从一个对象中删除与其他对象项目匹配的项目

Question

>我有一个对象结构是这样的，如下所示

"designProjects": [
{
"projectNumber": "number1",
"name": "test1"
},
{
"projectNumber": "number2",
"name": "test2"
},
{
"projectNumber": "number3",
"name": "test3"
},
]

我有另一个具有如下结构的对象

"allProjects": [
{
"project": {
"name": "test1",
"number": "number1"
},
"employee": {
"displayName": "name1"
},
"projectRoleName": "Editor"
},
{
"project": {
"name": "test2",
"number": "number2"
},
"employee": {
"displayName": "name2"
},
"projectRoleName": "Editor"
},
]

我正在查看如下所示的结果

"designProjects": [
{
"projectNumber": "number3",
"name": "test3"
},
]

这里的结果是designProjects只有一个，因为项目编号和名称与allprojects对象的数组project匹配。有没有办法在react js中实现这种结果. 任何建议或想法将非常感谢我，提前非常感谢

Answer 1

您只需将.filter与.some结合起来，如下所示：

let d = {
"designProjects": [
{
"projectNumber": "number1",
"name": "test1"
},
{
"projectNumber": "number2",
"name": "test2"
},
{
"projectNumber": "number3",
"name": "test3"
},
]
}
let a = {
"allProjects": [
{
"project": {
"name": "test1",
"number": "number1"
},
"employee": {
"displayName": "name1"
},
"projectRoleName": "Editor"
},
{
"project": {
"name": "test2",
"number": "number2"
},
"employee": {
"displayName": "name2"
},
"projectRoleName": "Editor"
},
]
};
console.log(
d.designProjects.filter((designProject) => {
return !a.allProjects.some((project) => designProject.projectNumber === project.project.number && designProject.name === project.project.name);
})
);

Answer 2

你可以结合过滤器和一些

filter用于根据条件返回新的过滤数组

some将是条件，一旦找到匹配项就返回

const designProjects = [{
"projectNumber": "number1",
"name": "test1"
},
{
"projectNumber": "number2",
"name": "test2"
},
{
"projectNumber": "number3",
"name": "test3"
},
];
const allProjects = [{
"project": {
"name": "test1",
"number": "number1"
},
"employee": {
"displayName": "name1"
},
"projectRoleName": "Editor"
},
{
"project": {
"name": "test2",
"number": "number2"
},
"employee": {
"displayName": "name2"
},
"projectRoleName": "Editor"
},
]
const cleaned = designProjects.filter((x) => {
return !allProjects.some(y => y.project.number === x.projectNumber);
});
console.info(cleaned);