如何删除JavaScript对象中未定义或空的所有属性?
(这个问题与数组的问题类似)
如何删除JavaScript对象中未定义或空的所有属性?
(这个问题与数组的问题类似)
当前回答
TypeScript的泛型函数
function cleanProps(object:Record<string, string>):Record<string, string> {
let cleanObj = {};
Object.keys(object).forEach((key) => {
const property = object[key];
cleanObj = property ? { ...cleanObj, [key]: property } : cleanObj;
});
return cleanObj;
}
export default cleanProps;
现在假设你有一个像下面这样的对象
interface Filters{
searchString: string;
location: string;
sector: string
}
const filters:Filters = {
searchString: 'cute cats',
location: '',
sector: 'education',
};
您可以按照如下方式使用该函数
const result = cleanProps(filters as Record<keyof Filters, string>);
console.log(result); // outputs: { searchString: 'cute cats', sector: 'education' }
其他回答
如果你想要4行纯ES7解决方案:
const clean = e => e instanceof Object ? Object.entries(e).reduce((o, [k, v]) => {
if (typeof v === 'boolean' || v) o[k] = clean(v);
return o;
}, e instanceof Array ? [] : {}) : e;
或者如果你喜欢更易读的版本:
function filterEmpty(obj, [key, val]) {
if (typeof val === 'boolean' || val) {
obj[key] = clean(val)
};
return obj;
}
function clean(entry) {
if (entry instanceof Object) {
const type = entry instanceof Array ? [] : {};
const entries = Object.entries(entry);
return entries.reduce(filterEmpty, type);
}
return entry;
}
这将保留布尔值,也将清理数组。它还通过返回一个清理过的副本来保存原始对象。
这可以使用递归来解决。JavaScript对象可以是一个数组,也可以有一个包含空值的数组作为值。
function removeNullValues(obj) {
// Check weather obj is an array
if (Array.isArray(obj)) {
// Creating copy of obj so that index is maintained after splice
obj.slice(0).forEach((val) => {
if (val === null) {
obj.splice(obj.indexOf(val), 1);
} else if (typeof val === 'object') {
// Check if array has an object
removeNullValues(val);
}
});
} else if (typeof obj === 'object') {
// Check for object
Object.keys(obj).forEach((key) => {
if (obj[key] === null) {
delete obj[key];
} else if (typeof obj[key] === 'object') {
removeNullValues(obj[key]);
}
});
}
return obj;
}
您可能正在寻找delete关键字。
var obj = { };
obj.theProperty = 1;
delete obj.theProperty;
ES6 arrow function and ternary operator:
Object.entries(obj).reduce((acc, entry) => {
const [key, value] = entry
if (value !== undefined) acc[key] = value;
return acc;
}, {})
const obj = {test:undefined, test1:1 ,test12:0, test123:false};
const newObj = Object.entries(obj).reduce((acc, entry) => {
const [key, value] = entry
if (value !== undefined) acc[key] = value;
return acc;
}, {})
console.log(newObj)
ES6+的最短一行
过滤所有错误值("",0,false, null, undefined)
Object.entries(obj).reduce((a,[k,v]) => (v ? (a[k]=v, a) : a), {})
过滤空值和未定义值:
Object.entries(obj).reduce((a,[k,v]) => (v == null ? a : (a[k]=v, a)), {})
仅过滤null
Object.entries(obj).reduce((a,[k,v]) => (v === null ? a : (a[k]=v, a)), {})
过滤器仅未定义
Object.entries(obj).reduce((a,[k,v]) => (v === undefined ? a : (a[k]=v, a)), {})
递归解决方案:过滤null和undefined
对象:
const cleanEmpty = obj => Object.entries(obj)
.map(([k,v])=>[k,v && typeof v === "object" ? cleanEmpty(v) : v])
.reduce((a,[k,v]) => (v == null ? a : (a[k]=v, a)), {});
对于对象和数组:
const cleanEmpty = obj => {
if (Array.isArray(obj)) {
return obj
.map(v => (v && typeof v === 'object') ? cleanEmpty(v) : v)
.filter(v => !(v == null));
} else {
return Object.entries(obj)
.map(([k, v]) => [k, v && typeof v === 'object' ? cleanEmpty(v) : v])
.reduce((a, [k, v]) => (v == null ? a : (a[k]=v, a)), {});
}
}