如果数据库中存在新的随机值，如何进行验证和插入

Question

This is my code

$lclReferCode = rand(100000,999999);

function random(){
   $lclReferCode1 = rand(100000,999999);
   return $lclReferCode1;
}

$lclReferCode = random();
$lclQuery = "SELECT * FROM `sign_up` WHERE us_refer_code = '$lclReferCode'";

$qryResult = mysql_query($lclQuery);

if(mysql_num_rows($qryResult) == 0){

$lclQuery = "INSERT INTO sign_up(us_refer_code) values('" . $lclReferCode . "')";


            $qryResult = mysql_query($lclQuery);

            }else{

              random();
             }

Here my intention is to insert unique values not the duplicates. Before inserting I am checking the database if this random number is present already or not if it is present I have to generate new values again I have to generate values and match if not present then the insert query executes. can anyone help me.

Answer 1

Like this (untested):

function setRandom(\PDO $PDO){
    $stmt = $PDO->prepare("SELECT * FROM `sign_up` WHERE us_refer_code = :us_refer_code");
    try{
        $stmt->execute([':us_refer_code' => rand(100000,999999)]);
    }catch(\PDOException $e){
        if($e->getCode() == 23000){ //if I recall right this is duplicate key
            setRandom($PDO); //<-- recursive.
        }else{
            //if not duplicate key re-throw the error.
            throw new \PDOException($e->getMessage(), $e->getCode(), $e);
        }
    }
}

But you have to use PDO, with the error mode set to exception, and with the database field defined as a unique index.

It might not be a bad idea to build in a recursion limit, as you could get in a situation where it never finds a value not entered in the DB. That would look like this.

function setRandom(\PDO $PDO, $limit=0){
    if($limit > 1000) throw new \Exception(__FUNCTION__.' recursion limit exceeded');
    $stmt = $PDO->prepare("SELECT * FROM `sign_up` WHERE us_refer_code = :us_refer_code");
    try{
        $stmt->execute([':us_refer_code' => rand(100000,999999)]);
    }catch(\PDOException $e){
        if($e->getCode() == 23000){ //if I recall right this is duplicate key
            setRandom($PDO, ++$limit);//<-- recursive
        }else{
            //if not duplicate key re-throw the error.
            throw new \PDOException($e->getMessage(), $e->getCode(), $e);
        }
    }
}

This would give it 1000 tries before throwing an exception. It's never good to have code that has a real possibility of creating infinite recursion.

On my sites I have a error handler that shoots me an email if I have any uncaught exceptions. So I would know right away. Of coarse this is optional and some of this is tightly coupled to your particular use case.

Otherwise you will have to use a select inside of a while loop. like this (Also untested).

/**
*  return false or the random value to use.
* @return false|int 
*/
function ckRandom($lclReferCode){
    $lclReferCode = rand(100000,999999);
    $lclQuery = "SELECT * FROM `sign_up` WHERE us_refer_code = '$lclReferCode'";
    $qryResult = mysql_query($lclQuery);
    if(mysql_num_rows($qryResult) == 0){
        //no result means the number is not used yet.
        return $lclReferCode;
    }else{
        return false;
    }
}

//$limit = 0;
$lclReferCode = false;
while( false === $lclReferCode){
    $lclReferCode = ckRandom();
    //if($limt > 1000) throw new \Exception(' recursion limit exceeded');
    //++$limit;
}
//if $lclReferCode IS NOT false then it is the last random number searched that had no results in the DB
$lclQuery = "INSERT INTO sign_up(us_refer_code) values('" . $lclReferCode . "')";

See the idea is the while loop will trap execution until the result returns something other then 0, if it's 0 then no row exists. (you have this backwards).

And again, this is prone to infinite recursion. I added the "limit" stuff in comments.

That said, I would strongly suggest doing away with any mysql_ functions as they wont work post PHP7.

Answer 2

you need to try like this

function random(){
   $lclReferCode = rand(100000,999999);
   $lclQuery = "SELECT * FROM `sign_up` WHERE us_refer_code = '$lclReferCode'";
   $qryResult = mysql_query($lclQuery);
   $checkRow=mysql_num_rows($qryResult);
   if ($checkRow>0) {
        return random();
   }else{
        $lclQuery = "INSERT INTO sign_up(us_refer_code) values('" . $lclReferCode . "')";
        $qryResult = mysql_query($lclQuery);
        return true;
    }
}

if (random()===true) {
    echo 'inserted succesfully';
}

remember this is demonstration.i recommend to you use mysqli or pdo because mysql is not valid in php7