Generate random new random string if exist on sql

Question

I have a question, how can I create a random string in PHP and check on a SQL if the string exist, if exist, create a new one, but if it doesnt exist use it

I have this code to create the random string:

$original_string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
$random_string = get_random_string($original_string, 14);
echo $random_string;

And the MySQL I would like to check if the random string is already been use is this one:

CREATE TABLE `en_videos` (
  `id` int(11) NOT NULL auto_increment,
  `name` text,
  `code` text,
  `description` text,
  `hits` int(11) default NULL,
  `program` int(11) default NULL,
  `lenght` int(11) default NULL,
  `hd` int(11) NOT NULL default '1',
  `subtitles` int(11) NOT NULL default '1',
  `type` text NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=1 DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

I would like to check if the random string exists on the code field

Daniel Li · Accepted Answer · 2012-09-24 17:45:55Z

1

Try:

$original_string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
$random_string = '';

do {
    $random_string = get_random_string($original_string, 14);
} while ($mysqli->query('SELECT 1 FROM `en_videos`
             WHERE `code` = "' . $random_string . '"')->fetch_row());

echo $random_string;

This assume you're using the mysqli library. If not, use whatever fetches rows as your condition handler.

Assuming you're using the mysql_* functions (in which case, I suggest you look into other mysql libraries):

$original_string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
$random_string = '';

do {
    $random_string = get_random_string($original_string, 14);
} while ($mysql_num_rows(mysql_query('SELECT 1 FROM `en_videos`
             WHERE `code` = "' . $random_string . '"')));

echo $random_string;

edited Sep 24, 2012 at 17:45

answered Sep 24, 2012 at 17:36

Daniel Li

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2 Comments

Ricardo Capistran Over a year ago

I try using the code, and didnt work :( enteratenorte.com/admon/test.php

Daniel Li Over a year ago

Try the second example (assuming you're using mysql_*)

juergen d · Accepted Answer · 2012-09-24 17:35:51Z

0

select count(*) as found_num
from en_videos
where `code` = '$random_string'

returns 0 if the string was not found and > 0 if it was.

answered Sep 24, 2012 at 17:35

juergen d

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