本人菜鸟一枚，写发low,大佬指点~~
一、动态向Url中添加参数
/** * 向 url 添加 query 参数 * * @param {string} targetUrl * @param {object} params */
function addQueryParam(targetUrl, params){ //Todo
}//输出
addQueryParam('https://cloud.tencent.com#hash', {a: 1,b: 1,c: 1})// 返回 https://cloud.tencent.com?a=1&b=1&c=1#hashconsole.log(addQueryParam('https://cloud.tencent.com?a=1&b=1#hash', { b: 1}))// 返回 https://cloud.tencent.com?a=1&b=1#hashconsole.log(addQueryParam('https://cloud.tencent.com?a=1&b=1#hash', { b: 2}))// 返回 https://cloud.tencent.com?a=1&b=2#hashaddQueryParam('https://cloud.tencent.com#hash', {a: 1,b: 1,c: 1})// 返回 https://cloud.tencent.com?a=1&b=1&c=1#hashconsole.log(addQueryParam('https://cloud.tencent.com?a=1&b=1#hash', { b: 1}))// 返回 https://cloud.tencent.com?a=1&b=1#hashconsole.log(addQueryParam('https://cloud.tencent.com?a=1&b=1#hash', { b: 2}))// 返回 https://cloud.tencent.com?a=1&b=2#hash
解法：
function toObj(params) { let newObj = params.reduce((acc, cur) => { var [key, value] = cur.split('='); acc[key] = (value - 0) || undefined; return acc; }, {}) return newObj};function toStr(params) { var arr = []; for (let [key, value] of Object.entries(params)) { arr.push(`${key}=${value}`) } return arr.join('&');}
function addQueryParam(targetUrl, params) { // TODO let newUrl; let a = targetUrl.slice(0, targetUrl.indexOf('?') + 1); let b = targetUrl.slice(targetUrl.indexOf('?') + 1, targetUrl.lastIndexOf('#')); let hash = targetUrl.slice(targetUrl.lastIndexOf('#')); if (targetUrl.indexOf('?') !== -1) { if (b) { c = b.split('&'); params = Object.assign(toObj(c), params); } } let toQ = toStr(params);
if (a) { newUrl = `${a}${toQ}${hash}` } else { newUrl = `${b}?${toQ}${hash}` }
return newUrl}
二、金额千分位
function addComm(num){
//Todo
}
addComm(1234)
//输出 1,234解法一：
function addComm(num) { //TODO var c = (num.toString().indexOf ('.') !== -1) ? num.toLocaleString() : num.toString().replace(/(\d)(?=(?:\d{3})+$)/g, '$1,'); return c;}解法二：浮点值情况
function thousands(num,end){ var splits=[],res=[]; var splits = num.toFixed(end).toString().split("."); splits[0].split("").reverse().map(function(item,i){ if(i%3 == 0 && i!=0){ res.push(","); } res.push(item); }); return res.reverse().join("")+(splits.length>1 ? "."+splits[1] : "");}console.log(thousands(1234.666666666,2))
// 输入 1,234.67三、Key首字符大写
const camelCasedData = { age: 18, gender: "female", experiences: [{ from: "2009-09", to: "2013-06", exp: "School" }, { from: "2013-09", to: "2020-02", exp: "Job", } ],}
const camelToPascal = (input) => { // TODO};
const pascalToCamel = (input) => { // TODO};
const pascalCased = camelToPascal(camelCasedData);console.log(pascalCased);/* 期望输出：{ Age: 18, Gender: "female", Experiences: [ { From: "2009-09", To: "2013-06", Exp: "School" }, { From: "2013-09", To: "2020-02", Exp: "Job" }, ]}*/const camelCasedData2 = pascalToCamel(pascalCased);console.log(camelCasedData2)// 应该和 camelCasedData 一样
解法：
const camelToPascal = (input) => { // TODO let newObj = {}; let upperKey = (key) => key.replace(key[0], key[0].toUpperCase()); for (let [key, value] of Object.entries(input)) { if (Array.isArray(value) && value.length > 0) { newObj[upperKey(key)] = value.map((item) => { if (JSON.stringify(item) !== "{}") { return camelToPascal(item) } }); } else { newObj[upperKey(key)] = value; } } return newObj;};
const pascalToCamel = (input) => { // TODO let newObj = {}; let upperKey = (key) => key.replace(key[0], key[0].toLowerCase()); for (let [key, value] of Object.entries(input)) { if (Array.isArray(value) && value.length > 0) { newObj[upperKey(key)] = value.map((item) => { if (JSON.stringify(item) !== "{}") { return pascalToCamel(item) } }); } else { newObj[upperKey(key)] = value; } } return newObj;};
四、add(1)(2)(3)输出值等价于add(1,2,3)
function add(){
}
add(1)(2)(3) //6
add(1,2,3) //6解法一：
function add () { let sum = Array.from(arguments).reduce((acc, curr) => acc + curr, 0);
function innerAdd() { let _sum = Array.from(arguments).reduce((acc, curr)=> acc + curr, 0); sum += _sum; return innerAdd } innerAdd.toString = function(){ return sum }; return innerAdd}var rusut=add(1)(2)(3).toString();console.log(rusut)五、随机输出数字的顺序
Array.prototype.shuffle = function() { var input = this; for (var i = input.length-1; i >=0; i--) { var randomIndex = Math.floor(Math.random()*(i+1)); console.log(randomIndex) var itemAtIndex = input[randomIndex]; input[randomIndex] = input[i]; input[i] = itemAtIndex; } return input;}[1,2,3,4,5,6,7,8].shuffle();//[4, 6, 3, 2, 5, 1, 7, 8] // 每次结果都是随机的解法：
/** * Fisher–Yates shuffle */Array.prototype.shuffle = function() { var input = this; for (var i = input.length-1; i >=0; i--) { var randomIndex = Math.floor(Math.random()*(i+1)); console.log(randomIndex) var itemAtIndex = input[randomIndex]; input[randomIndex] = input[i]; input[i] = itemAtIndex; } return input;}
六、展开多维数组
[1, [2, [3, 4], 5], 6]
//输出
[ 1, 2, 3, 4, 5, 6 ]解法一：
function * flatten(array) { for (const item of array) { if (Array.isArray(item)) { yield * flatten(item); } else { yield item; } }}const flattened = [...flatten([1, [2, [3, 4], 5], 6])];解法二：
function flatten2(input) { const stack = [...input]; const res = []; while (stack.length) { // 使用 pop 从 stack 中取出并移除值 const next = stack.pop(); if (Array.isArray(next)) { // 使用 push 送回内层数组中的元素，不会改动原始输入 stack.push(...next); } else { res.push(next); } } // 反转恢复原数组的顺序 return res.reverse();}解法三：
function flatDeep(arr, d = 1) { let newA = null; if (d > 0) { newA = arr.reduce(function(acc, val) { if (Array.isArray(val)) { return acc.concat(flatDeep(val, d - 1)); } else { return acc.concat(val) } }, []) } else { newA = arr.slice(); } return newA};console.log(flatDeep([1, [2, [3, 4], 5], 6],Infinity))解法四：
function flatten(arr, d = 1) { // TODO return d > 0 ? arr.reduce( (acc, val) => acc.concat( Array.isArray(val) ? flatten(val, d - 1) : val ), []) : arr.slice();}let a = flatten([1, [2, [3, 4], 5], 6], Infinity);后续有其他的继续添加，有其他解法的，大佬留言哈，学习一哈
