Reputation: 1719
I'm working on a game which involves vehicles at some point. I have a MySQL table named "vehicles" containing the data about the vehicles, including the column "plate" which stores the License Plates for the vehicles.
Now here comes the part I'm having problems with. I need to find an unused license plate before creating a new vehicle - it should be an alphanumeric 8-char random string. How I achieved this was using a while loop in Lua, which is the language I'm programming in, to generate strings and query the DB to see if it is used. However, as the number of vehicles increases, I expect this to become even more inefficient it is right now. Therefore, I decided to try and solve this issue using a MySQL query.
The query I need should simply generate a 8-character alphanumeric string which is not already in the table. I thought of the generate&check loop approach again, but I'm not limiting this question to that just in case there's a more efficient one. I've been able to generate strings by defining a string containing all the allowed chars and randomly substringing it, and nothing more.
Any help is appreciated.
Upvotes: 160
Views: 379722
Reputation: 29759
I wouldn't bother with the likelihood of collision. Just generate a random string and check if it exists. If it does, try again and you shouldn't need to do it more than a couple of times unless you have a huge number of plates already assigned.
Another solution for generating an 8-character long pseudo-random string in pure (My)SQL:
SELECT LEFT(UUID(), 8);
You can try the following (pseudo-code):
DO
SELECT LEFT(UUID(), 8) INTO @plate;
INSERT INTO plates (@plate);
WHILE there_is_a_unique_constraint_violation
-- @plate is your newly assigned plate number
Since this post has received an unexpected level of attention, let me highlight ADTC's comment: the above piece of code is quite dumb and produces sequential digits.
For slightly less stupid randomness, try something like this instead:
SELECT LEFT(MD5(RAND()), 8)
And for true (cryptographically secure) randomness, use RANDOM_BYTES()
rather than RAND()
(but then I would consider moving this logic up to the application layer).
Upvotes: 188
Reputation: 103
I see a lot of answers suggesting fairly sequential number-generators (UUID
) or deterministic (and broken) hashes (MD5
), while MySQL can very well generate a completely random hexadecimal string by itself!
SELECT HEX(RANDOM_BYTES(4));
The 4
here represents the number of bytes, hexadecimal will produce twice the number of characters. So 4 bytes = 8 character string.
If more characters are wanted and the special characters +
and /
are fine, base64 can be used, as well:
SELECT TO_BASE64(RANDOM_BYTES(6));
Base64 encodes every 3 bytes as 4 characters, so 6 bytes = 8 characters, of a wider character set.
Note: These are cryptographically secure, too!
Upvotes: 0
Reputation: 1189
Create a random string
Here's a MySQL function to create a random string of a given length.
DELIMITER $$
CREATE DEFINER=`root`@`%` FUNCTION `RandString`(length SMALLINT(3)) RETURNS varchar(100) CHARSET utf8
begin
SET @returnStr = '';
SET @allowedChars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
SET @i = 0;
WHILE (@i < length) DO
SET @returnStr = CONCAT(@returnStr, substring(@allowedChars, FLOOR(RAND() * LENGTH(@allowedChars) + 1), 1));
SET @i = @i + 1;
END WHILE;
RETURN @returnStr;
END
DELIMITER ;
Usage SELECT RANDSTRING(8)
to return an 8 character string.
You can customize the @allowedChars
.
Uniqueness isn't guaranteed - as you'll see in the comments to other solutions, this just isn't possible. Instead you'll need to generate a string, check if it's already in use, and try again if it is.
