shirts
    -> (<BR>
    -> id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
    -> ticketnumber INT,
    -> colors VARCHAR(30)
    -> );<BR>
Query OK, 0 rows affected (0.19 sec)

lwdba@localhost (DB test) :: INSERT INTO shirts (ticketnumber,colors) VALUES
    -> (32423,'1,2,5,12,15'),
    -> (32424,'1,5,12,15,30'),
    -> (32425,'2,5,11,15,28'),
    -> (32426,'1,2,7,12,15'),
    -> (32427,'2,4,8,12,15');
Query OK, 5 rows affected (0.06 sec)
Records: 5  Duplicates: 0  Warnings: 0

lwdba@localhost (DB test) :: SELECT * FROM shirts WHERE LOCATE(CONCAT(',', 1 ,','),CONCAT(',',colors,',')) > 0;
+----+--------------+--------------+
| id | ticketnumber | colors       |
+----+--------------+--------------+
|  1 |        32423 | 1,2,5,12,15  |
|  2 |        32424 | 1,5,12,15,30 |
|  4 |        32426 | 1,2,7,12,15  |
+----+--------------+--------------+
3 rows in set (0.00 sec)

Give it a Try !!!

Answer 5

If the set of colors is more or less fixed, the most efficient and also most readable way would be to use string constants in your app and then use MySQL's SET type with FIND_IN_SET('red',colors) in your queries. When using the SET type with FIND_IN_SET , MySQL uses one integer to store all values and uses binary "and" operation to check for presence of values which is way more efficient than scanning a comma-separated string.

In SET('red','blue','green') , 'red' would be stored internally as 1 , 'blue' would be stored internally as 2 and 'green' would be stored internally as 4 . The value 'red,blue' would be stored as 3 ( 1|2 ) and 'red,green' as 5 ( 1|4 ).

Answer 6

select * from shirts where find_in_set('1',colors) <> 0

对我有用

Answer 7

If you're using MySQL, there is a method REGEXP that you can use...

http://dev.mysql.com/doc/refman/5.1/en/regexp.html#operator_regexp

So then you would use:

SELECT * FROM `shirts` WHERE `colors` REGEXP '\b1\b'

Answer 8

You should actually fix your database schema so that you have three tables:

shirt: shirt_id, shirt_name
color: color_id, color_name
shirtcolor: shirt_id, color_id

Then if you want to find all of the shirts that are red, you'd do a query like:

SELECT *
FROM shirt, color
WHERE color.color_name = 'red'
  AND shirt.shirt_id = shirtcolor.shirt_id
  AND color.color_id = shirtcolor.color_id

Answer 9

You can achieve this by following function.

Run following query to create function.

DELIMITER ||
CREATE FUNCTION `TOTAL_OCCURANCE`(`commastring` TEXT, `findme`     VARCHAR(255)) RETURNS int(11)
NO SQL
-- SANI: First param is for comma separated string and 2nd for string to find.
return ROUND (   
    (
        LENGTH(commastring)
        - LENGTH( REPLACE ( commastring, findme, "") ) 
    ) / LENGTH(findme)        
);

And call this function like this

msyql> select TOTAL_OCCURANCE('A,B,C,A,D,X,B,AB', 'A');

Answer 10

1. For MySQL:

SELECT FIND_IN_SET(5, columnname) AS result 
FROM table

2.For Postgres SQL :

SELECT * 
FROM TABLENAME f
WHERE 'searchvalue' = ANY (string_to_array(COLUMNNAME, ','))

Example

select * 
from customer f
where '11' = ANY (string_to_array(customerids, ','))

Answer 11

所有的答案都不是真的正确，试试这个：

select * from shirts where 1 IN (colors);