“能力越大,责任越大。”— 本叔叔,蜘蛛侠 (2002)就像蜘蛛侠必须掌握他新发现的能力一样,开发人员需要掌握 javascript 强大的数组方法才能高效、负责任地进行编码。 让我们深入研究一些必须知道的数组方法! 1. 查找find() 方法返回满足所提供的测试函数的第一个数组元素的值。arr.find(callback(element, index, arr),thisarg)立即学习“Java免费学习笔记(深入)”;返回数组中满足给定函数的第一个元素的值。如果没有元素满足该函数,则返回未定义const cuties = [ { name: "wanda maximoff", age: 31 }, { name: "natasha romanoff", age: 32 }, { name: "jane foster", age: 27 }, { name: "gwen stacy", age: 26 },];// find method returns the value of the first element // in the array that satisfies the given function // or else returns undefinedlet cuty = cuties.find(({ age }) => age >= 30);// output: { name: 'wanda maximoff', age: 31 }console.log(cuty);登录后复制 2. 查找索引findindex() 方法返回满足所提供的测试函数的第一个数组元素的索引,否则返回 -1。arr.findindex(callback(element, index, arr),thisarg)返回数组中满足给定函数的第一个元素的索引。如果没有元素满足该函数,则返回-1。const cuties = [ { name: "wanda maximoff", age: 31 }, { name: "natasha romanoff", age: 32 }, { name: "jane foster", age: 27 }, { name: "gwen stacy", age: 26 },];// findindex method returns the index of the first array element // that satisfies the provided test function or else returns -1.let cutyindex = cuties.findindex(({ age }) => age >= 30);// output: 0console.log(cutyindex);登录后复制 3.索引indexof() 方法返回数组元素出现的第一个索引,如果未找到,则返回 -1。arr.indexof(searchelement, fromindex)如果元素在数组中至少出现一次,则返回该元素的第一个索引。如果在数组中未找到该元素,则返回-1。const productprices = [5, 12, 3, 20, 5, 2, 50];// indexof() returns the first index of occurance // of an array element, or?-1?if it is not found.let firstindex = productprices.indexof(20);console.log(firstindex); // 3let secondindex = productprices.indexof(5);console.log(secondindex); // 0// the second argument specifies the search start indexlet thirdindex = productprices.indexof(5, 1);console.log(thirdindex); // 4// indexof returns -1 if not foundlet notfoundindex = productprices.indexof(15);console.log(notfoundindex); // -1登录后复制 4. 排序sort() 方法按特定顺序(升序或降序)对数组的项目进行排序。arr.sort(comparefunction)对数组元素进行排序后返回数组(这意味着它更改了原始数组并且不进行复制)。const avengers = ["captain", "tony", "thor", "natasha", "bruce", "clint"];// modifies the array in placeavengers.sort(); // [ 'bruce', 'captain', 'clint', 'natasha', 'thor', 'tony' ]console.log(avengers); const nums = [1000, 50, 2, 7, 14];// number is converted to string and sortednums.sort(); // output: [ 1000, 14, 2, 50, 7 ]console.log(nums) // sort nums in ascending order by providing compare functionnums.sort((a, b) => a - b);// output: [ 2, 7, 14, 50, 1000 ]console.log(nums);登录后复制 5. 包括includes() 方法检查数组是否包含指定元素。arr.includes(valuetofind, fromindex)includes() 方法返回:如果在数组 searchvalue 中的任何位置找到,则为 true如果在数组 searchvalue 中找不到任何位置,则返回 falseconst avengers = ["captain", "tony", "thor", "natasha", "bruce", "clint"];// includes() method returns true if an array contains // a specified element or else returns false.let check1 = avengers.includes("thor");console.log(check1); // true // second argument specifies position to start the searchlet check2 = avengers.includes("thor", 3);console.log(check2); // false// the search starts from the 4th-to-last element ("hulk") // and checks the rest of the array for "thor". let check3 = avengers.includes("thor", -4);console.log(check3); // true登录后复制 6. foreachforeach() 方法为每个数组元素执行提供的函数。