Question

我想根据categories数组中的条件从otherCategories数组中过滤项目。

如果otherCategories包含一个对象，其中title匹配categories.subCategory[i].title中的一个标题，而name匹配categories.subCategory[i].details.name，则仅过滤该对象，例如， categories。

var categories = [
{
    title:"item1",
    subCategory:[
        {
            title:"subCat1",
            details:{
                name:"detail1",
                email:"test@test.com"
            }
        },
        {
            title:"subCat2",
            details:{
                name:"detail2",
                email:"test@test.com"
            }
        }
    ]
},
{
    title:"item2",
    subCategory:[
        {
            title:"subCat1",
            details:{
                name:"detail3",
                email:"test@test.com"
            }
        },
        {
            title:"subCat2",
            details:{
                name:"detail2",
                email:"test@test.com"
            }
        }
    ]
}
]

var otherCategories = [
{
    title:"subCat1",
    name:"detail1"        
}
]

预期结果

categories = [
{
    title:"item1",
    subCategory:[
        {
            title:"subCat1",
            details:{
                name:"detail1",
                email:"test@test.com"
            }
        },
        {
            title:"subCat2",
            details:{
                name:"detail2",
                email:"test@test.com"
            }
        }
    ]
}]

Answer 1

使用Array.reduce，Array.filter和Array.some

将otherCategories数组转换为标题为键，名称为值的对象
过滤categories数组，其中有些subCategory存在匹配的值

var categories = [{title:"item1",subCategory:[{title:"subCat1",details:{name:"detail1",email:"test@test.com"}},{title:"subCat2",details:{name:"detail2",email:"test@test.com"}}]},{title:"item2",subCategory:[{title:"subCat1",details:{name:"detail3",email:"test@test.com"}},{title:"subCat2",details:{name:"detail2",email:"test@test.com"}}]}];
var otherCategories = [{title:"subCat1",name:"detail1"}];
var obj = otherCategories.reduce((a,c) => Object.assign(a,{[c.title]:c.name}), {});

categories = categories.filter(v => v.subCategory.some(o => obj[o.title] === o.details.name));
console.log(categories);

Answer 2

另一种通常适用于示例中的简单对象的方法是将@Entity @Table(name = "eq_panel_category_option") public class PanelCategoryOption { @EmbeddedId private PanelCategoryOptionId id; }数组转换为“字符串化”对象的数组，然后通过比较“字符串化”来过滤@Embeddable public class PanelCategoryOptionId implements Serializable { @ManyToOne(fetch = FetchType.LAZY) @JoinColumns({ @JoinColumn(name = "panel_id", referencedColumnName = "panel_id"), @JoinColumn(name = "category_id", referencedColumnName = "category_id")}) private PanelCategory panelCategory; @ManyToOne(fetch = FetchType.LAZY) @JoinColumns({ @JoinColumn(name = "category_option_id", referencedColumnName = "id", insertable = false, updatable = false), @JoinColumn(name = "category_id", referencedColumnName = "category_id", insertable = false, updatable = false)}) private CategoryOption categoryOption; }所需otherCategories键值对的版本到转换后的categories数组。

但是，值得注意的是，JavaScript无法保证对象属性顺序（尽管许多浏览器会保留属性顺序）。这意味着该方法可能在某些情况下不起作用，并且类似@NikhilAggarwal所建议的方法更为稳定。

例如：

subCategory

Answer 3

您可以通过map子类别filtering对结果进行分类：

  function matches(sub, filters) {
    return filters.some(filter => filter.title === sub.title && filter.name === sub.name);
 }

 const result = categories.map(({ title, subCategories }) => ({ title, subCategories: subCategories.filter(sub => matches(sub, otherCategories)) }));

根据另一个对象数组中的条件过滤对象数组

3 个答案: