假设我们只得到
var obj = {};
var propName = "foo.bar.foobar";
我们如何将属性设置obj.foo.bar.foobar为某个值（比如“hello world”）？所以我想实现这一点，而我们只有字符串中的属性名称：
obj.foo.bar.foobar = "hello world";
问问题
30661 次
15 回答
88
function assign(obj, prop, value) {
if (typeof prop === "string")
prop = prop.split(".");
if (prop.length > 1) {
var e = prop.shift();
assign(obj[e] =
Object.prototype.toString.call(obj[e]) === "[object Object]"
? obj[e]
: {},
prop,
value);
} else
obj[prop[0]] = value;
}
var obj = {},
propName = "foo.bar.foobar";
assign(obj, propName, "Value");
于 2012-12-05T09:21:36.877 回答
12
由于这个问题似乎是由不正确的答案来回答的，所以我只会参考类似问题的正确答案
function setDeepValue(obj, value, path) {
if (typeof path === "string") {
var path = path.split('.');
}
if(path.length > 1){
var p=path.shift();
if(obj[p]==null || typeof obj[p]!== 'object'){
obj[p] = {};
}
setDeepValue(obj[p], value, path);
}else{
obj[path[0]] = value;
}
}
采用：
var obj = {};
setDeepValue(obj, 'Hello World', 'foo.bar.foobar');
于 2012-12-05T09:21:15.793 回答
5
编辑：我创建了一个jsPerf.com 测试用例来将接受的答案与我的版本进行比较。事实证明，我的版本更快，尤其是当您深入时。
var nestedObjectAssignmentFor = function(obj, propString, value) {
var propNames = propString.split('.'),
propLength = propNames.length-1,
tmpObj = obj;
for (var i = 0; i <= propLength ; i++) {
tmpObj = tmpObj[propNames[i]] = i !== propLength ? {} : value;
}
return obj;
}
var obj = nestedObjectAssignment({},"foo.bar.foobar","hello world");
</p>
</p>
于 2012-12-05T09:30:18.573 回答
4
所有解决方案在设置时都会覆盖任何原始数据，因此我对以下内容进行了调整，也将其变成了一个对象：
var obj = {}
nestObject.set(obj, "a.b", "foo");
nestObject.get(obj, "a.b"); // returns foo
var nestedObject = {
set: function(obj, propString, value) {
var propNames = propString.split('.'),
propLength = propNames.length-1,
tmpObj = obj;
for (var i = 0; i <= propLength ; i++) {
if (i === propLength){
if(tmpObj[propNames[i]]){
tmpObj[propNames[i]] = value;
}else{
tmpObj[propNames[i]] = value;
}
}else{
if(tmpObj[propNames[i]]){
tmpObj = tmpObj[propNames[i]];
}else{
tmpObj = tmpObj[propNames[i]] = {};
}
}
}
return obj;
},
get: function(obj, propString){
var propNames = propString.split('.'),
propLength = propNames.length-1,
tmpObj = obj;
for (var i = 0; i <= propLength ; i++) {
if(tmpObj[propNames[i]]){
tmpObj = tmpObj[propNames[i]];
}else{
break;
}
}
return tmpObj;
}
};
还可以将函数更改为 Oject.prototype 方法，将 obj 参数更改为：
Object.prototype = { setNested = function(){ ... }, getNested = function(){ ... } }
{}.setNested('a.c','foo')
于 2014-03-13T15:27:48.503 回答
3
这是我刚刚从几个线程+一些自定义代码编译的获取和设置函数。
它还将创建现场不存在的键。
function setValue(object, path, value) {
var a = path.split('.');
var o = object;
for (var i = 0; i < a.length - 1; i++) {
var n = a[i];
if (n in o) {
o = o[n];
} else {
o[n] = {};
o = o[n];
}
}
o[a[a.length - 1]] = value;
}
function getValue(object, path) {
var o = object;
path = path.replace(/\[(\w+)\]/g, '.$1');
path = path.replace(/^\./, '');
var a = path.split('.');
while (a.length) {
var n = a.shift();
if (n in o) {
o = o[n];
} else {
return;
}
}
return o;
}
于 2013-11-27T10:50:37.967 回答
3
这是一个返回更新后的对象
function deepUpdate(value, path, tree, branch = tree) {
const last = path.length === 1;
branch[path[0]] = last ? value : branch[path[0]];
return last ? tree : deepUpdate(value, path.slice(1), tree, branch[path[0]]);
}
const path = 'cat.dog';
const updated = deepUpdate('a', path.split('.'), {cat: {dog: null}})
// => { cat: {dog: 'a'} }
于 2016-07-22T15:28:55.977 回答
3
这是一个使用参考的简单函数。
function setValueByPath (obj, path, value) {
var ref = obj;
path.split('.').forEach(function (key, index, arr) {
ref = ref[key] = index === arr.length - 1 ? value : {};
});
return obj;
}
于 2016-10-19T15:32:29.077 回答
3
您可以拆分路径并检查以下元素是否存在。如果没有将对象分配给新属性。
然后返回属性的值。
最后赋值。
function setValue(object, path, value) {
var fullPath = path.split('.'),
way = fullPath.slice(),
last = way.pop();
way.reduce(function (r, a) {
return r[a] = r[a] || {};
}, object)[last] = value;
}
var object = {},
propName = 'foo.bar.foobar',
value = 'hello world';
setValue(object, propName, value);
console.log(object);于 2017-03-30T12:14:50.270 回答
3
很直接的一个。
这个实现应该非常高效。它避免了递归和函数调用，同时保持了简单性。
/**
* Set the value of a deep property, creating new objects as necessary.