arr.foreach(callback(currentvalue), thisarg)? foreach() 不会对没有值的数组元素执行回调。? 返回未定义。const nums = [120, 150, 80, , 200];// foreach() method executes a provided // function for each array element which // have values. it returns undefined./*num 0: 120num 1: 150num 2: 80num 4: 200*/nums.foreach((value, index) => { console.log('num ' + index + ': ' + value);});登录后复制 7. 切片slice() 方法将数组的一部分的浅拷贝返回到新的数组对象中。arr.slice(开始, 结束)? 返回包含提取元素的新数组。const fruits = ["apple", "banana", "orange", "grape", "mango"];// slicing the array (from start to end)let slicedfruits1 = fruits.slice();console.log(slicedfruits1); // [ 'apple', 'banana', 'orange', 'grape', 'mango' ]// slicing from the third elementlet slicedfruits2 = fruits.slice(2);console.log(slicedfruits2); // [ 'orange', 'grape', 'mango' ]// slicing from the second element to fourth elementlet slicedfruits3 = fruits.slice(1, 4);console.log(slicedfruits3); // [ 'banana', 'orange', 'grape' ]// slicing the array from start to second-to-lastlet slicedfruits4 = fruits.slice(0, -1);console.log(slicedfruits4); // [ 'apple', 'banana', 'orange', 'grape' ]// slicing the array from third-to-lastlet slicedfruits5 = fruits.slice(-3);console.log(slicedfruits5); // [ 'orange', 'grape', 'mango' ]登录后复制 8. 拼接splice() 方法修改数组(添加、删除或替换元素)。arr.splice(start, deletecount, item1, ..., itemn)? 返回包含已删除元素的数组。let animals = ["dog", "cat", "elephant", "lion"];// replacing "elephant" & "lion" with "tiger" & "giraffe"let removedanimals1 = animals.splice(2, 2, "tiger", "giraffe");console.log(removedanimals1); // [ 'elephant', 'lion' ]console.log(animals); // [ 'dog', 'cat', 'tiger', 'giraffe' ]// adding elements without deleting existing elementslet removedanimals2 = animals.splice(1, 0, "elephant", "lion");console.log(removedanimals2); // []console.log(animals); // [ 'dog', 'elephant', 'lion', 'cat', 'tiger', 'giraffe' ]// removing 3 elementslet removedanimals3 = animals.splice(2, 3);console.log(removedanimals3); // [ 'lion', 'cat', 'tiger' ]console.log(animals); // [ 'dog', 'elephant', 'giraffe' ]登录后复制 9. 每个every() 方法检查所有数组元素是否通过给定的测试函数。arr.every(callback(currentvalue), thisarg)every() 方法返回:true - 如果所有数组元素都通过给定的测试函数(回调返回真值)。false?- 如果任何数组元素未通过给定的测试函数。const nums1 = [ 1 , 2 , 3 , 4 , 5];// every()?method returns true if all the array // elements pass the given test function or else// returns falselet result1 = nums1.every(element => element element <h3> 10.<strong>一些</strong></h3><p>some() 方法测试是否有任何数组元素通过给定的测试函数。</p><p>arr.some(callback(currentvalue), thisarg)</p>登录后复制如果数组元素通过给定的测试函数,则返回 true(回调返回真值)。否则返回 falseconst nums1 = [ 8 , 2 , 7 , 9 , 6];// some()?method returns true if any of the array // elements pass the given test function or else // returns falselet result1 = nums1.some(element => element element <p>请继续关注我们系列的第 2 部分,我们将深入探讨更重要的 javascript 数组方法!快乐学习!</p> 登录后复制以上就是每个开发人员都应该掌握的 JavaScript 数组方法(第 1 部分)的详细内容,更多请关注我的其它相关文章!
标签:console,log,开发人员,age,JavaScript,element,returns,数组 From: https://www.cnblogs.com/aow054/p/18434407