Question

我想基于另一个JavaScript数组在JavaScript中对数组进行排序。

First arr = ["Name", "Address1", "Address2", "City", "State", "Zip Code", "Formatted Address"];

Second arr = [Name: "kd", State: "test", Address1: "tt"]

现在我想按照第一个arr的顺序对第二个arr进行排序。我想要这种结果：

[Name: "kd", Address1: "tt", State: "test"]

Answer 1

您可以通过对第一个数组进行操作来对第二个数组进行排序，如下所示：

Firstarr = ["Name", "Address1", "Address2", "City", "State", "Zip Code", "Formatted Address"];
Secondarr = {Name: "kd", State: "test", Address1: "tt"};

var finalArr = {};

$.each(Firstarr, function(i, value){
      if(typeof Secondarr[value] !== "undefined"){
         finalArr[value] =  Secondarr[value];
      }
});

Answer 2

您必须将JSON格式用于第二个数组。

var reference = ["Name", "Address1", "Address2", "City", "State", "Zip Code", "Formatted Address"];
var toSort = { "Name": "kd","State": "test", "Address1": "tt"};
var sorted={};
for(var i=0;i<reference.length;i++)
{
if(toSort.hasOwnProperty(reference[i]))
  {
    sorted[reference[i]]=toSort[reference[i]];
  }
}
console.log(sorted);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 3

这是一个没有jQuery的js中的简单算法。 orderArray [i]可以被称为键，js中的关联数组实际上是对象，但这只是语义。

＆＃13;

:-use_module(library(clpfd)).

zebra_prob:-

% left ------> right
House=[Color1,Color2,Color3,Color4,Color5],
Pet=[Pet1,Pet2,Pet3,Pet4,Pet5],
Race=[Race1,Race2,Race3,Race4,Race5],
Drink=[Bev1,Bev2,Bev3,Bev4,Bev5],
Smoke=[Cig1,Cig2,Cig3,Cig4,Cig5],

% Race 1:English 2:Spaniard 3:Ukrainian 4:Norwegian 5:Japanese
% HouseColor 1:red 2:green 3:ivory 4:yellow 5:blue
% Pet 1:dog 2:snails 3:fox 4:horse 5:zebra
% Drink 1:coffee 2:tea 3:milk 4:orange juice 5:water
% Smoke 1:Old Gold 2:Kool 3:Chesterfields 4:Lucky Strike 5:Parliaments

all_different(House), %house
all_different(Pet), %pet
all_different(Race), %country
all_different(Drink), %drink
all_different(Smoke), %smoke

House ins 1..5,
Pet ins 1..5,
Race ins 1..5,
Drink ins 1..5,
Smoke ins 1..5,

% The Englishman(1) lives in the red house(1).
Race1 #= 1 #<==> Color1 #= 1,
Race2 #= 1 #<==> Color2 #= 1,
Race3 #= 1 #<==> Color3 #= 1,
Race4 #= 1 #<==> Color4 #= 1,
Race5 #= 1 #<==> Color5 #= 1,

% The Spaniard(2) owns the dog(1).
Race1 #= 2 #<==> Pet1 #= 1,
Race2 #= 2 #<==> Pet2 #= 1,
Race3 #= 2 #<==> Pet3 #= 1,
Race4 #= 2 #<==> Pet4 #= 1,
Race5 #= 2 #<==> Pet5 #= 1,

% Coffee(1) is drunk in the green house(2).
Bev1 #= 1 #<==> Color1 #= 2,
Bev2 #= 1 #<==> Color2 #= 2,
Bev3 #= 1 #<==> Color3 #= 2,
Bev4 #= 1 #<==> Color4 #= 2,
Bev5 #= 1 #<==> Color5 #= 2,

% The Ukrainian(3) drinks tea(2).
Race1 #= 3 #<==> Bev1 #= 2,
Race2 #= 3 #<==> Bev2 #= 2,
Race3 #= 3 #<==> Bev3 #= 2,
Race4 #= 3 #<==> Bev4 #= 2,
Race5 #= 3 #<==> Bev5 #= 2,

% The green house(2) is immediately to the right of the ivory house(3).
Color1 #= 3 #<==> Color2 #= 2 ,
Color2 #= 3 #<==> Color3 #= 2 ,
Color3 #= 3 #<==> Color4 #= 2 ,
Color4 #= 3 #<==> Color5 #= 2 ,

% So, green house(2) is not leftmost.
Color1 #\= 2,

% The Old Gold(1) smoker owns snails(2).
Cig1 #= 1 #<==> Pet1 #= 2,
Cig2 #= 1 #<==> Pet2 #= 2,
Cig3 #= 1 #<==> Pet3 #= 2,
Cig4 #= 1 #<==> Pet4 #= 2,
Cig5 #= 1 #<==> Pet5 #= 2,

% Kools(2) are smoked in the yellow(4) house.
Cig1 #= 2 #<==> Color1 #= 4,
Cig2 #= 2 #<==> Color2 #= 4,
Cig3 #= 2 #<==> Color3 #= 4,
Cig4 #= 2 #<==> Color4 #= 4,
Cig5 #= 2 #<==> Color5 #= 4,

% Milk(3) is drunk in the middle house.
Bev3 #= 3,

% The Norwegian(4) lives in the first house. (I assume that "first" means leftmost)
Race1 #= 4,