* @param {Object} obj The object to set the value on.
* @param {String|String[]} path The property to set.
* @param {*} value The value to set.
* @return {Object} The object at the end of the path.
* @author github.com/victornpb
* @see https://stackoverflow.com/a/46060952/938822
* @example
* setDeep(obj, 'foo.bar.baz', 'quux');
*/
function setDeep(obj, path, value) {
const props = typeof path === 'string' ? path.split('.') : path;
for (var i = 0, n = props.length - 1; i < n; ++i) {
obj = obj[props[i]] = obj[props[i]] || {};
}
obj[props[i]] = value;
return obj;
}
/*********************** EXAMPLE ***********************/
const obj = {
hello : 'world',
};
setDeep(obj, 'root', true);
setDeep(obj, 'foo.bar.baz', 1);
setDeep(obj, ['foo','quux'], '');
console.log(obj);
// ⬇︎ Click "Run" below to see output于 2017-09-05T18:07:44.597 回答
1
我正在寻找一个不会覆盖现有值并且易于阅读并且能够提出这个答案的答案。把这个留在这里，以防它帮助其他有同样需求的人
function setValueAtObjectPath(obj, pathString, newValue) {
// create an array (pathComponents) of the period-separated path components from pathString
var pathComponents = pathString.split('.');
// create a object (tmpObj) that references the memory of obj
var tmpObj = obj;
for (var i = 0; i < pathComponents.length; i++) {
// if not on the last path component, then set the tmpObj as the value at this pathComponent
if (i !== pathComponents.length-1) {
// set tmpObj[pathComponents[i]] equal to an object of it's own value
tmpObj[pathComponents[i]] = {...tmpObj[pathComponents[i]]}
// set tmpObj to reference tmpObj[pathComponents[i]]
tmpObj = tmpObj[pathComponents[i]]
// else (IS the last path component), then set the value at this pathComponent equal to newValue
} else {
// set tmpObj[pathComponents[i]] equal to newValue
tmpObj[pathComponents[i]] = newValue
}
}
// return your object
return obj
}
于 2018-10-19T08:29:02.683 回答
1
与 Rbar 的答案相同，在使用redux reducers时非常有用。我也使用 lodash clone 而不是扩展运算符来支持数组：
export function cloneAndPatch(obj, path, newValue, separator='.') {
let stack = Array.isArray(path) ? path : path.split(separator);
let newObj = _.clone(obj);
obj = newObj;
while (stack.length > 1) {
let property = stack.shift();
let sub = _.clone(obj[property]);
obj[property] = sub;
obj = sub;
}
obj[stack.shift()] = newValue;
return newObj;
}
于 2019-01-05T07:12:52.547 回答
1
Object.getPath = function(o, s) {
s = s.replace(/\[(\w+)\]/g, '.$1'); // convert indexes to properties
s = s.replace(/^\./, ''); // strip a leading dot
var a = s.split('.');
for (var i = 0, n = a.length; i < n; ++i) {
var k = a[i];
if (k in o) {
o = o[k];
} else {
return;
}
}
return o;
};
Object.setPath = function(o, p, v) {
var a = p.split('.');
var o = o;
for (var i = 0; i < a.length - 1; i++) {
if (a[i].indexOf('[') === -1) {
var n = a[i];
if (n in o) {
o = o[n];
} else {
o[n] = {};
o = o[n];
}
} else {
// Not totaly optimised
var ix = a[i].match(/\[.*?\]/g)[0];
var n = a[i].replace(ix, '');
o = o[n][ix.substr(1,ix.length-2)]
}
}
if (a[a.length - 1].indexOf('[') === -1) {
o[a[a.length - 1]] = v;
} else {
var ix = a[a.length - 1].match(/\[.*?\]/g)[0];
var n = a[a.length - 1].replace(ix, '');
o[n][ix.substr(1,ix.length-2)] = v;
}
};
于 2019-11-17T10:39:44.113 回答
1
这是一个简单的方法，它使用一个范围Object递归地设置路径的正确道具。
function setObjectValueByPath(pathScope, value, obj) {
const pathStrings = pathScope.split('/');
obj[pathStrings[0]] = pathStrings.length > 1 ?
setObjectValueByPath(
pathStrings.splice(1, pathStrings.length).join('/'),
value,
obj[pathStrings[0]]
) :
value;
return obj;
}
于 2020-03-13T12:03:12.780 回答
0
一个简单而简短的怎么样？
Object.assign(this.origin, { [propName]: value })
于 2022-01-18T17:12:58.520 回答