展平嵌套 json 对象
flattening nested json object
[
{
"children": [
{
"children": [
{
"dateAdded": 1493033302670,
"id": "1534",
"index": 0,
"parentId": "1",
"title": "data1",
"url": "data2"
},
{
"children": [
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "data3",
"url": "data4"
}
],
"dateAdded": 1490363326576,
"dateGroupModified": 1490363326576,
"id": "1401",
"index": 1,
"parentId": "1",
"title": "daily"
},
{
"children": [
{
"dateAdded": 1481787664555,
"id": "1429",
"index": 0,
"parentId": "1407",
"title": "data56",
"url": "data"
},
{
"dateAdded": 1483365608504,
"id": "1430",
"index": 1,
"parentId": "1407",
"title": "data34",
"url": "data55"
}
]
}
]
}
]
}
]
这是 Chrome 书签数据的表示。
如果对象有url 属性,则表示这是一个书签。如果没有 url 属性 就是一个文件夹。
是树状结构
我想用名为 type 的附加 属性 创建展平对象。喜欢:
[
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "title",
"url": "some url",
"type": "bookmark"
},
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1402",
"title": "title2",
"url": "some url2"
"type": "folder"
}
]
提前致谢。
我创建了一个循环访问包含对象的数组的函数。如果给定的对象有一个名为 children
的 属性,该函数将调用自身。如果没有,则将其推送到新数组 flattenedBookmarks
.
解决方案
var flattenedBookmarks = [];
flattenBookmarks(bookmarks);
function flattenBookmarks(bookmarks) {
for (var i = 0; i < bookmarks.length; i++) {
var potentialBookmark = bookmarks[i];
if (potentialBookmark.hasOwnProperty("url")) {
potentialBookmark.type = "bookmark";
} else {
potentialBookmark.type = "folder";
}
if (potentialBookmark.hasOwnProperty("children")) {
flattenBookmarks(potentialBookmark.children);
if (potentialBookmark.hasOwnProperty("dateGroupModified")) {
flattenedBookmarks.push(potentialBookmark);
}
} else {
flattenedBookmarks.push(potentialBookmark);
}
}
}
您可能应该从函数返回展平的数组,而不是将其存储在新的全局数组中 flattenedBookmarks
,但至少这会让您入门。
您可以使用迭代和递归方法来获取平面数据。
function flatten(array) {
var result = [];
array.forEach(function iter(o) {
var temp = {},
keys = Object.keys(o);
if (keys.length > 1) {
keys.forEach(function (k) {
if (k !== 'children') {
temp[k] = o[k];
}
});
temp.type = 'url' in o ? 'bookmark' : 'folder';
result.push(temp);
}
Array.isArray(o.children) && o.children.forEach(iter);
});
return result;
}
var data = [{ children: [{ children: [{ dateAdded: 1493033302670, id: "1534", index: 0, parentId: "1", title: "data1", url: "data2" }, { children: [{ dateAdded: 1489571506844, id: "1451", index: 0, parentId: "1401", title: "data3", url: "data4" }], dateAdded: 1490363326576, dateGroupModified: 1490363326576, id: "1401", index: 1, parentId: "1", title: "daily" }, { children: [{ dateAdded: 1481787664555, id: "1429", index: 0, parentId: "1407", title: "data56", url: "data" }, { dateAdded: 1483365608504, id: "1430", index: 1, parentId: "1407", title: "data34", url: "data55" }] }] }] }];
console.log(flatten(data));
.as-console-wrapper { max-height: 100% !important; top: 0; }
示例显示了如何操作
data = [
{
"children": [
{
"children": [
{
"dateAdded": 1493033302670,
"id": "1534",
"index": 0,
"parentId": "1",
"title": "data1",
"url": "data2"
},
{
"children": [
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "data3",
"url": "data4"
}
],
"dateAdded": 1490363326576,
"dateGroupModified": 1490363326576,
"id": "1401",
"index": 1,
"parentId": "1",
"title": "daily"
},
{
"children": [
{
"dateAdded": 1481787664555,
"id": "1429",
"index": 0,
"parentId": "1407",
"title": "data56",
"url": "data"
},
{
"dateAdded": 1483365608504,
"id": "1430",
"index": 1,
"parentId": "1407",
"title": "data34",
"url": "data55"
}
]
}
]
}
]
}
];
data2 = [];
function search(data) {
for (n in data) {
if (typeof data[n] == 'object') {
if (data[n].id != undefined) {
if (data[n].url != undefined) {
data[n].type="folder";
} else {
data[n].type="bookmark";
}
data2.push(data[n]);
}
search(data[n]);
}
}
}
search(data);
console.log(data2);
[
{
"children": [
{
"children": [
{
"dateAdded": 1493033302670,
"id": "1534",
"index": 0,
"parentId": "1",
"title": "data1",
"url": "data2"
},
{
"children": [
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "data3",
"url": "data4"
}
],
"dateAdded": 1490363326576,
"dateGroupModified": 1490363326576,
"id": "1401",
"index": 1,
"parentId": "1",
"title": "daily"
},
{
"children": [
{
"dateAdded": 1481787664555,
"id": "1429",
"index": 0,
"parentId": "1407",
"title": "data56",
"url": "data"
},
{
"dateAdded": 1483365608504,
"id": "1430",
"index": 1,
"parentId": "1407",
"title": "data34",
"url": "data55"
}
]
}
]
}
]
}
]
这是 Chrome 书签数据的表示。
如果对象有url 属性,则表示这是一个书签。如果没有 url 属性 就是一个文件夹。
是树状结构
我想用名为 type 的附加 属性 创建展平对象。喜欢:
[
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "title",
"url": "some url",
"type": "bookmark"
},
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1402",
"title": "title2",
"url": "some url2"
"type": "folder"
}
]
提前致谢。
我创建了一个循环访问包含对象的数组的函数。如果给定的对象有一个名为 children
的 属性,该函数将调用自身。如果没有,则将其推送到新数组 flattenedBookmarks
.
