沒有推入陣

Question

創建通過數字的指數陣列我有一個數組

var nums = [1,2,4]; 

而且我還有一個陣列坐滿了人

var people = [ 
    { name: 'Adam',  email: 'adam@email.com',  age: 12, country: 'United States' }, 
    { name: 'Amalie', email: 'amalie@email.com', age: 12, country: 'Argentina' }, 
    { name: 'Estefanía', email: 'estefania@email.com', age: 21, country: 'Argentina' }, 
    { name: 'Adrian', email: 'adrian@email.com', age: 21, country: 'Ecuador' }, 
    { name: 'Wladimir', email: 'wladimir@email.com', age: 30, country: 'Ecuador' }, 
]; 

我希望創建一個基於關使用的變量nums變量充當people變量的索引。

// With the nums array I take each value and set it as the value of the new variable 
// This is my expected output. Although this is line of code is not possible since the nums variable will be unique each time the code run. 
var select_people = [people[1], people[2], people[4]]; 

我無法創建一個空數組，然後將每個元素都推入select_people數組中，像這樣。

// This will not do for me 
var select_people = []; 

for(var i = 0; i < nums.length; i++) { 
    select_people.push(people[nums[i]]) 
} 

我的問題是這樣的。如何編寫此代碼，以便我可以分配select_people變量而不必將值推入數組中？

Answer 1

如果這是你想要簡潔，那麼你可以嘗試：

var selectPeople = people.filter(function(k, i) { return nums.indexOf(i) >= 0; }); 

同樣，你可以做（​​我其實更喜歡這個）：

var selectPeople = nums.map(function(k) { return people[k]; }); 

注意：這隻適用於現代瀏覽器。

但是，我想不到很多情況下使用push不是最好的選擇。

如果它是一個命名衝突，你可以隨時將它包裝在一個臨時的功能（這在所有瀏覽器上運行）：

var selectPeople = (function() { 
    var temp = []; 
    for (var i = 0; i < nums.length; ++i) temp.push(people[nums[i]]); 
    return temp; 
})(); 

這基本上消除了任何命名衝突（或者，例如衝突，其中selectPeople不是一個真正的數組，因爲它缺少push方法）。

Answer 2

您還沒有初始化你i變量在你for循環：

這應該工作：

var select_people = []; 

for(var i = 0; i < nums.length; i++) { 
    select_people.push(people[nums[i]]) 
} 

Answer 3

另一種獲得相同結果的方法。

var people = [ 
 
    { name: 'Adam',  email: 'adam@email.com',  age: 12, country: 'United States' }, 
 
    { name: 'Amalie', email: 'amalie@email.com', age: 12, country: 'Argentina' }, 
 
    { name: 'Estefanía', email: 'estefania@email.com', age: 21, country: 'Argentina' }, 
 
    { name: 'Adrian', email: 'adrian@email.com', age: 21, country: 'Ecuador' }, 
 
    { name: 'Wladimir', email: 'wladimir@email.com', age: 30, country: 'Ecuador' }, 
 
]; 
 
var nums = [1,2,4]; 
 
var j = []; 
 
for(var i = 0, l = nums.length; i < l; i++) { 
 
    j.push(JSON.stringify(people[nums[i]])); 
 
} 
 
j = '[' + j.join(',') + ']'; 
 
var selectPeople = JSON.parse(j); 
 
console.log(selectPeople);

Answer 4

for(x=0;x<nums.length;x++){ 
    alert(people[nums[x]]['name']); 
    // or you can define your select_people here with people[nums[x]] 
    // yes, you get people[1], people[2] and people[4] 
    // but, your first people is "Adam", and people[1] is "Amalie" 
} 

，所以如果你想採取第一人看重 「NUMS」， 「1」，只是做

for(x=0;x<nums.length;x++){ 
    alert(people[nums[x]-1]['name']); 
    // or you can define your select_people here with people[nums[x]-1] 
    // yes, you get people[0], people[1] and people[3] 
    // your first people is "Adam", and people[0] is "Adam" 
}