CODEWITHSUNDEEP
javascriptnode.jsnode-mysql
JonasGame
JonasGame

Reputation: 25

return value from function is always undefined and order

I try to make a function which generate an aleatory number and verify if it doesn't already used, and return it.

But I got a undefined result, and I got in first the result of the function against the console.log which is in this function, whereas it should be the reverse.

// function main()
console.log('The result of the function in the main() is ' + Bank_generateAccountNumber());

// function Bank_generateAccountNumber()
function Bank_generateAccountNumber()
{
    var account_number = Math.floor((Math.random() * 8999) + 1000);

    console.log('Bank_generateAccountNumber trying with this number: ' + account_number); 
    bdd.query('SELECT * FROM bank_accounts WHERE account_number = ?', gm.mysql.escape(account_number), function(e, d, f) 
    {
        if(!d.id) 
        {
            console.log("this number is available ! " + account_number);
            return account_number;
        }

        console.log("this number is already used ! " + account_number);
        Bank_generateAccountNumber();
        return 0;
    });  
}

I'm seeing when I'm writing this, even if i'm not connected to mysql, i got "the result of the function in the main() is undefined", and after got a error because "d.id" is not defined.

error with a image

I want get the console.log (which is in the function) in first, and after get the result of the function.

Do you have any idea? Thank you

Upvotes: 0

Views: 97

Answers (1)

Johnny Tordgeman
Johnny Tordgeman

Reputation: 148

I modified your function to use a deferred promise using the Q library. As Naresh Walia wrote in the comments you have more then one library to do so:

var q = require('q');

Bank_generateAccountNumber().then(function(response) {
  console.log('The result of the function in the main() is ' + response);
})

function Bank_generateAccountNumber() {
  var response = q.defer();
  var account_number = Math.floor((Math.random() * 8999) + 1000);

  console.log('Bank_generateAccountNumber trying with this number: ' + account_number);
  bdd.query('SELECT * FROM bank_accounts WHERE account_number = ?', gm.mysql.escape(account_number), function(e, d, f) {
    if (!d.id) {
      console.log("this number is available ! " + account_number);
      response.resolve(account_number);
    }

    console.log("this number is already used ! " + account_number);
    Bank_generateAccountNumber();
  });

  return response.promise;
}

Upvotes: 1

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