根据以数组形式存在于对象值内的值搜索对象数组

Posted 2023-03-15

【中文标题】根据以数组形式存在于对象值内的值搜索对象数组

【英文标题】：searching an array of objects on the basis of value present inside an value of an object in form of array

【发布时间】：2021-10-24 21:52:37

【问题描述】：

我有一个对象，其中有两个字段 filter1 和 filter2 具有数组形式的值

let filter = filter1:["mine","your"]: filter2:["C","D"] //值不固定

数据是对象数组的形式

let data = [
 id:1, filter1:["mine"], filter2:["E","C"],
 id:2, filter1:["mine"], filter2:["E","C","F"],
 id:3, filter1:["your"], filter2:["C"],
 id:3, filter1:["your"], filter2:["D","C"],
 id:5, filter1:["other"], filter2:["F"],
...
]

我必须过滤掉那些在特定键中存在任何字段的对象 例如，如果filter 是filter1:["mine"]: filter2:["F","D"]，它将首先在数据对象的filter1 中搜索filter1 的任何元素，然后搜索存在于数据对象的filter2 中的filter2 的任何元素，如果其中有任何一个，则返回该对象找到了

few example

filter1:["mine"]: filter2:["F","D"] 的结果

result = [
 id:1, filter1:["mine"], filter2:["E","C"], //since filter1 "mine"
 id:2, filter1:["mine"], filter2:["E","C","F"], //since filter1 "mine"
 id:3, filter1:["your"], filter2:["D","C"], //since from "F" and "D" from filter2 "D" is present
 id:5, filter1:["other"], filter2:["F"], //since "F" from filter2 is present
]

filter1:["your"]: filter2:["F","G"] 的结果

result = [
 id:2, filter1:["mine"], filter2:["E","C","F"], //since "F" from filter2 is present
 id:3, filter1:["your"], filter2:["D","C"], //since filter1 is "your"
 id:5, filter1:["other"], filter2:["F"], //since "F" from filter2 is present
]

filter1:[]: filter2:["D"] 的结果

result = [
 id:3, filter1:["your"], filter2:["D","C"], //since filter2 has "D"
]

【问题讨论】：

你可以使用 Array.prototype.filter()。

【参考方案1】：

您可以使用.filter()、.some() 和.includes() 的组合：

const data = [
   id:1, filter1:["mine"], filter2:["E","C"],
   id:2, filter1:["mine"], filter2:["E","C","F"],
   id:3, filter1:["your"], filter2:["C"],
   id:3, filter1:["your"], filter2:["D","C"],
   id:5, filter1:["other"], filter2:["F"]
];

const search = ( filter1, filter2 ) => 
  data.filter(item => 
    item.filter1.some(fItem => filter1.includes(fItem)) ||
    item.filter2.some(fItem => filter2.includes(fItem))
);

const result = search( filter1:["mine"], filter2:["F","D"] );
console.log(result);

【讨论】：

如果filter1不是这样的数组形式 const data = [ id:1, filter1:"mine", filter2:["E","C"], id ：2，过滤器1：“我的”，过滤器2：[“E”，“C”，“F”]，];那么使用 some 会抛出错误如何处理

在我的回答中查看concat() 的使用来解决这个问题。

是的，@pilchard 是正确的。将item.filter1.some 替换为[].concat(item.filter1).some，并对下一行执行相同操作。然后它将处理包含字符串而不是数组的数据。

【参考方案2】：

您可以通过在传递的过滤器对象的Object.entries() 上调用some() 来概括solution proposed by RoMilton，然后使用嵌套的some() 调用迭代每个key 和filter_array。

如果您还 Array#concat() 将当前迭代的数据元素属性放入数组中，您可以在过滤器对象中包含非数组属性，即在这种情况下为 id。

const data = [
   id: 1, filter1: ["mine"], filter2: ["E", "C"] ,
   id: 2, filter1: ["mine"], filter2: ["E", "C", "F"] ,
   id: 3, filter1: ["your"], filter2: ["C"] ,
   id: 3, filter1: ["your"], filter2: ["D", "C"] ,
   id: 5, filter1: ["other"], filter2: ["F"] 
];

const search = (array, filter_object) =>
  array.filter(item =>
    Object.entries(filter_object).some(([key, filter_array]) =>
      [].concat(item[key]).some(fitem => filter_array.includes(fitem)))
  );

const filter =  filter1: ["mine"], filter2: ["F", "D"] ;
const result = search(data, filter);

console.log(...result.map(( id ) => ( id )));

const filter2 =  id: [5], filter1: ["mine"] 
const result2 = search(data, filter2);

console.log(...result2.map(( id ) => ( id )));

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concat() 的这种用法也可以应用于硬编码解决方案，以允许数据数组中可能不是数组的属性。

const data = [
   id: 1, filter1: "mine", filter2: ["E", "C"] ,
   id: 2, filter1: ["mine"], filter2: ["E", "C", "F"] ,
   id: 3, filter1: ["your"], filter2: ["C"] ,
   id: 3, filter1: ["your"], filter2: ["D", "C"] ,
   id: 5, filter1: ["other"], filter2: ["F"] 
];

const search = ( filter1, filter2 ) => (
  data.filter(item =>
    [].concat(item.filter1).some(fItem => filter1.includes(fItem)) ||
    [].concat(item.filter2).some(fItem => filter2.includes(fItem))
  ));

const result = search( filter1: ["mine"], filter2: ["F", "D"] );
console.log(result);