% The man who smokes Chesterfields(3) lives in the house next to the man with the fox(3).
(Cig1 #= 3 #/\ Pet2 #= 3) #\/
(Cig2 #= 3 #/\ Pet3 #= 3) #\/
(Cig3 #= 3 #/\ Pet4 #= 3) #\/
(Cig4 #= 3 #/\ Pet5 #= 3) #\/
(Pet1 #= 3 #/\ Cig2 #= 3) #\/
(Pet2 #= 3 #/\ Cig3 #= 3) #\/
(Pet3 #= 3 #/\ Cig4 #= 3) #\/
(Pet4 #= 3 #/\ Cig5 #= 3) ,

% Kools(2) are smoked in the house next to the house where the horse(4) is kept.
(Cig1 #= 2 #/\ Pet2 #= 4) #\/
(Cig2 #= 2 #/\ Pet3 #= 4) #\/
(Cig3 #= 2 #/\ Pet4 #= 4) #\/
(Cig4 #= 2 #/\ Pet5 #= 4) #\/
(Pet1 #= 4 #/\ Cig2 #= 2) #\/
(Pet2 #= 4 #/\ Cig3 #= 2) #\/
(Pet3 #= 4 #/\ Cig4 #= 2) #\/
(Pet4 #= 4 #/\ Cig5 #= 2) ,

% The Lucky Strike(4) smoker drinks orange juice(4).
Cig1 #= 4 #<==> Bev1 #= 4,
Cig2 #= 4 #<==> Bev2 #= 4,
Cig3 #= 4 #<==> Bev3 #= 4,
Cig4 #= 4 #<==> Bev4 #= 4,
Cig5 #= 4 #<==> Bev5 #= 4,

% The Japanese(5) smokes Parliaments(5).
Race1 #= 5 #<==> Cig1 #= 5,
Race2 #= 5 #<==> Cig2 #= 5,
Race3 #= 5 #<==> Cig3 #= 5,
Race4 #= 5 #<==> Cig4 #= 5,
Race5 #= 5 #<==> Cig5 #= 5,

% The Norwegian(4) lives next to the blue house(5).
(Race1 #= 4 #/\ Color2 #= 5) #\/
(Race2 #= 4 #/\ Color3 #= 5) #\/
(Race3 #= 4 #/\ Color4 #= 5) #\/
(Race4 #= 4 #/\ Color5 #= 5) #\/
(Color1 #= 5 #/\ Race2 #= 4) #\/
(Color2 #= 5 #/\ Race3 #= 4) #\/
(Color3 #= 5 #/\ Race4 #= 4) #\/
(Color4 #= 5 #/\ Race5 #= 4) ,

label(House),
label(Pet),
label(Race),
label(Smoke),
label(Drink),

write('house:'),write(House),nl,
write('pet:'),write(Pet),nl,
write('race:'),write(Race),nl,
write('drink:'),write(Drink),nl,
write('smoke:'),write(Smoke),nl,nl.
　

[3] 23 ?- zebra_prob.
house:[4,5,1,3,2]
pet:[3,4,2,1,5]
race:[4,3,1,2,5]
drink:[5,2,3,4,1]
smoke:[2,3,1,4,5]

true ;
false.



% Race  1:English 2:Spaniard 3:Ukrainian 4:Norwegian 5:Japanese
% HouseColor 1:red 2:green 3:ivory 4:yellow 5:blue
% Pet 1:dog 2:snails 3:fox 4:horse 5:zebra
% Drink 1:coffee 2:tea 3:milk 4:orange juice 5:water
% Smoke 1:Old Gold 2:Kool 3:Chesterfields 4:Lucky Strike 5:Parliaments

Now, who drinks water? Who owns the zebra?
5:water is drunk by 4:Norwegian
5:zebra is tamed by 5:Japanese

＆＃13;

Answer 4

正如我已经评论过的那样，第二个看起来更像是一个Object，而不是一个数组。它必须初始化为var second = {Name：“kd”，State：“test”，Address1：“tt”};

您的最终输出将位于一个Array（）中，该Array包含在名为finlObj的Object中。请找到以下代码。您可以优化代码。但是为了你的理解，我做了一个精心设计的方式：

 var second = {Name: "kd", State: "test", Address1: "tt"};
     var first = ["Name", "Address1", "Address2", "City", "State", "Zip Code", "Formatted Address"];
     var finlObj ={specList:[]};

     for(var elem in first){
           if(second[first[elem]] !== undefined && second[first[elem]] !== null){


     finlObj.specList[finlObj.specList.length] = second[first[elem]];
     }
     }

     for(var el in finlObj.specList){
     alert(finlObj.specList[el]);

     }

Answer 5

这不会奏效。在JavaScript中，由于数组是一个对象，它将接受非整数键，如您在此处所示，但是运行时或提供的使用数组的方法都不允许您对其顺序进行任何控制。这完全取决于解释器的实现。

同样，JavaScript对象是不可移植的，虽然它们通常在调试工具中显示时按字母顺序显示，但这通常是浏览器提供的便利。

您需要的是Map。您可以使用此处显示的表示对其进行排序。

如何基于另一个JavaScript数组在JavaScript中对数组进行排序？

5 个答案: