Question

我正在使用fetch和promise来对api进行2次调用并返回数据。我在承诺中获取数据没有问题，我无法在它之外访问它。我原以为return apiData会将它返回使用？我想我错过了像.finalise或.success这样的东西，让它超出承诺范围？

基本上我正试图弄清楚如何访问promise之外的JSON对象，其中包括所有数据，而不仅仅是JSON结构。

var api1 = fetch('api.example1.com/search').then(function(response){ 
         return response.json()
});
var api2 = fetch('api.example2.com/search').then(function(response){
         return response.json()
});

var apiData = {"api1":{},"api2":{}};
Promise.all([api1,api2]).then(function(values){
    apiData.api1 = values[0];
    apiData.api2 = values[1];
    console.log(JSON.stringify(apiData, null, 4)); 
    //this displays all the data as it's still within the promise
    return apiData;
});
console.log(JSON.stringify(apiData, null, 4)); 
    //this doesn't get the values of the api1 and api2 data as it's outside of the
    //promise and only displays the object structure

Answer 1

您始终可以全局访问apiData。只需确保在分配数据后访问它。您之所以没有获取数据，是因为您在承诺返回前致电console.log。

试试这个：

var apiData = {"api1":{},"api2":{}};
Promise.all([api1,api2]).then(function(values){
        apiData.api1 = values[0];
        apiData.api2 = values[1];
        console.log(JSON.stringify(apiData, null, 4)); 
        //this displays all the data as it's still within the promise
        return apiData;
    })
    .then(function(){
        console.log(JSON.stringify(apiData, null, 4)); // You still reference apiData globally
            //this doesn't get the values of the api1 and api2 data as it's outside of the
            //promise and only displays the object structure
    });

甚至这个（仅用于演示，不要使用它）：

var apiData = {"api1":{},"api2":{}};
Promise.all([api1,api2]).then(function(values){
    apiData.api1 = values[0];
    apiData.api2 = values[1];
    console.log(JSON.stringify(apiData, null, 4)); 
    //this displays all the data as it's still within the promise
    return apiData;
});
setTimeout(function(){
    console.log(JSON.stringify(apiData, null, 4)); // should be now able to log the data here 
        //this doesn't get the values of the api1 and api2 data as it's outside of the
        //promise and only displays the object structure
}, 1000); // assume the fetch calls finish in 1s

访问承诺之外的数据

1 个答案: