2017-07-18 75 views
1

我无法从单独的数组更新对象值。使用JavaScript中的单个数组更新数组对象

例子:

mainArray = [ 
    { "name": "bob", "complete": "25" }, 
    { "name": "john", "complete": "50" }, 
    { "name": "mike", "complete": "75" },   
]; 

colorArray = ["#ff0000", "#00ff00", "#0000ff"]; 

我需要创建一个新的磁盘阵列,结合这些值。

for (i = 0, ilen = mainArray.length; ilen > i; i++) { 
    newArray.push({ 
     name: mainArray[i].name, 
     complete: mainArray[i].complete, 
     color: '', 
    }); 
} 

无论我做什么,我要么只得到#0000ff,要么根本得不到任何工作。失败的尝试:

for (j = 0, jlen = newArray.length; jlen > j; j++) { 
    for (k = 0, km = colorArray.length; km > k; k++) { 
     newArray[j].color = colorArray[k]; 
    } 
} 

目标是要取回:

newArray = [ 
    { "name": "bob", "complete": "25", "color": "#ff0000" }, 
    { "name": "john", "complete": "50", "color": "#00ff00" }, 
    { "name": "mike", "complete": "75", "color": "#0000ff" }   
]; 

什么是做这种正确的方法是什么?

回答

4

只需根据colorArray的索引设置每个人的color密钥。

var mainArray = [ 
    { "name": "bob", "complete": "25" }, 
    { "name": "john", "complete": "50" }, 
    { "name": "mike", "complete": "75" },   
]; 
var colorArray = ["#ff0000", "#00ff00", "#0000ff"]; 
var newArray = []; 

for (var i = 0; i < mainArray.length; i++) { 
    newArray.push({ 
    name: mainArray[i].name, 
    complete: mainArray[i].complete, 
    color: colorArray[i] 
    }); 
} 
+0

这很有效,谢谢。我仍然不明白在这种情况下JavaScript如何通过索引来“循环”我的颜色数组。我应该在文档/参考文献中看到什么? – DusmaN

+1

当你使用像'i'这样的索引的循环时,不要认为它是循环遍历数组,而是将它想象成循环遍历一系列值。在这种情况下,通过循环第一次,'i'为0,然后1,然后2,然后停止,因为比较'3

+0

@FelixGuo很好的解释 – m0meni

0

使用Array.mapObject.assign功能更强大的方法看起来像

var mainArray = [ 
 
    { "name": "bob", "complete": "25" }, 
 
    { "name": "john", "complete": "50" }, 
 
    { "name": "mike", "complete": "75" },   
 
]; 
 
var colorArray = ["#ff0000", "#00ff00", "#0000ff"]; 
 

 
var newArray = mainArray.map((x, i) => 
 
    Object.assign({}, x, {color: colorArray[i]}) 
 
) 
 
    
 
console.log(newArray);

0

您有以下不必要的嵌套循环:

for (j = 0, jlen = newArray.length; jlen > j; j++) { 
    for (k = 0, km = colorArray.length; km > k; k++) { 
     newArray[j].color = colorArray[k]; 
    } 
} 

您可以将其更改为:

for (j = 0, jlen = newArray.length; jlen > j; j++) { 
    newArray[j].color = colorArray[j]; 
} 
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