INSERT if doesn't already exist in database

Question

I created a function that looks for all rows (records) in a specific column, generates a random number and then checks if it already exists withing that column, if the generated random number exists restart the function (recursive) if it doesn't exist then insert it as new.

$verifs = mysqli_query($conlink, 'SELECT verif FROM memebers');
$records = array();
while($record = mysqli_fetch_array($verifs,MYSQLI_NUM)) {
    $records[] = $record[0];
}
function randomize(){
    $rand= rand(1,5);
    if(in_array($rand, $GLOBALS['records'])){
        randomize();
    }else{
        return $rand;
    }
}
mysqli_query($conlink, "INSERT INTO `memebers`(`verif`) VALUES (".randomize().")");

i expected that a number MUST be added EACH time i refresh my php page
and it was the case, but when i reach 3 or 4 records, refreshing the page doesn't add any ! so the return statement must prevent the function from being recursive. am i having anything wrong with my logic thanks

are you sure $records is present in $GLOBALS? Try doing a `print_r($GLOBALS['records'])` and let us know what output you get. — Webeng, May 29 '16 at 12:16
records is actually present, but empty at first, and starts having values the more i refresh the page, tried `in_array($rand, global $records)` but i got a syntax error unexpected global, i think $GLOBALS is being as expected — Simo Simontchensé, May 29 '16 at 12:27
Then im pretty sure it has to do with the randomize function. It's been a while since I used callbacks. My answer might get the job done, but if it doesn't I'll go do some reading to see what's up. Let me know if it works or not — Webeng, May 29 '16 at 12:29

score 1 · Accepted Answer · edited May 23 '17 at 12:24

It appears to me that you are not using the callback function correctly. If you didn't know, when you use a function within a function, you actually have to pass that function inside the parameters. Functions that have a function in the parameter are called callback functions and you can read more about them in the following post: What is a callback function and how do I use it with OOP

I would then change your randomize function to the following code:

function randomize($callbackRandomize){
    $rand= rand(1,5);
    if(in_array($rand, $GLOBALS['records'])){
        return $callbackRandomize($callbackRandomize);
    }else{
        return $rand;
    }
}

and also change your last line of code to:

mysqli_query($conlink, "INSERT INTO `memebers`(`verif`) VALUES (".randomize('randomize').")");

Let me know if that worked for you.

INSERT if doesn't already exist in database

1 Answers1