我正在嘗試拼合一個對象，其中的鍵將是葉節點的完整路徑。我可以遞歸地識別哪些是葉節點，但是試圖構建整個路徑。json對象的完整路徑
樣品輸入：
{
one: 1,
two: {
three: 3
},
four: {
five: 5,
six: {
seven: 7
},
eight: 8
},
nine: 9
}
輸出：
{
one: 1,
'two.three': 3,
'four.five': 5,
'four.six.seven': 7,
'four.eight': 8,
nine: 9
}
你可以使用遞歸計算策略，並收集對象的鑰匙。該建議也查找數組。
function getFlatObject(object) {
function iter(o, p) {
if (Array.isArray(o)){
o.forEach(function (a, i) {
iter(a, p.concat(i));
});
return;
}
if (o !== null && typeof o === 'object') {
Object.keys(o).forEach(function (k) {
iter(o[k], p.concat(k));
});
return;
}
path[p.join('.')] = o;
}
var path = {};
iter(object, []);
return path;
}
var obj = { one: 1, two: { three: 3 }, four: { five: 5, six: { seven: 7 }, eight: 8 }, nine: 9 },
path = getFlatObject(obj);
\t
console.log(path);使用最新的JS功能，如對象傳播和Object.entries它應該是很容易的：
function flatObj(obj, path = []) {
let output = {};
Object.entries(obj).forEach(([ key, value ]) => {
const nextPath = [ ...path, key ];
if (typeof value !== 'object') {
output[nextPath.join('.')] = value;
return;
}
output = {
...output,
...flatObj(value, nextPath)
};
});
}
請注意，此代碼可能不是最優的一個，因爲它拷貝每次我們想要合併它的對象。更多地將它看作是它看起來像什麼的主旨，而不是一個完整的最終解決方案。
部分解決： 給輸入的完整路徑的功能，它爲您提供了相應的輸出
var obj = {
one: 1,
two: {
three: 3
},
four: {
five: 5,
six: {
seven: 7
},
eight: 8
},
nine: 9
};
function deepFind(obj, path) {
var paths = path.split('.')
, current = obj
, i;
for (i = 0; i < paths.length; ++i) {
if (current[paths[i]] == undefined) {
return undefined;
} else {
current = current[paths[i]];
}
}
return current;
}
console.log(deepFind(obj, 'four.six.seven'))
var obj = {
one: 1,
two: {
three: 3
},
four: {
five: 5,
six: {
seven: 7
},
eight: 8
},
nine: 9
};
function flatten(obj) {
var flatObj = {}
function makeFlat(obj, path) {
var keys = Object.keys(obj);
if (keys.length) {
keys.forEach(function (key) {
makeFlat(obj[key], (path ? path + "." : path) + key);
})
} else {
flatObj[path] = obj;
}
}
makeFlat(obj, "");
return flatObj;
}
console.log(flatten(obj));非花哨做法，內部使用遞歸。
var x = { one:1,two:{three:3},four:{five: 5,six:{seven:7},eight:8},nine:9};
var res = {};
var constructResultCurry = function(src){ return constructResult(res,src); }
function constructResult(target, src) {
if(!src) return;
target[src.key] = src.val;
}
function buildPath(key, obj, overAllKey) {
overAllKey += (overAllKey ? "." : "") + key;
if(typeof obj[key] != "object") return { key : overAllKey, val : obj[key] };
Object.keys(obj[key]).forEach(function(keyInner) {
constructResultCurry(buildPath(keyInner, obj[key], overAllKey));
});
}
Object.keys(x).forEach(function(k){
constructResultCurry(buildPath(k, x, ""));
});
console.log(res);你可以簡單地做如下;
var obj = {one: 1, two: {three: 3}, four: {five: 5, six: {seven: 7}, eight: 8}, nine: 9},
flatObj = (o,p="") => { return Object.keys(o)
.map(k => o[k] === null ||
typeof o[k] !== "object" ? {[p + (p ? ".":"") + k]:o[k]}
: flatObj(o[k],p + (p ? ".":"") + k))
.reduce((p,c) => Object.assign(p,c));
};
console.log(flatObj(obj));
你可以看到這樣的回答：http://stackoverflow.com/questions/19098797/fastest-way-to-flatten-un-flatten-nested-json-objects 希望它可以幫你。 –
你爲什麼要這樣？ – Victor
@Victor我正在做api調用，如api.moviestore.com/movies?where[movie.name:eq]=gravity。我需要扁平化一個對象來構造過濾器查詢。 – Sayem