Check if the random string is already in use
If we want to keep the collision checking code out of the app, we can create a trigger:
DELIMITER $$
CREATE TRIGGER Vehicle_beforeInsert
BEFORE INSERT ON `Vehicle`
FOR EACH ROW
BEGIN
SET @vehicleId = 1;
WHILE (@vehicleId IS NOT NULL) DO
SET NEW.plate = RANDSTRING(8);
SET @vehicleId = (SELECT id FROM `Vehicle` WHERE `plate` = NEW.plate);
END WHILE;
END;$$
DELIMITER ;
Upvotes: 56
Reputation: 111
This work form me, generate 6 digit number and update in MySQL:
Generate:
SELECT SUBSTRING(MD5(RAND()) FROM 1 FOR 6)
Update:
UPDATE table_name
SET column_name = SUBSTRING(MD5(RAND()) FROM 1 FOR 6)
WHERE id = x12
Upvotes: 10
Reputation: 142298
UPPER(HEX(UUID_SHORT()))
gives you a 16-character alphanumeric string that is unique. It has some unlikely caveats, see https://dev.mysql.com/doc/refman/5.7/en/miscellaneous-functions.html#function_uuid-short
The "next" value is often predictable:
mysql> SELECT UPPER(HEX(UUID_SHORT()));
+--------------------------+
| UPPER(HEX(UUID_SHORT())) |
+--------------------------+
| 161AA3FA5000006 |
+--------------------------+
mysql> SELECT UPPER(HEX(UUID_SHORT()));
+--------------------------+
| UPPER(HEX(UUID_SHORT())) |
+--------------------------+
| 161AA3FA5000007 |
+--------------------------+
Converting to BASE64 can get the string down to 11 characters:
https://dev.mysql.com/doc/refman/8.0/en/string-functions.html#function_to-base64
mysql> SELECT TO_BASE64(UNHEX(HEX(UUID_SHORT())));
+-------------------------------------+
| TO_BASE64(UNHEX(HEX(UUID_SHORT()))) |
+-------------------------------------+
| AWGqP6UAABA= |
+-------------------------------------+
That's 12 chars, stripping off the '=' gives you 11.
These may make it unsuitable for your use: The "next" plate is somewhat predictable. There can be some punctuation marks (+
,/
) in the string. Lower case letters are likely to be included.
Upvotes: 3
Reputation: 883
SQL Triggers are complex and resource-intensive. Against a MySQL "Trigger"-based solutions, here is a simpler solution.
This will be lighter and more efficient on DB Server.
Here's a sample (pseudo-) code in PHP:
function refercode()
{
$string = '';
$characters = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
$max = strlen($characters) - 1;
for ($i = 0; $i < 8; $i++) {
$string .= $characters[mt_rand(0, $max)];
}
$refer = "select * from vehicles where refer_code = '".$string."' ";
$coderefertest = mysqli_query($con,$refer);
if(mysqli_num_rows($coderefertest)>0)
{
return refercode();
}
else
{
return $string;
}
}
$refer_by = refercode();
Upvotes: 2
Reputation: 285
Simple and efficient solution to get a random 10 characters string with uppercase and lowercase letters and digits :
select substring(base64_encode(md5(rand())) from 1+rand()*4 for 10);
Upvotes: 2
Reputation: 105
To create a random 10 digit alphanumeric, excluding lookalike chars 01oOlI:
LPAD(LEFT(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(TO_BASE64(UNHEX(MD5(RAND()))), "/", ""), "+", ""), "=", ""), "O", ""), "l", ""), "I", ""), "1", ""), "0", ""), "o", ""), 10), 10, 0)
This is exactly what I needed to create a voucher code. Confusing characters are removed to reduce errors when typing it into a voucher code form.
Hopes this helps somebody, based on Jan Uhlig's brilliant answer.
Please see Jan's answer for a breakdown on how this code works.
Upvotes: 2
Reputation: 65274
This problem consists of two very different sub-problems:
While randomness is quite easily achieved, the uniqueness without a retry loop is not. This brings us to concentrate on the uniqueness first. Non-random uniqueness can trivially be achieved with AUTO_INCREMENT
. So using a uniqueness-preserving, pseudo-random transformation would be fine:
RAND(N)
itself!A sequence of random numbers created by the same seed is guaranteed to be
INT32
So we use @AndreyVolk's or @GordonLinoff's approach, but with a seeded RAND
:
e.g. Assumin id
is an AUTO_INCREMENT
column:
INSERT INTO vehicles VALUES (blah); -- leaving out the number plate
SELECT @lid:=LAST_INSERT_ID();
UPDATE vehicles SET numberplate=concat(
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@lid)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed:=round(rand(@seed)*4294967296))*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand(@seed)*36+1, 1)
)
WHERE id=@lid;
Upvotes: 96
Reputation: 471
I was looking for something similar and I decided to make my own version where you can also specify a different seed if wanted (list of characters) as parameter:
CREATE FUNCTION `random_string`(length SMALLINT(3), seed VARCHAR(255)) RETURNS varchar(255) CHARSET utf8
NO SQL
BEGIN
SET @output = '';
IF seed IS NULL OR seed = '' THEN SET seed = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'; END IF;
SET @rnd_multiplier = LENGTH(seed);
WHILE LENGTH(@output) < length DO
# Select random character and add to output
SET @output = CONCAT(@output, SUBSTRING(seed, RAND() * (@rnd_multiplier + 1), 1));
END WHILE;
RETURN @output;
END
Can be used as:
SELECT random_string(10, '')
Which would use the built-in seed of upper- and lowercase characters + digits. NULL would also be value instead of ''.
