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使用filter过滤参数
let a = [0];
a = a.filter(item => {
return item
});
console.log(a); //[] 0也会被过滤掉
let b = [0,1,2,undefined];
b = b.filter(item => {
return item
});
console.log(b);//[1, 2]
reduce谜之操作
const arr = [{id: 1,type: 'A',total: 3}, {id: 2,type: 'B',total: 5}, {id: 3,type: 'E',total: 7}];
const re = arr.reduce((sum,{total})=>{
return sum + total;
},0);
const re1 = arr.reduce((str, { id, type,total }) => {
return str + `id:${id},type:${type},total:${total};`;
}, '');
const re2 = arr.reduce((res, { id, type, total }) => {
res[id] = {
type,
total
};
return res;
}, {});
const re3 = arr.reduce((res, { id, type, total },index) => {
res[index] = type;
return res;
}, []);
console.log(re); //结果为total的累加
console.log(re1); //id:1,type:A,total:3;id:2,type:B,total:5;id:3,type:E,total:7;字符串
console.log(re2); //{1: {type: "A", total: 3}, 2: {…}, 3: {…}}
console.log(re3); //["A", "B", "E"]
const re3 = arr.reduce((
第一个参数:累加器累加回调的返回值-res,
第二个参数:数组中正在处理的元素-{ id, type, total },
第三个参数可选:当前元素的索引-index,
第四个参数可选:调用reduce()的数组-array) => {
res[index] = type;
return res;
}, []);