根据其他对象数组过滤对象数组

Question

大家好，我有两个对象数组，如下所示

const selectedCodes = [
  { id: 1, modifiedObject: null, originalObject: {id: 23, name: 'test'}},
  { id: 2, modifiedObject: {id: 24, name: 'test2'}, originalObject: null },
  ....
  ....
];

另一个对象数组，如下所示

const originalCodes = [
  { id: 23, name: 'test' },
  { id: 24, name: 'test2'},
  { id: 25, name: 'test3' },
  { id: 26, name: 'test4' }
];

我正在寻找 originalCodes 应该返回这两项的结果

const originalCodes = [
      { id: 25, name: 'test3' },
      { id: 26, name: 'test4' }
    ];

在这里，我想根据 selectedCodes 数组 modifiedObject/originalObject Id 中可用的 id 过滤 originalCodes 数组的结果，并且每个索引始终只有一个对象，修改后的对象和原始对象都会出现。

任何人都可以让我知道或任何关于如何实现结果的想法，非常感谢提前。

我在下面尝试过

const removedArrays = selectCodes.reduce((acc, item) => {
   console.log(item)// here i need to verify  with other array of objects
},[{}])

Answer 1

它并不完全清楚你在寻找什么。听起来您想找到originalCodes 中不存在的selectedCodes。如果是这样，那么以下内容可能对您有用：

const selectedCodes = [
  { id: 1, modifiedObject: null, originalObject: {id: 23, name: 'test'}},
  { id: 2, modifiedObject: {id: 24, name: 'test2'}, originalObject: null }
];

const originalCodes = [
  { id: 23, name: 'test' },
  { id: 24, name: 'test2'},
  { id: 25, name: 'test3' },
  { id: 26, name: 'test4' }
];

const filteredCodes = originalCodes.filter(function(item) {
  return !(selectedCodes.find(element => (element.originalObject && element.originalObject.id) == item.id) || selectedCodes.find(element => (element.modifiedObject && element.modifiedObject.id) == item.id));
});

console.log(filteredCodes);

Answer 2

您可以将对象字符串化为 exclude 对象并过滤 originalCodes。

这种方法要求所有对象的所有属性都具有相同的顺序。

const
    selectedCodes = [{ id: 1, modifiedObject: null, originalObject: { id: 23, name: 'test' } }, { id: 2, modifiedObject: { id: 24, name: 'test2' }, originalObject: null }],
    originalCodes = [{ id: 23, name: 'test' }, { id: 24, name: 'test2'}, { id: 25, name: 'test3' }, { id: 26, name: 'test4' }],
    exclude = Object.fromEntries(selectedCodes.map(o => [JSON.stringify(o.originalObject || o.modifiedObject), true])),
    result = originalCodes.filter(o => !exclude[JSON.stringify(o)]);

console.log(result);

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Answer 3

const selectedCodes=[{id:1,modifiedObject:null,originalObject:{id:23,name:"test"}},{id:2,modifiedObject:{id:24,name:"test2"},originalObject:null}],originalCodes=[{id:23,name:"test"},{id:24,name:"test2"},{id:25,name:"test3"},{id:26,name:"test4"}];

let result = originalCodes.filter(({id}) => 
    !selectedCodes.find(e => 
    (e.originalObject?e.originalObject.id:e.modifiedObject.id) === id ))


console.log(result)

Answer 4

如果想法是过滤掉在originalObject 或modifiedObject 中作为name 属性存在的originalCodes 项目，则解决方案可能很简单：

const selectedCodes = [{id:1,modifiedObject:null,originalObject:{id:23,name:"test"}},{id:2,modifiedObject:{id:24,name:"test2"},originalObject:null}],
      originalCodes = [{id:23,name:"test"},{id:24,name:"test2"},{id:25,name:"test3"},{id:26,name:"test4"}],
      
      
      result = originalCodes.filter(({id}) => 
                  !selectedCodes.some(({originalObject, modifiedObject}) => {
                    const {id: id1} = (originalObject||{}),
                          {id: id2} = (modifiedObject||{})
                    return id == id1 || id == id2
                  })
               )
      
console.log(result)

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