But one could specify a custom seed while calling:
SELECT random_string(10, '1234')
Upvotes: 0
Reputation: 1228
Generate 8 characters key
lpad(conv(floor(rand()*pow(36,6)), 10, 36), 8, 0);
How do I generate a unique, random string for one of my MySql table columns?
Upvotes: 1
Reputation: 5617
This function generates a Random string based on your input length and allowed characters like this:
SELECT str_rand(8, '23456789abcdefghijkmnpqrstuvwxyz');
function code:
DROP FUNCTION IF EXISTS str_rand;
DELIMITER //
CREATE FUNCTION str_rand(
u_count INT UNSIGNED,
v_chars TEXT
)
RETURNS TEXT
NOT DETERMINISTIC
NO SQL
SQL SECURITY INVOKER
COMMENT ''
BEGIN
DECLARE v_retval TEXT DEFAULT '';
DECLARE u_pos INT UNSIGNED;
DECLARE u INT UNSIGNED;
SET u = LENGTH(v_chars);
WHILE u_count > 0
DO
SET u_pos = 1 + FLOOR(RAND() * u);
SET v_retval = CONCAT(v_retval, MID(v_chars, u_pos, 1));
SET u_count = u_count - 1;
END WHILE;
RETURN v_retval;
END;
//
DELIMITER ;
This code is based on shuffle string function sends by "Ross Smith II"
Upvotes: 2
Reputation: 1146
An easy way that generate a unique number
set @i = 0;
update vehicles set plate = CONCAT(@i:=@i+1, ROUND(RAND() * 1000))
order by rand();
Upvotes: 0
Reputation: 2476
Here's another method for generating a random string:
SELECT SUBSTRING(MD5(RAND()) FROM 1 FOR 8) AS myrandomstring
Upvotes: 26
Reputation: 720
For generate random string, you can use:
SUBSTRING(MD5(RAND()) FROM 1 FOR 8)
You recieve smth like that:
353E50CC
Upvotes: 7
Reputation: 31
DELIMITER $$
USE `temp` $$
DROP PROCEDURE IF EXISTS `GenerateUniqueValue`$$
CREATE PROCEDURE `GenerateUniqueValue`(IN tableName VARCHAR(255),IN columnName VARCHAR(255))
BEGIN
DECLARE uniqueValue VARCHAR(8) DEFAULT "";
WHILE LENGTH(uniqueValue) = 0 DO
SELECT CONCAT(SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1),
SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789', RAND()*34+1, 1)
) INTO @newUniqueValue;
SET @rcount = -1;
SET @query=CONCAT('SELECT COUNT(*) INTO @rcount FROM ',tableName,' WHERE ',columnName,' like ''',@newUniqueValue,'''');
PREPARE stmt FROM @query;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
IF @rcount = 0 THEN
SET uniqueValue = @newUniqueValue ;
END IF ;
END WHILE ;
SELECT uniqueValue;
END$$
DELIMITER ;
Use this stored procedure and use it everytime like
Call GenerateUniqueValue('tableName','columnName')
Upvotes: 0
Reputation: 5840
If you dont have a id or seed, like its its for a values list in insert:
REPLACE(RAND(), '.', '')
Upvotes: 3
Reputation: 91
For a String consisting of 8 random numbers and upper- and lowercase letters, this is my solution:
LPAD(LEFT(REPLACE(REPLACE(REPLACE(TO_BASE64(UNHEX(MD5(RAND()))), "/", ""), "+", ""), "=", ""), 8), 8, 0)
Explained from inside out:
RAND
generates a random number between 0 and 1MD5
calculates the MD5 sum of (1), 32 characters from a-f and 0-9UNHEX
translates (2) into 16 bytes with values from 00 to FFTO_BASE64
encodes (3) as base64, 22 characters from a-z and A-Z and 0-9 plus "/" and "+", followed by two "="REPLACE
s remove the "/", "+" and "=" characters from (4)LEFT
takes the first 8 characters from (5), change 8 to something else if you need more or less characters in your random stringLPAD
inserts zeroes at the beginning of (6) if it is less than 8 characters long; again, change 8 to something else if neededUpvotes: 9
Reputation: 714
8 letters from the alphabet - All caps:
UPDATE `tablename` SET `tablename`.`randomstring`= concat(CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25)))CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25))),CHAR(FLOOR(65 + (RAND() * 25))));
Upvotes: 5
Reputation: 63
