Question

如何允许条件承诺触发/不触发并仍然正确解决？

让我解释一下背景。我有三个要查找的值。起初我做了

var CustomerReceived = myAjaxLibCall("Customer", "ID", report.CustomerID);
var VehicleReceived  = myAjaxLibCall("Vehicle", "ID", report.VehicleID);
var DealerReceived  = myAjaxLibCall("Dealer", "ID", report.DealerID);

$.when(CustomerReceived,
       VehicleReceived,
       DealerReceived)
.then((Customer,
       Vehicle,
       Dealer) => {
    //do some code here
});

这很好用。但是，AJAX调用很慢并且需要花费很多时间。当用户在创建新报告后到达此点时，他们已经填充了每个项目的列表，这意味着我可以这样做

var Customer = customerList.findbyID(report.CustomerID);
var Vehicle = vehicleList.findbyID(report.VehicleID);
var Dealer = dealerList.findbyID(report.DealerID);

//do some code here

然后，事实证明，这些列表中的每一个都可能被填充，而且可能没有，有时一个或两个是，而其他则不是。理想情况下，如果相应的本地列表不存在，我只想执行每个ajax调用。

但这样做的最佳方法是什么？这是我尝试过的，但它不起作用：

var CustomerReceived = customerList.length === 0
                     ? myAjaxLibCall("Customer", "ID", report.CustomerID)
                     : customerList.findbyID(report.CustomerID);

var VehicleReceived  = vehicleList.length === 0
                     ? myAjaxLibCall("Vehicle", "ID", report.VehicleID)
                     : vehicleList.findbyID(report.VehicleID);

var DealerReceived  = dealerList.length === 0
                    ? myAjaxLibCall("Dealer", "ID", report.DealerID)
                    : dealerList.findbyID(report.DealerID);

$.when(CustomerReceived,
       VehicleReceived,
       DealerReceived)
.then((Customer,
       Vehicle,
       Dealer) => {
    //do some code here
});

Answer 1

等待

您可以使用await关键字：

async function someStuff()
{
    var CustomerReceived = customerList.length === 0
                         ? myAjaxLibCall("Customer", "ID", report.CustomerID)
                         : customerList.findbyID(report.CustomerID);

    var VehicleReceived  = vehicleList.length === 0
                         ? myAjaxLibCall("Vehicle", "ID", report.VehicleID)
                         : vehicleList.findbyID(report.VehicleID);

    var DealerReceived  = dealerList.length === 0
                        ? myAjaxLibCall("Dealer", "ID", report.DealerID)
                        : dealerList.findbyID(report.DealerID);

    const Customer = await CustomerReceived;
    const Vehicle = await VehicleReceived;
    const Dealer = await DealerReceived;

    //do some code here
}

根据MDN：

如果await运算符后面的表达式的值不是a 承诺，它已转换为已解决的承诺。

没有等待

如果您不想使用async关键字，可以使用Promise.all()，但需要将promises / functions存储到数组中

var CustomerReceived = customerList.length === 0
                     ? myAjaxLibCall("Customer", "ID", report.CustomerID)
                     : customerList.findbyID(report.CustomerID);

var VehicleReceived  = vehicleList.length === 0
                     ? myAjaxLibCall("Vehicle", "ID", report.VehicleID)
                     : vehicleList.findbyID(report.VehicleID);

var DealerReceived  = dealerList.length === 0
                    ? myAjaxLibCall("Dealer", "ID", report.DealerID)
                    : dealerList.findbyID(report.DealerID);

Promise.all([ CustomerReceived, VehicleReceived, DealerReceived ])
    .then(([ Customer, Vehicle, Dealer ]) => {
        //do some code here
    });

仍然根据MDN：

如果所有传入的承诺都履行或不承诺，那么 Promise.all返回的promise是异步完成的。在所有例如，返回的promise通过包含all的数组来实现迭代的值作为参数传递（也是非承诺值）。

如何混合承诺和未承诺的功能？

1 个答案:

等待

没有等待