从现有的 JSON 构建新的 JSON

Question

我想从现有的JSON 中构建一个新的JSON。来源有部分和量规，我不再需要列表。名为“items”的新对象应该有一个项目数组。

最终的 JSON 应该按属性 'name' 排序，看起来像

{
  "items": [
    {
      "id": 10000006,
      "name": "Boah"
    },
    {
      "id": 10000013,
      "name": "Gut"
    },
    {
      "id": 10000003,
      "name": "Ipsum"
    },
    {
      "id": 10000001,
      "name": "Lorem"
    },
    {
      "id": 10000005,
      "name": "Lorum"
    },
    {
      "id": 10000004,
      "name": "Name"
    },
    {
      "id": 10000002,
      "name": "Stet"
    }
  ]
}

为了构建新的 JSON，我得到了这个来源：

{
  "sections": [
    {
      "name": "FooBar",
      "rubrics": [
        {
          "name": "Foo",
          "items": [
            {
              "id": 10000001,
              "name": "Lorem"
            },
            {
              "id": 10000002,
              "name": "Stet"
            },
            {
              "id": 10000003,
              "name": "Ipsum"
            }
          ]
        },
        {
          "name": "Bar",
          "items": [
            {
              "id": 10000004,
              "name": "Name"
            },
            {
              "id": 10000005,
              "name": "Lorum"
            },
            {
              "id": 10000006,
              "name": "Boah"
            }
          ]
        }
      ]
    },
    {
      "name": "BlahBloob",
      "rubrics": [
        {
          "name": "Bla",
          "items": [
            {
              "id": 10000013,
              "name": "Gut"
            }
          ]
        },
        {
          "name": "Bloob",
          "items": [
            {
              "id": 10000014,
              "name": "Name"
            },
            {
              "id": 10000015,
              "name": "Lorem"
            }
          ]
        }
      ]
    }
  ]
}

你怎么看？我怎样才能用普通的JavaScript 或TypeScript 做到这一点？

感谢您阅读并有时间回答我的问题。并感谢您提前回复。

Answer 1

给你。您只需要遍历源的每个部分的每个量规即可获取项目。最后，按项目对项目列表进行排序，就完成了。

此示例使用 ES6 语法，但如果需要，可以轻松将其转换为 ES5。

function extractItems(source) {
  const items = [];

  for (const section of source.sections) {
    for (const rubric of section.rubrics) {
      items.push(...rubric.items);
    }
  }

  items.sort((a, b) => a.name.localeCompare(b.name));

  return { items };
}

Answer 2

更实用的方法是使用 ma​​p 和 reduce 来挑选和合并它们。

data.sections
  .map(section => section.rubrics) // get rubrics
  .reduce((a, b) => a.concat(b)) // merge rubrics
  .map(rubric => rubric.items) // get items from each rubric
  .reduce((a, b) => a.concat(b)) // merge items
  .sort((a, b) => a.name.localeCompare(b.name)); // sort

Answer 3

function(oldObj) {
    var newObj = {
        "items": []
    };

    oldObj.sections.forEach(function(section) {
        section.rubrics.forEach(function(rubric) {
            rubric.items.forEach(function(item) {
                newObj.items.push(item);
            });
        });
    });

    newObj.items = newObj.items.sort(function(a, b) {
        if (a.name < b.name) { return -1; }
        if (a.name > b.name) { return 1;  }
        return 0;
    });
    return newObj;
}

并且只需使用JSON.parse() 和JSON.stringify() 将JSON 转换为对象。

Answer 4

也许对你有帮助

var data ={
	"sections": [
{
"name": "FooBar",
"rubrics": [{"name": "Foo", "items": [{"id": 10000001,"name": "Lorem"}, {"id": 10000002,"name": "Stet"}, {"id": 10000003,"name": "Ipsum"}]
}, {
			"name": "Bar",
			"items": [{
				"id": 10000004,
				"name": "Name"
			}, {
				"id": 10000005,
				"name": "Lorum"
			}, {
				"id": 10000006,
				"name": "Boah"
			}]
		}]
	}, {
		"name": "BlahBloob",
		"rubrics": [{
			"name": "Bla",
			"items": [{
				"id": 10000013,
				"name": "Gut"
			}]
		}, {
			"name": "Bloob",
			"items": [{
				"id": 10000014,
				"name": "Name"
			}, {
				"id": 10000015,
				"name": "Lorem"
			}]
		}]
	}]
};
var itemObj = {};
var itemArr = [];
var sections = data.sections;
for(var i=0;i<sections.length;i++)
{	
  for(var j=0;j<sections[i].rubrics.length;j++){    
    for(var k=0;k<sections[i].rubrics[j].items.length;k++){      
      var itemObj;
      itemObj['id'] = sections[i].rubrics[j].items[k].id;
      itemObj['name'] = sections[i].rubrics[j].items[k].name;      
      itemArr.push(itemObj);
    }
  }
}
var finalObj = {"items":itemArr};
console.log(finalObj);

JSFiddle