Angular过滤器，删除使用id的项

Question

我们如何在angular中过滤值或使用id删除值?我想从对象数组中删除id等于5、12和9的数据或对象。

除此之外，我们还使用多个id进行过滤，例如id等于5、12和9。

索像

let newArr = data。过滤器…

#数据

[
{
"id": 12,
"name": "Architect",
"isShow": true,
"transactionRoleId": 12
},
{
"id": 11,
"name": "Construction Project Director",
"isShow": true,
"transactionRoleId": 11
},
{
"id": 9,
"name": "COVP",
"isShow": true,
"transactionRoleId": 9
},
{
"id": 6,
"name": "DVP Real Estate",
"isShow": true,
"transactionRoleId": 6
},
{
"id": 10,
"name": "Estimator",
"isShow": true,
"transactionRoleId": 10
},
{
"id": 15,
"name": "Finance Director",
"isShow": true,
"transactionRoleId": 15
},
{
"id": 19,
"name": "Local Broker",
"isShow": true,
"transactionRoleId": 19
},
{
"id": 20,
"name": "Master Broker",
"isShow": true,
"transactionRoleId": 20
},
{
"id": 8,
"name": "MPR Director",
"isShow": true,
"transactionRoleId": 8
},
{
"id": 3,
"name": "MPR Manager",
"isShow": true,
"transactionRoleId": 3
},
{
"id": 17,
"name": "Project Manager",
"isShow": true,
"transactionRoleId": 17
},
{
"id": 7,
"name": "RE Associate",
"isShow": true,
"transactionRoleId": 7
},
{
"id": 4,
"name": "RE Director",
"isShow": true,
"transactionRoleId": 4
},
{
"id": 1,
"name": "RE Finance",
"isShow": true,
"transactionRoleId": 1
},
{
"id": 13,
"name": "RE Lead",
"isShow": true,
"transactionRoleId": 13
},
{
"id": 5,
"name": "REPM",
"isShow": true,
"transactionRoleId": 5
},
{
"id": 16,
"name": "Store Development Manager",
"isShow": true,
"transactionRoleId": 16
},
{
"id": 2,
"name": "Store Planner",
"isShow": true,
"transactionRoleId": 2
},
{
"id": 14,
"name": "SVP Property and Store Planning",
"isShow": true,
"transactionRoleId": 14
}
]

Answer 1

试试这个:

const idsToExlude = [5, 12, 9];
const newData = data.filter((obj) => idsToExlude.indexOf(obj.id) === -1);
console.log(newData);