return 函数的值始终未定义且顺序

return value from function is always undefined and order

我尝试创建一个生成随机数的函数并验证它是否尚未使用,然后return它。

但是我得到了一个未定义的结果,我首先得到了这个函数中针对 console.log 的函数的结果,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了,即使我没有连接到 mysql,我也收到了 "the result of the function in the main() is undefined",并且在收到错误消息后因为 "d.id" 是未定义。

我想先得到console.log(在函数中),然后得到函数的结果。

你有什么想法吗? 谢谢

我修改了您的函数以使用 Q 库使用延迟承诺。正如 Naresh Walia 在评论中所写,你有不止一个图书馆可以这样做:

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;
}