javascript - 函数的返回值总是未定义且顺序

Question

我尝试创建一个生成随机数字的函数，并验证它是否尚未使用，然后返回它。

但是我得到了一个未定义的结果，并且我首先根据该函数中的 console.log 获得了该函数的结果，而它应该是相反的。

// 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;
    });  
}

我在写这篇文章时看到，即使我没有连接到 mysql，我也会收到“main() 中函数的结果未定义”，并且在收到错误后，因为“d. id”未定义。

我想首先获取console.log(位于函数中)，然后获取函数的结果。

你有什么想法吗？谢谢

Answer 1

我修改了您的函数以使用 Q 库的延迟 promise 。正如 Naresh Walia 在评论中所写，您有多个库可以这样做:

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;
}