Hướng dẫn mysql random string

I woudn'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 that 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 a 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 (cryptograpically secure) randomness, use RANDOM_BYTES() rather than RAND() (but then I would consider moving this logic up to the application layer).

answered May 24, 2013 at 15:16

RandomSeedRandomSeed

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6

This problem consists of two very different sub-problems:

• the string must be seemingly random
• the string must be unique

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:

• Hash has been suggested by @paul
• AES-encrypt fits also
• But there is a nice one: RAND(N) itself!

A sequence of random numbers created by the same seed is guaranteed to be

• reproducible
• different for the first 8 iterations
• if the seed is an 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;

Yves M.

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answered May 24, 2013 at 15:21

Eugen RieckEugen Rieck

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12

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))

answered May 24, 2013 at 15:05

paulpaul

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4

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

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 ;

answered Nov 30, 2016 at 10:41

Paddy MannPaddy Mann

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2

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.

answered May 24, 2013 at 15:07

Gordon LinoffGordon Linoff

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2

Here's another method for generating a random string:

SELECT SUBSTRING(MD5(RAND()) FROM 1 FOR 8) AS myrandomstring

answered Dec 19, 2017 at 10:10

beingalexbeingalex

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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;

answered May 24, 2013 at 15:11

Andrey VolkAndrey Volk

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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.

(sqlfiddle demo)

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.

answered Sep 10, 2016 at 0:02

Paul SpiegelPaul Spiegel

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8

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:

1. RAND generates a random number between 0 and 1
2. MD5 calculates the MD5 sum of (1), 32 characters from a-f and 0-9
3. UNHEX translates (2) into 16 bytes with values from 00 to FF
4. TO_BASE64 encodes (3) as base64, 22 characters from a-z and A-Z and 0-9 plus "/" and "+", followed by two "="
5. the three REPLACEs remove the "/", "+" and "=" characters from (4)
6. LEFT takes the first 8 characters from (5), change 8 to something else if you need more or less characters in your random string
7. LPAD inserts zeroes at the beginning of (6) if it is less than 8 characters long; again, change 8 to something else if needed

answered Aug 22, 2017 at 14:08

Jan UhligJan Uhlig

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2

For generate random string, you can use:

SUBSTRING(MD5(RAND()) FROM 1 FOR 8)

You recieve smth like that:

353E50CC

answered Apr 14, 2018 at 5:26

Nikita G.Nikita G.

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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 )

slfan

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answered Oct 5, 2016 at 20:37

0

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))));

answered Oct 12, 2016 at 4:29

TV-C-1-5TV-C-1-5

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If you dont have a id or seed, like its its for a values list in insert:

REPLACE(RAND(), '.', '')

answered Sep 19, 2017 at 17:36

ekernerekerner

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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.

answered May 24, 2013 at 15:12

1

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"

answered May 21, 2019 at 12:52

Mahoor13Mahoor13

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1

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.

answered Feb 17, 2020 at 10:35

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);

answered Mar 1, 2020 at 21:18

AntaresAntares

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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.

answered Dec 6, 2021 at 20:03

Rick JamesRick James

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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%.

answered May 24, 2013 at 15:18

τεκτεκ

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SQL Triggers are complex and resource-intensive. Against a MySQL "Trigger"-based solutions, here is a simpler solution.

1. Create a UNIQUE INDEX on the MySQL table column which will hold the vehicle registration plate string. This will ensure only unique values go in.
2. Simply generate the standard alphanumeric random string in Lua (or any other programming language like ASP, PHP, Java, etc.)
3. Execute the INSERT statement with the generated string, and have error-catching code to parse the failure (in case of the UNIQUE INDEX violation)
4. If the INSERT fails, generate a new random string and re-insert. Length of 8 chars in itself is pretty difficult to repeat, and once found in table generating another one will be next to impossible to be another repeat.

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();

answered Nov 27, 2021 at 13:43

AquaholicAquaholic

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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

marc_s

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answered Dec 10, 2021 at 10:51

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')

answered Apr 13, 2018 at 7:00

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();

answered Oct 12, 2018 at 7:31

GautierGautier

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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')

answered Sep 11, 2019 at 9:47