Question

这里我附上了我的json。

"mainSteps": [
  {
    "id": "9b3b64b4-d8a5-46d5-b464-066dc5c45dc3",
    "name": "Main Step 1",
    "steps": [
      {
        "name": "sub step 1.1"
      },
      {
        "name": "sub step 1.2"
      }
    ]
  },
  {
    "name": "Main step 2"
    "steps": [
      {
        "name": "sub step 2.1"
      },
      {
      "name": "sub step 2.2"
      }
    ],
  },
  {
    "name": "Main Step 3",
    "steps": [
      {
        "name": "sub step 3.1"
      },
      {
        "name": "sub step 3.2"
      }
    ],
  }
]

我正在寻找像 - ＆gt;这样的输出[主要步骤1，子步骤1.1，子步骤1.2]，[主要步骤2，子步骤2.1，子步骤2.2]，[主要步骤3，子步骤3.1，子步骤3.2]。我花了一整天的时间来输出这个输出，但得到的输出像[[主要步骤1，主要步骤2，主要步骤3，子步骤1.1，子步骤1.2 ....]就像我得到不同的格式但我无法得到我提到的实际输出，有人可以澄清我。

	var dataProcess = {
        				parentProcess:[],
        				subProcess:[]
        		                   };   		                   
   var steps = mainData.steps; // Steps Having the Full json data                   
        		var proc = [];
                 $scope.getSteps = function(steps) {
                    for (var i=0;i< steps.length;i++) {
                	       dataProcess.parentProcess.push(steps[i].name);
                	     for(var j=i;j<steps[i].steps.length;j++){
                		  dataProcess.subProcess.push(steps[i].steps[j].name);
                	   }
                   }

这是我试过的方式之一，

Answer 1

如果您需要ES5语法：

var details = mainSteps.map(function(step) {
  return [ step.name ].concat((step.steps || []).map(function(substep){
    return substep.name;
  })
});

ES6语法：

var details = mainSteps.map(step =< [step.name].concat((step.steps || []).map(sub => sub.name));

如果你需要比一层深度更多的递归，你可以使用一个函数作为顶级映射器，它自己调用。

Answer 2

可能你可以这样做;

＆＃13;

var mainSteps = [
  {
    "id": "9b3b64b4-d8a5-46d5-b464-066dc5c45dc3",
    "name": "Main Step 1",
    "steps": [
      {
        "name": "sub step 1.1"
      },
      {
        "name": "sub step 1.2"
      }
    ]
  },
  {
    "name": "Main step 2",
    "steps": [
      {
        "name": "sub step 2.1"
      },
      {
      "name": "sub step 2.2"
      }
    ],
  },
  {
    "name": "Main Step 3",
    "steps": [
      {
        "name": "sub step 3.1"
      },
      {
        "name": "sub step 3.2"
      }
    ],
  }
],
mapped = mainSteps.map(e => [e.name, e.steps[0].name, e.steps[1].name]);
console.log(mapped);

＆＃13;

Answer 3

使用Array.map和Array.concat函数的简单解决方案：

// supposing "obj" is your initial object
var dataProcess = obj.mainSteps.map(function (o) {
    return [o.name].concat(o.steps.map(function(v){ return v.name; }));
});

console.log(JSON.stringify(dataProcess, 0, 4));

输出：

[
    [
        "Main Step 1",
        "sub step 1.1",
        "sub step 1.2"
    ],
    [
        "Main step 2",
        "sub step 2.1",
        "sub step 2.2"
    ],
    [
        "Main Step 3",
        "sub step 3.1",
        "sub step 3.2"
    ]
]

DEMO link

Answer 4

这样，它就可以使用不同长度的数组和子数组：

var results = mainSteps.map(x => [x.name].concat(x.steps.map(y => y.name)));

如何将两个不同的数组对象放在一起

4 个答案: