Question

我从ajax电话回来了这个json：

{"datasets":[
    {"data":[0,0,2,0,1,0,0,0,0,0,5,2], "service_ID": 1},
    {"data":[0,0,0,0,0,0,0,0,0,0,1,3], "service_ID": 2},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,2], "service_ID": 3},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,4], "service_ID": 4},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,1], "service_ID": 5}
   ]
}

data对象中的每个元素都是一月份的值，从1月到12月（12个元素的数组）

我需要在datasets内循环以获取data对象中的每个值并得到总和，在此示例中为：2+1+5+2+1+3+2+4+1 = 21

我已经尝试了

var total = 0;

          $.each(datasets, function(index, data) {
                $.each(data, function(index, month) {
                    $.each(month, function(index, value) {
                        total = total + value;
                    });
                });
          });

但我收到TypeError: invalid 'in' operand a错误。

获得我需要的最佳方法是什么？

Answer 1

使用reduce

obj.datasets.reduce( function( a, b ){  //
   return a + b.data.reduce( (c, d) => c + d , 0); // 0 is the accumulator for data property of each item in the array 
} , 0); //0 is the accumulator which is the final output

甚至更短

obj.datasets.reduce( ( a, b ) => a + b.data.reduce( ( c, d ) => c + d , 0)  , 0 );

<强>演示

var obj = {"datasets":[
    {"data":[0,0,2,0,1,0,0,0,0,0,5,2], "service_ID": 1},
    {"data":[0,0,0,0,0,0,0,0,0,0,1,3], "service_ID": 2},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,2], "service_ID": 3},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,4], "service_ID": 4},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,1], "service_ID": 5}
   ]
};

var output = obj.datasets.reduce( function( a, b ){ 
   return a + b.data.reduce( (c, d) => c + d , 0);
} , 0);

console.log( output );

Answer 2

以下是对reduce的快速解释。它的名字就是这里的线索，因为它基本上采用了一系列单独的值（一个数组）并将它们缩减为单个值。在数组中得到整数之和实际上是一个非常好的例子。

reduce接受一个接受两个参数的回调 - 第一个是累加器，所有其他值都添加到的值，第二个是迭代中的当前值。这里我们还将累加器的初始值设置为0。

＆＃13;

const arr = [1, 2, 3, 4];
const out = arr.reduce((acc, curr) => acc + curr, 0);
console.log(out); // 10

＆＃13;

在gurvinder的回答中（我已经更改了变量名称以使其更容易理解）所发生的一切都是发生了几次求和操作。简而言之，我们用reduce迭代数据集数组并获取每个数据数组中的整数之和（再次使用reduce），并将每个arrSum添加到主累加器中，sum。

obj.datasets.reduce((sum, obj) => {
   return sum + obj.data.reduce((arrSum, int) => arrSum + int , 0);
}, 0);

但还有更多！

这是一个简单的函数，它返回两个数字的总和：

＆＃13;

const sum = (x, y) => x + y;
console.log(sum(1, 2)); // 3

＆＃13;

现在，因为JavaScript是一种函数式语言（函数是第一类对象），我们可以使用该函数来分离reducer的和函数。

＆＃13;

const arr = [1, 2, 3, 4];
const sum = (acc, curr) => acc + curr;
const out = arr.reduce(sum, 0);
console.log(out);

＆＃13;

这有助于我们编写更多功能代码，并且通常允许更好的代码重用。

例如，这里重写了gurvinders代码：

const sum = (sum, value) => sum + value;

const out = data.datasets.reduce((total, obj) => { 
  return sum(total, obj.data.reduce(sum), 0);
}, 0);

希望这是对reduce的有用介绍。

Answer 3

试试这段代码：

var dataset = {"datasets":[
    {"data":[0,0,2,0,1,0,0,0,0,0,5,2], "service_ID": 1},
    {"data":[0,0,0,0,0,0,0,0,0,0,1,3], "service_ID": 2},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,2], "service_ID": 3},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,4], "service_ID": 4},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,1], "service_ID": 5}
   ]
}

var total = 0;

$.each(dataset['datasets'], function(index1, data) {
  $.each(data['data'], function(index2, value) {

    total = total + value;

  });
});

console.log(total)

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

Answer 4

var example = {"datasets":[
    {"data":[0,0,2,0,1,0,0,0,0,0,5,2], "service_ID": 1},
    {"data":[0,0,0,0,0,0,0,0,0,0,1,3], "service_ID": 2},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,2], "service_ID": 3},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,4], "service_ID": 4},
    {"data":[0,0,0,0,0,0,0,0,0,0,0,1], "service_ID": 5}
   ]
}

var sum = 0;
example.datasets.forEach(set => {
    set.data.forEach(data => {
        sum += data;        
  })
})
console.log(sum) // 21

在javascript或jquery中迭代多级对象

4 个答案: