更改對象數組的順序

Question

想想看，我有如下一個對象數組：

var array = [ 
    { name: "A", position: "0" }, 
    { name: "B", position: "1" }, 
    { name: "C", position: "2" }, 
    { name: "D", position: "3" }, 
    { name: "E", position: "4" }, 
    { name: "F", position: "5" } 
]; 

比方說，用戶拖動有4位的元素，把它有位置1

var replacedItem = { name: "E", position: "4" }; 
var destinationItem = { name: "B", position: "1" }; 

的元素

如何可以重新排列使用javascript具有包含下列值的陣列元素的位置：

var array = [ 
    { name: "A", position: "0" }, 
    { name: "E", position: "1" }, 
    { name: "B", position: "2" }, 
    { name: "C", position: "3" }, 
    { name: "D", position: "4" }, 
    { name: "F", position: "5" } 
]; 

個謝謝，

Answer 1

1 - destionNationItemIndex

4 - replacedItemIndex

你可以試試這個： -

var removed = array.splice(4, 1); 
array.splice(1, 0, removed[0]); 
for (var i = 1; i <= 4; i++) { 
    array[i].position = (parseInt(array[i - 1].position) + 1).toString(); 
} 

打印你所需要的： -

var array = [ 
    { name: "A", position: "0" }, 
    { name: "E", position: "1" }, 
    { name: "B", position: "2" }, 
    { name: "C", position: "3" }, 
    { name: "D", position: "4" }, 
    { name: "F", position: "5" } 
] 

Answer 2

可以使用splice分兩步移動元素：

var a = [ ... ]; 
var temp = a.splice(4, 1); // remove element at index 4 
a.splice(1, 0, temp[0]); // insert the removed element at index 1 

splice()返回被刪除的元素的數組（在這種情況下，1個元素的數組），這就是爲什麼你需要下標temp。

既然你只需要左右移動name屬性，一個方法是將數組分成兩個陣列，重新排列對應name屬性陣列和重組：

var fromIndex = 4, toIndex = 1; 
// Step 1: collect names and positions in separate arrays 
var names = [], positions = []; 
array.forEach(function(elt) { 
     this.names.push(elt.name); 
     this.positions.push(elt.position); 
    }, {names: names, positions: positions}); 
// Step 2: do a circular permutation of the names between the indexes (inclusive) 
var temp = names.splice(fromIndex, 1); 
names.splice(toIndex, 0, temp[0]); 
// Step 3: recombine names and positions into an array of objects 
var n = array.length; 
array.length = 0; 
for (var i = 0; i < n; ++i) { 
    array.push({name: names[i], position: positions[i]}); 
} 

Answer 3

這是最簡單的邏輯實現的是：

var array = [ 
    { name: "A", position: "0" }, 
    { name: "B", position: "1" }, 
    { name: "C", position: "2" }, 
    { name: "D", position: "3" }, 
    { name: "E", position: "4" }, 
    { name: "F", position: "5" } 
]; 

var replacedItem = { name: "E", position: "4" }; 
var destinationItem = { name: "B", position: "1" }; 

for(var i=0; i<array.length; i++){ 
    if(array[i].position == replacedItem.position) 
     array[i].position = destinationItem.position; 
    else if(array[i].position == destinationItem.position) 
     array[i].position = replacedItem.position; 
} 

新陣列將是：

var array = [ 
     { name: "A", position: "0" }, 
     { name: "B", position: "4" }, 
     { name: "C", position: "2" }, 
     { name: "D", position: "3" }, 
     { name: "E", position: "1" }, 
     { name: "F", position: "5" } 
    ]; 

你只需要交換他們的位置。

DEMO