解决方案
var flattenedBookmarks = [];
flattenBookmarks(bookmarks);
function flattenBookmarks(bookmarks) {
for (var i = 0; i < bookmarks.length; i++) {
var potentialBookmark = bookmarks[i];
if (potentialBookmark.hasOwnProperty("url")) {
potentialBookmark.type = "bookmark";
} else {
potentialBookmark.type = "folder";
}
if (potentialBookmark.hasOwnProperty("children")) {
flattenBookmarks(potentialBookmark.children);
if (potentialBookmark.hasOwnProperty("dateGroupModified")) {
flattenedBookmarks.push(potentialBookmark);
}
} else {
flattenedBookmarks.push(potentialBookmark);
}
}
}
您可能应该从函数返回展平的数组,而不是将其存储在新的全局数组中 flattenedBookmarks
,但至少这会让您入门。
您可以使用迭代和递归方法来获取平面数据。
function flatten(array) {
var result = [];
array.forEach(function iter(o) {
var temp = {},
keys = Object.keys(o);
if (keys.length > 1) {
keys.forEach(function (k) {
if (k !== 'children') {
temp[k] = o[k];
}
});
temp.type = 'url' in o ? 'bookmark' : 'folder';
result.push(temp);
}
Array.isArray(o.children) && o.children.forEach(iter);
});
return result;
}
var data = [{ children: [{ children: [{ dateAdded: 1493033302670, id: "1534", index: 0, parentId: "1", title: "data1", url: "data2" }, { children: [{ dateAdded: 1489571506844, id: "1451", index: 0, parentId: "1401", title: "data3", url: "data4" }], dateAdded: 1490363326576, dateGroupModified: 1490363326576, id: "1401", index: 1, parentId: "1", title: "daily" }, { children: [{ dateAdded: 1481787664555, id: "1429", index: 0, parentId: "1407", title: "data56", url: "data" }, { dateAdded: 1483365608504, id: "1430", index: 1, parentId: "1407", title: "data34", url: "data55" }] }] }] }];
console.log(flatten(data));
.as-console-wrapper { max-height: 100% !important; top: 0; }
示例显示了如何操作
data = [
{
"children": [
{
"children": [
{
"dateAdded": 1493033302670,
"id": "1534",
"index": 0,
"parentId": "1",
"title": "data1",
"url": "data2"
},
{
"children": [
{
"dateAdded": 1489571506844,
"id": "1451",
"index": 0,
"parentId": "1401",
"title": "data3",
"url": "data4"
}
],
"dateAdded": 1490363326576,
"dateGroupModified": 1490363326576,
"id": "1401",
"index": 1,
"parentId": "1",
"title": "daily"
},
{
"children": [
{
"dateAdded": 1481787664555,
"id": "1429",
"index": 0,
"parentId": "1407",
"title": "data56",
"url": "data"
},
{
"dateAdded": 1483365608504,
"id": "1430",
"index": 1,
"parentId": "1407",
"title": "data34",
"url": "data55"
}
]
}
]
}
]
}
];
data2 = [];
function search(data) {
for (n in data) {
if (typeof data[n] == 'object') {
if (data[n].id != undefined) {
if (data[n].url != undefined) {
data[n].type="folder";
} else {
data[n].type="bookmark";
}
data2.push(data[n]);
}
search(data[n]);
}
}
}
search(data);
console.log(data2);