var listToDelete = ['abc', 'efg'];
var arrayOfObjects = [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'}, // delete me
{id:'hij',name:'ge'}] // all that should remain
如何通过匹配对象属性从数组中删除对象?
请仅使用原生JavaScript。
我在使用拼接时遇到问题,因为每次删除都会缩短长度。 在orignal索引上使用克隆和拼接仍然会让你遇到缩短长度的问题。
答案 0 :(得分:134)
我假设你使用了splice
这样的东西?
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
你需要做的就是修复这个bug,下次再减量i
,然后(向后循环也是一个选项):
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
i--;
}
}
为避免线性时间删除,您可以在阵列上编写要保持的数组元素:
var end = 0;
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) === -1) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
要避免在现代运行时中进行线性时间查找,可以使用哈希集:
const setToDelete = new Set(listToDelete);
let end = 0;
for (let i = 0; i < arrayOfObjects.length; i++) {
const obj = arrayOfObjects[i];
if (setToDelete.has(obj.id)) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
可以包含在一个很好的函数中:
const filterInPlace = (array, predicate) => {
let end = 0;
for (let i = 0; i < array.length; i++) {
const obj = array[i];
if (predicate(obj)) {
array[end++] = obj;
}
}
array.length = end;
};
const toDelete = new Set(['abc', 'efg']);
const arrayOfObjects = [{id: 'abc', name: 'oh'},
{id: 'efg', name: 'em'},
{id: 'hij', name: 'ge'}];
filterInPlace(arrayOfObjects, obj => !toDelete.has(obj.id));
console.log(arrayOfObjects);
如果您不需要这样做,那就是Array#filter
:
const toDelete = new Set(['abc', 'efg']);
const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));
答案 1 :(得分:61)
您可以通过其中一个属性删除项目,而无需使用任何第三方库,如下所示:
var removeIndex = array.map(function(item) { return item.id; })
.indexOf("abc");
~removeIndex && array.splice(removeIndex, 1);
答案 2 :(得分:25)
findIndex适用于现代浏览器:
var myArr = [{id:'a'},{id:'myid'},{id:'c'}];
var index = arr.findIndex(function(o){
return o.id === 'myid';
})
if (index !== -1) myArr.splice(index, 1);
答案 3 :(得分:24)
如果您想修改现有阵列本身,那么我们必须使用拼接。以下是使用 findWhere 下划线/ lodash的更好/可读方式:
var items= [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'},
{id:'hij',name:'ge'}];
items.splice(_.indexOf(items, _.findWhere(items, { id : "abc"})), 1);
(没有lodash /下划线)
从ES5开始,我们在数组上有findIndex
方法,所以很容易没有lodash / underscore
items.splice(items.findIndex(function(i){
return i.id === "abc";
}), 1);
(几乎所有现代浏览器都支持ES5)
答案 4 :(得分:9)
如果您只想将其从现有阵列中删除而不是创建新阵列,请尝试:
var items = [{Id: 1},{Id: 2},{Id: 3}];
items.splice(_.indexOf(items, _.find(items, function (item) { return item.Id === 2; })), 1);
答案 5 :(得分:5)
请只使用原生JavaScript。
作为替代方案,使用ECMAScript 5的更多“功能”解决方案,您可以使用:
var listToDelete = ['abc', 'efg'];
var arrayOfObjects = [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'}, // delete me
{id:'hij',name:'ge'}]; // all that should remain
arrayOfObjects.reduceRight(function(acc, obj, idx) {
if (listToDelete.indexOf(obj.id) > -1)
arrayOfObjects.splice(idx,1);
}, 0); // initial value set to avoid issues with the first item and
// when the array is empty.
console.log(arrayOfObjects);
[ { id: 'hij', name: 'ge' } ]
根据the definition of 'Array.prototype.reduceRight' in ECMA-262:
reduceRight不会直接改变名为的对象,但是对象可能会被callbackfn 的调用所突变。
因此,这是reduceRight
的有效用法。
答案 6 :(得分:4)
通过递减i
来反向循环以避免问题:
for (var i = arrayOfObjects.length - 1; i >= 0; i--) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
或使用filter
:
var newArray = arrayOfObjects.filter(function(obj) {
return listToDelete.indexOf(obj.id) === -1;
});
答案 7 :(得分:2)
var arrayOfObjects = [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'}, // delete me
{id:'hij',name:'ge'}] // all that should remain
根据你的回答将是这样的。当你单击某个特定对象时,在param中发送删除我的函数的索引。这个简单的代码就像魅力一样。
function deleteme(i){
if (i > -1) {
arrayOfObjects.splice(i, 1);
}
}
答案 8 :(得分:2)
使用Set和ES6过滤器检查出来。
let result = arrayOfObjects.filter( el => (-1 == listToDelete.indexOf(el.id)) );
console.log(result);
这是JsFiddle: https://jsfiddle.net/jsq0a0p1/1/
答案 9 :(得分:1)
带有过滤器和indexOf
withLodash = _.filter(arrayOfObjects, (obj) => (listToDelete.indexOf(obj.id) === -1));
withoutLodash = arrayOfObjects.filter(obj => listToDelete.indexOf(obj.id) === -1);
带有过滤器和包含项
withLodash = _.filter(arrayOfObjects, (obj) => (!listToDelete.includes(obj.id)))
withoutLodash = arrayOfObjects.filter(obj => !listToDelete.includes(obj.id));
答案 10 :(得分:0)
如果您喜欢简短的自我描述性参数,或者您不想使用splice
并使用直接过滤器,或者您只是像我这样的SQL人员:
function removeFromArrayOfHash(p_array_of_hash, p_key, p_value_to_remove){
return p_array_of_hash.filter((l_cur_row) => {return l_cur_row[p_key] != p_value_to_remove});
}
示例用法:
l_test_arr =
[
{
post_id: 1,
post_content: "Hey I am the first hash with id 1"
},
{
post_id: 2,
post_content: "This is item 2"
},
{
post_id: 1,
post_content: "And I am the second hash with id 1"
},
{
post_id: 3,
post_content: "This is item 3"
},
];
l_test_arr = removeFromArrayOfHash(l_test_arr, "post_id", 2); // gives both of the post_id 1 hashes and the post_id 3
l_test_arr = removeFromArrayOfHash(l_test_arr, "post_id", 1); // gives only post_id 3 (since 1 was removed in previous line)
答案 11 :(得分:0)
要通过给定数组中的ID删除对象;
const hero = [{'id' : 1, 'name' : 'hero1'}, {'id': 2, 'name' : 'hero2'}];
//remove hero1
const updatedHero = hero.filter(item => item.id !== 1);
答案 12 :(得分:0)
您可以使用filter
。如果条件为true,则此方法始终返回元素。因此,如果要按ID删除,则必须保留所有与给定ID不匹配的元素。这是一个示例:
arrayOfObjects = arrayOfObjects.filter(obj => obj.id!= idToRemove)
答案 13 :(得分:-1)
var apps = [{id:34,name:'My App',another:'thing'},{id:37,name:'My New App',another:'things'}]
var removeIndex = apps.map(function(item) { return item.id; }).indexOf(37)
apps.splice(removeIndex, 1);