Question

我在Parse Cloud Code中复制数组时遇到问题。在下面的findAllAssociatedBrands函数中，我使用Parse Queries来获取成功回调的数组。在成功回调时，日志语句：console.log（JSON.stringify（brandList [i]））;打印数组元素，然后我复制新数组中的所有元素。在for循环之后，log语句将新数组显示为[]，并且其中没有元素。我错过了什么？

function findAllAssociatedBrands(beaconObj) {

var typeOfBeacon = beaconObj.get("type");
var beaconBrandRelation = beaconObj.relation("brand");
var query = beaconBrandRelation.query();
var listOfAllBeaconAssociatedBrands = [];

query.find({
    success: function(brandList) {
        for (var i = 0; i < brandList.length; i++) {
            console.log(JSON.stringify(brandList[i]));
            listOfAllBeaconAssociatedBrands[i] = brandList[i];
        }

        console.log(listOfAllBeaconAssociatedBrands);
    },
    error: function(error) {
        console.log("Beacon Brand Relation Error: " + error.code + "  " + error.message);
    }
});

return listOfAllBeaconAssociatedBrands;
 }

Answer 1

query.find（）是一个异步函数，所以findAllAssociatedBrands（）函数的最后一行在query.find（）之后立即执行，所以query.find（）通常甚至没有足够的时间来运行成功回调所以你必须绑“return listOfAllBeaconAssociatedBrands;”到成功回调结束。

function findAllAssociatedBrands(beaconObj) {

var typeOfBeacon = beaconObj.get("type");
var beaconBrandRelation = beaconObj.relation("brand");
var query = beaconBrandRelation.query();
var listOfAllBeaconAssociatedBrands = [];

query.find({
    success: function(brandList) {
        for (var i = 0; i < brandList.length; i++) {
            console.log(JSON.stringify(brandList[i]));
            listOfAllBeaconAssociatedBrands[i] = brandList[i];
        }

        console.log(listOfAllBeaconAssociatedBrands);
return listOfAllBeaconAssociatedBrands;
    },
    error: function(error) {
        console.log("Beacon Brand Relation Error: " + error.code + "  " + error.message);
    }
});


 }

在Parse Cloud代码中未复制数组

1 个答案: