There's a mysql database that stores ids and names, when users are creating names with a form they can create existent names since unique ids are the ids such as:
d49f2c32f9107c87c7bb7b44f1c8f297 name
2fa9c810fbe082900975f827e8ed9408 name
what i want to do is saving the second "name" -> "name(1)" when inserting into database.
So far what I've got is this as the idea
lets say the name entered is 'name'
$input = 'name';
select the name we want to check from mysql database
mysql_query(SELECT * FROM `table` WHERE `name` = '$input');
if the result exists, then insert as $input.'(1)'
my question is what if name exists, and name(1) also exists, then how can i add the name(2) there...
$i = 1;
$sourceName = $name;
while( sql "SELECT COUNT(*) FROM table WHERE name = '$name'" ) {
$name = $sourceName.' ('.$i.')';
$i++;
}
At this point you have the final $name (with $i counting the iteration)
You could return the number of people with that name in the database, then add 1 to that number.
SELECT COUNT(id) FROM table WHERE name LIKE '$input(%)');
Something like this should do the trick:
SELECT * FROM table WHERE name = '$input' OR name LIKE '$input(%)'
Note that for this to work, you'd need to escape any percent signs in $input in the LIKE clause, otherwise they'll be treated as wildcards.
Use a regex that checks for integers between two parentheses at the end of the string. If there exists an integer, add 1 to it.
You could also attempt to do it the other way around, make name field unique and try to input it in a while loop, if it fails add ($i) and do $i++ every iteration.
//edit: The problem with using solutions that do a like comparison is that you will get false positives, for instance $hi% will also count hippie. Which gives you unnecessary (1) additions!