I Use data from another column to generate a "hash" or unique string
UPDATE table_name SET column_name = Right( MD5(another_column_with_data), 8 )
Upvotes: 5
Reputation: 31792
You can generate a random alphanumeric string with:
lpad(conv(floor(rand()*pow(36,8)), 10, 36), 8, 0);
You can use it in a BEFORE INSERT
trigger and check for a duplicate in a while loop:
CREATE TABLE `vehicles` (
`plate` CHAR(8) NULL DEFAULT NULL,
`data` VARCHAR(50) NOT NULL,
UNIQUE INDEX `plate` (`plate`)
);
DELIMITER //
CREATE TRIGGER `vehicles_before_insert` BEFORE INSERT ON `vehicles`
FOR EACH ROW BEGIN
declare str_len int default 8;
declare ready int default 0;
declare rnd_str text;
while not ready do
set rnd_str := lpad(conv(floor(rand()*pow(36,str_len)), 10, 36), str_len, 0);
if not exists (select * from vehicles where plate = rnd_str) then
set new.plate = rnd_str;
set ready := 1;
end if;
end while;
END//
DELIMITER ;
Now just insert your data like
insert into vehicles(col1, col2) values ('value1', 'value2');
And the trigger will generate a value for the plate
column.
That works this way if the column allows NULLs. If you want it to be NOT NULL you would need to define a default value
`plate` CHAR(8) NOT NULL DEFAULT 'default',
You can also use any other random string generating algorithm in the trigger if uppercase alphanumerics isn't what you want. But the trigger will take care of uniqueness.
Upvotes: 18
Reputation: 22001
What about calculating the MD5 (or other) hash of sequential integers, then taking the first 8 characters.
i.e
MD5(1) = c4ca4238a0b923820dcc509a6f75849b => c4ca4238
MD5(2) = c81e728d9d4c2f636f067f89cc14862c => c81e728d
MD5(3) = eccbc87e4b5ce2fe28308fd9f2a7baf3 => eccbc87e
etc.
caveat: I have no idea how many you could allocate before a collision (but it would be a known and constant value).
edit: This is now an old answer, but I saw it again with time on my hands, so, from observation...
Chance of all numbers = 2.35%
Chance of all letters = 0.05%
First collision when MD5(82945) = "7b763dcb..." (same result as MD5(25302))
Upvotes: 62
Reputation: 440
Taking into account the total number of characters that you require, you would have a very small chance of generating two exactly similar number plates. Thus you could probably get away with generating the numbers in LUA.
You have 36^8 different unique numberplates (2,821,109,907,456, that's a lot), even if you already had a million numberplates already, you'd have a very small chance of generating one you already have, about 0.000035%
Of course, it all depends on how many numberplates you will end up creating.
Upvotes: 2
Reputation: 3104
If you're OK with "random" but entirely predictable license plates, you can use a linear-feedback shift register to choose the next plate number - it's guaranteed to go through every number before repeating. However, without some complex math, you won't be able to go through every 8 character alphanumeric string (you'll get 2^41 out of the 36^8 (78%) possible plates). To make this fill your space better, you could exclude a letter from the plates (maybe O), giving you 97%.
Upvotes: 1
Reputation: 3549
You may use MySQL's rand() and char() function:
select concat(
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97),
char(round(rand()*25)+97)
) as name;
Upvotes: 17
Reputation: 1269753
Here is one way, using alpha numerics as valid characters:
select concat(substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1),
substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', rand()*36+1, 1)
) as LicensePlaceNumber;
Note there is no guarantee of uniqueness. You'll have to check for that separately.
Upvotes: 29