如何显示JSON对象的ID和子代？

Question

标题听起来不太清楚，但是我试图在ID s的警报中显示树的JSON ，如果ID “有”孩子，我将添加孩子一词，当“组”完成后，我将添加| 表示小组结束了，问题是我找不到添加方法| 当小组结束时。

因此，这棵树：

[
  {
    "name": "node1",
    "id": 1,
    "is_open" :true,
    "children": [
      {
        "name": "child2",
        "id": 3
      },
      {
        "name": "child2",
        "id": 7
      },
      {
        "name": "child1",
        "id": 2
      }
    ]
  },
  {
    "name": "node1",
    "id": 4,
    "is_open": true,
    "children": [
      {
        "name": "child2",
        "id": 5
      }
    ]
  }
]

将显示为：

1，children，3,7,2 | 4，children，5 |

因此，我知道ID:1有children ，他们是ID:3 ， ID:7和ID:2 ，由于| ，该组结束了| 管道符号等，但我不知道如何创建要检查的内容并添加| 。 你能帮助我吗？

提前致谢。

我的代码：

 var data = '[{"name":"node1","id":1,"is_open":true,"children":[{"name":"child2","id":3},{"name":"child2","id":7},{"name":"child1","id":2}]},{"name":"node1","id":4,"is_open":true,"children":[{"name":"child2","id":5}]}]'; var arrays = []; //convert you string to object //or you can simply change your first row. data = $.parseJSON(data); //function to loop the array function listNodes(inputVal) { if (jQuery.isArray(inputVal)) { $.each(inputVal, function(i, elem) { arrays.push(elem.id); console.log(elem); if (jQuery.isArray(elem.children)) { arrays.push('children'); listNodes(elem.children); } }); } } listNodes(data); alert(arrays); 

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

更新：

如果我有两个或两个以上的孩子水平怎么办？

var data = JSON.parse('[{"name":"node1","id":1,"is_open":true,"children":[{"name":"child2","id":7},{"name":"child1","id":2,"is_open":true,"children":[{"name":"child2","id":3}]}]},{"name":"node1","id":4,"is_open":true,"children":[{"name":"child2","id":5}]}]');

var result = data.map(function(item) {
    return [item.id].concat('children', item.children.map(function(subitem) {
        return subitem.id;
    }));
}).join('|');

alert(result);

Answer 1

您不需要jQuery。 对于脚本的简化版本，您可以为此使用带有concat的本机Array.prototype.map ：

 var data = JSON.parse('[{"name":"node1","id":1,"is_open":true,"children":[{"name":"child2","id":3},{"name":"child2","id":7},{"name":"child1","id":2}]},{"name":"node1","id":4,"is_open":true,"children":[{"name":"child2","id":5}]}]'); var result = data.map(function(item) { return [item.id].concat('children', item.children.map(function(subitem) { return subitem.id; })); }).join('|'); alert(result); 

或者，如果您想看看自己的代码在修复后的样子：

function listNodes(inputVal) {
    if (jQuery.isArray(inputVal)) {
        $.each(inputVal, function (i, elem) {
            arrays.push(elem.id, ',');
            if (jQuery.isArray(elem.children)) {
                arrays.push('children,', elem.children.map(function(el) { return el.id; }));
            }
            arrays.push('|');
        });
    }
    arrays = arrays.join('');
}

Answer 2

尝试这个：

 var data = [{"name":"node1","id":1,"is_open":true,"children":[{"name":"child2","id":3},{"name":"child2","id":7},{"name":"child1","id":2}]},{"name":"node1","id":4,"is_open":true,"children":[{"name":"child2","id":5}]}]; var str = ''; //1,children,3,7,2|4,children,5| $.each(data, function(i, item){ if(item.id){ str += item.id + ',children,' $.each(item.children, function(j, child){ if(child.id){ str += child.id; str += (j >= item.children.length-1) ? "|" : ","; } }); } }); alert(str); 

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

Answer 3

我已经为任何类型的json对象编写了一个递归函数，以递归方式打印..在Java中..您可以对其进行调整以使用它。

static void printRecursive(JSONObject obj) {   
for(Object key:obj.keySet()) {
       //System.out.println(obj.get(key.toString()).getClass().getSimpleName());
        if(obj.get(key.toString()) instanceof JSONArray) {
            JSONArray aobj = ((JSONArray)obj.get(key.toString())); 

            for(int i=0;i<aobj.length();i++) {
                System.out.println(key.toString());
                printRecursive(aobj.getJSONObject(i));
            }
        }
        else
        if(obj.get(key.toString()) instanceof JSONObject) {
            System.out.println(key.toString());
            printRecursive((JSONObject)obj.get(key.toString()));
        }
        else
            System.out.println(key.toString()+" -> "+obj.get(key.toString()));
    }
}