递归JavaScript函数返回数组

Question

我有一个对象数组。 这些对象可能具有称为子代的属性，而该子代又将包含对象的数组。 它实际上是一棵对象树。

var my_array_of_objects=[
  {
    "object_info": "blah blah",
    "children": [
      {
        "object_info": "blah blah",
        "children": [
          {
            "object_info": "blah blah"
          }
        ]
      }
    ]
  },
  {
    "object_info": "blah blah"
  },
  {
    "object_info": "blah blah"
  }
]

我正在尝试递归遍历my_array_of_objects以返回一个数组，该数组将删除树结构，并为每个对象提供一个Relationship_id和parent_relationship_id，但是我在递归发生之前设置了Relationship_id时遇到了问题，我看不到为什么...我认为这是一个范围界定问题？

  function sort_relationships(objects, return_objects, parent_relationship_id, relationship_id){
      $.each( objects, function( field_key, object ){
        relationship_id++;
        var new_object={};
        new_object.relationship_id=relationship_id;
        new_object.parent_relationship_id=parent_relationship_id;
        new_object.object_info=object.object_info;
        // PUSH OBJECT TO THE ARRAY TO BE RETURNED
        return_objects.push(new_object);
        // IF THE OBJECT HAS CHILDREN THEN RUN THE FUNCTION AGAIN WITH IT'S OBJECTS
        if(object.children instanceof Array){
          sort_relationships(object.children,return_objects,new_object.relationship_id,relationship_id);
        }
      });
      return return_objects;
  }

var my_new_array = sort_relationships（my_array_of_objects，[]，0,0）;

基于上面的代码，我希望每个新对象都有一个唯一的Relationship_id，但是它将返回以下内容。

[
  {
    "relationship_id": 1,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 2,
    "parent_relationship_id": 1,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 3,
    "parent_relationship_id": 2,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 2,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 3,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  }
]

我希望有以下几点：

[
  {
    "relationship_id": 1,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 2,
    "parent_relationship_id": 1,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 3,
    "parent_relationship_id": 2,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 4,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  },
  {
    "relationship_id": 5,
    "parent_relationship_id": 0,
    "object_info": "blah blah"
  }
]

Answer 1

您正在对每个递归调用本地的变量执行“ relationship_id++ ”，并且未在调用方处反映出来。 您要么需要return更新后的值，要么仅使用一个自由变量：

function sort_relationships(objects) {
    var return_objects = [];
    var relationship_id = 0;
    function recurse(objects, parent_relationship_id) {
        $.each(objects, function(field_key, object) {
            relationship_id++;
            var new_object = {
                relationship_id: relationship_id,
                parent_relationship_id: parent_relationship_id,
                object_info: object.object_info
            };
            // PUSH OBJECT TO THE ARRAY TO BE RETURNED
            return_objects.push(new_object);
            // IF THE OBJECT HAS CHILDREN THEN RUN THE FUNCTION AGAIN WITH IT'S OBJECTS
            if (Array.isArray(object.children)) {
                recurse(object.children, new_object.relationship_id);
            }
        });
    }
    recurse(objects, 0);
    return return_objects;
}

var my_new_array=sort_relationships(my_array_of_objects);

Answer 2

您可以为此使用reduce ，对于id（出问题的地方），可以将已累积对象的数量添加到父id中：

 function flatten(arr, parentId = 0) { if (!arr) return []; return arr.reduce ( function (acc, obj) { var rel = { object_info: obj.object_info, relationship_id: parentId + acc.length + 1, parent_relationship_id: parentId }; return acc.concat([rel], flatten(obj.children, rel.relationship_id)); }, []); } // sample data var my_array_of_objects=[{ "object_info": "blah blah", "children": [{ "object_info": "blah blah", "children": [{ "object_info": "blah blah" } ] } ] }, { "object_info": "blah blah" }, { "object_info": "blah blah" } ]; // convert var result = flatten(my_array_of_objects); // output console.log(result);