2016-01-18 70 views
1

比方說,我有一個這樣的數組:增加值到陣列中的另一個陣列

var someData = [ // count: 5 

    [ something, something, something ], // count: 3 
    [ something ], // count: 1 
    [ something, something, something, something ], // count: 4 
    [ ], // count: 0 
    [ something, something ] // count: 2 
]; 

我需要數組是這樣的:

var myArray = []; 

var firstCount = someData.lenght; // count: 5 

for(a = 0; a < firstCount; a++) { 

    allObjects[a] = firstObject; 


    var secondCount = someData[a].lenght; // each has different 

    for(b = 0; b < secondCount; b++) { 

     allObjects[a] = secondObject; 
    } 
} 

如何IM電影期待是:

allObjects = [ 

    [ firstObject, secondObject, secondObject ], // count: 3 
    [ firstObject ], // count: 1 
    [ firstObject, secondObject, secondObject, secondObject ], // count: 4 
    [ ], // count: 0 
    [ firstObject, secondObject ], // count: 2 
]; 

我可能會做一些(格式)失誤在這個例子中,但我希望我的目標很明確 - 我需要推動對象,以對象的數組,它是已經在陣列但似乎替換值而是增加新的價值。

我在這裏錯過了什麼?

+0

它不是完全清楚你想要做什麼,但一般'陣列[I] =「value''覆蓋值,如果要添加任一'推()'或'的unshift()' – adeneo

+2

請的具體例,沒有'something'和'secondObject' ,包括一個示例函數調用。 – Amadan

+0

=符號將您的對象分配給變量(allObjects並[a]),則希望將其推到陣列'allObjects並[a] .push(secondObject)' – David

回答

1

我可能是錯的,但是從你期望的結果,似乎想要把它推到數組的第二個索引。在這種情況下,你可以改變你的內部for循環是

var secondCount = someData[a].lenght; // each has different 

    for(b = 1; b < secondCount; b++) { 

     allObjects[a][b] = secondObject; 
    } 

注意的變化在B = 1和allObjects [A] [B]