我有一个目标数组[“apple”、“banana”、“orange”]，我想检查其他数组是否包含任何一个目标阵列元素。
例如：
["apple","grape"] //returns true;
["apple","banana","pineapple"] //returns true;
["grape", "pineapple"] //returns false;
如何在JavaScript中实现？
当前回答
嵌套调用.find（）的Array.filter（）将返回第一个数组中属于第二个数组的所有元素。检查返回数组的长度，以确定第一个数组中是否有第二个数组。
getCommonItems(firstArray, secondArray) {
return firstArray.filter((firstArrayItem) => {
return secondArray.find((secondArrayItem) => {
return firstArrayItem === secondArrayItem;
});
});
}
其他回答
带下划线js
var a1 = [1,2,3];
var a2 = [1,2];
_.every(a1, function(e){ return _.include(a2, e); } ); //=> false
_.every(a2, function(e){ return _.include(a1, e); } ); //=> true
您正在寻找两个数组之间的交集。你有两种主要的交叉点类型：“每个”和“一些”。让我举几个好例子：
每
let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia'];
// Find 'every' brand intersection.
// Meaning all elements inside 'brands2' must be present in 'brands1':
let intersectionEvery = brands2.every( brand => brands1.includes(brand) );
if (intersectionEvery) {
const differenceList = brands1.filter(brand => !brands2.includes(brand));
console.log('difference list:', differenceList);
const commonList = brands1.filter(brand => brands2.includes(brand));
console.log('common list:', commonList);
}
如果条件不满足（比如你在品牌中加入了“梅赛德斯”），那么“intersectionEvery”就不满足了——这将是错误的。
如果满足条件，它将把[“福特”、“大众”]列为区别，把[“起亚”、“奥迪”]列为了常见列表。
沙盒：https://jsfiddle.net/bqmg14t6/
SOME
let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia', 'Mercedes', 'Land Rover'];
// Find 'some' brand intersection.
// Meaning some elements inside 'brands2' must be also present in 'brands1':
let intersectionSome = brands2.some( brand => brands1.includes(brand) );
if (intersectionSome) {
const differenceList = brands1.filter(brand => !brands2.includes(brand));
console.log('difference list:', differenceList);
const commonList = brands1.filter(brand => brands2.includes(brand));
console.log('common list:', commonList);
}
我们在这里寻找一些常见的品牌，但不一定全部。
它将把[“福特”、“大众”]列为不同品牌，把[“起亚”、“奥迪”]列为了共同品牌。
沙盒：https://jsfiddle.net/zkq9j3Lh/
可以使用嵌套的Array.prototype.some调用。这有一个好处，即它将在第一场比赛中获胜，而不是其他将在整个嵌套循环中运行的解决方案。
eg.
var arr = [1, 2, 3];
var match = [2, 4];
var hasMatch = arr.some(a => match.some(m => a === m));
console.log("searching Array: "+finding_array);
console.log("searching in:"+reference_array);
var check_match_counter = 0;
for (var j = finding_array.length - 1; j >= 0; j--)
{
if(reference_array.indexOf(finding_array[j]) > 0)
{
check_match_counter = check_match_counter + 1;
}
}
var match = (check_match_counter > 0) ? true : false;
console.log("Final result:"+match);
ES6溶液：
let arr1 = [1, 2, 3];
let arr2 = [2, 3];
let isFounded = arr1.some( ai => arr2.includes(ai) );
与之不同：必须包含所有值。
let allFounded = arr2.every( ai => arr1.includes(ai) );
希望，会有所帮助。
