164

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.

4
  • How random do you need these to be? If someone receives a particular license plate, is it important or not whether they can work out the next or previous license plate that you handed out? Commented May 24, 2013 at 14:57
  • 1
    @YaK See my answer on how to avoid even the tiny possibility of collision Commented May 24, 2013 at 15:21
  • If HEX is enough, here is a short solution: SELECT HEX(RAND()*0xFFFFFFFF) It might be less than 8 chars, but you can add for "F"s and use LEFT() to be safe. Commented Feb 6, 2021 at 13:49
  • Please provide all your requirements: No dups? "Next name unpredictable? Cannot reconstruct original value? Max of 8 characters? Less than 8 is OK? Just capitals? Just letters? Commented Dec 6, 2021 at 20:21

26 Answers 26

192

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).

Sign up to request clarification or add additional context in comments.

8 Comments

Thanks for your solution i have one question regarding UUID. how many chance the id you generate 8 character is again repeat.
@user3099183 Officially, "very low". 16^8 is about 4 billion possible strings.
Please note that the first 8 characters of UUID is not random, but sequential, since it is based on the timestamp.
I want to generate a 9-character long random string, and when I use 9 in your code SELECT LEFT(UUID(), 9);, there is always - at the end of generated string as the ninth character. It is constant. Why?
@MartinAJ because the string is a uuid. You can easily replace the hyphens e.g. SELECT LEFT(REPLACE(UUID(), '-', ''), 16);
|
97

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;

12 Comments

I was just wondering. Why do you use those numbers ..4294967296))*36+1 ?
4294967296=2^32 - The seed must be an INT32, the rand() function gives a value between 0 and 1, so 4294967296*rand(somthing) will give us a random 32bit int, that is suitable as the next seed.
This is a bit old but I'd like to note that I had to add FLOOR() around the second substring parameters: substring('ABC … 789', floor(rand(@seed:= … )*36+1), 1), On some occasions, substring was trying to pick character 36.9, which when rounded up to 37, would result in no character being chosen.
You cannot call a string random if it is reproducible. And duplicates are also possible, because you are using floor(). This sqlfiddle shows, that duplicates are created for three character long strings.
@EugenRieck I don't understand how you get your numbers ("first 2^32 iterations"). But i need only one example of duplicates to disprove this concept. For the IDs 193844 and 775771 your algorithm will generate the same string T82X711 (demo).
|
62

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))

4 Comments

Good idea, could use it on the primary key. Will keep this in mind for future projects, thanks!
There is a chance that this could result with only numbers
It is not random at all.
You could make it more "random" if instead of using the auto incremental id, you use the datetime in which the insert was done, which is also unique.
60

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 ;

2 Comments

this should be the accepted answer, clear and to the point, worked just fine for me, thanks @paddy-mann
This work better than the accepted solution!, Thanks a lot
29

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.

2 Comments

use floor(rand()*36+1) instead. Otherwise some results will be 'short'.
there should be 35+1 not 36+1 ! Otherwise you will get some some strings with 8 chars and some of them with 7
26

Here's another method for generating a random string:

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

Comments

18

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.

8 Comments

Amazing! This is exactly what I wanted. I only would like to understand how it gets converted to a string. Why is there a pow, what to do, to add numbers and so on. Anyhow, thank you.
@Akxe conv() can be used to convert a number to an alphanumeric string. pow(36,8)-1 is the numeric representation of ZZZZZZZZ. So we generate a random integer between 0 and '36^8-1' (from 0 to 2821109907455) and convert it to an alphanumeric string between 0 and ZZZZZZZZ unsing conv(). lapad() will fill the string with zeros until it has a length of 8.
You sir are a genius. I imagine adding small letters is impossible due to non-continuity of chars? (91-96) Not that I need it, just curious...
@Akxe conv() only supports a base up to 36 (10 digits + 26 uppercase letters). If you want to include lowercase letters, you will need another way to convert a number to a string.
Caveat: doesn't work for str_len > 13. From 14 on, you always get '3W5E11264SGSF'. ;-)
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17

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;

Comments

10

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

Comments

9

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

1 Comment

Great, exactly what I was looking for to create a token-like ID natively in MySQL
7

For generate random string, you can use:

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

You recieve smth like that:

353E50CC

Comments

5

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 )

Comments

5

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))));

Comments

4

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

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

Comments

3

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();

Comments

3
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.

Comments

2

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.

1 Comment

True, I'll keep doing it in the actual game instead of SQL. Thank you very much.
2

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"

2 Comments

This function will not generate random unique value.
Faisal: It seems to produce the right amount of letters and numbers to me, so what makes you say that?
2

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.

Comments

2

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);

Comments

2

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!

Comments

1

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%.

Comments

1

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?

Comments

0
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')

Comments

0

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();

Comments

0

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')

Comments

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