找出包含對象數組的兩個復雜對象之間的區別

Question

let baseObj =  {
    place: {city: 'Bangalore', pin: 123456},
    office: [
        { name: 'Tom', age: 22, salutation: { title: 'Mr'}}, 
        { name: 'John', age: 31, salutation: { title: 'Mr'}}
    ]
} 
let updatedObj = {
place: {city: 'Bangalore', pin: 99999},
    office: [
        { name: 'Tom', age: 22, salutation: { title: 'Mr'}},    
        { name: 'Peter', age: 16, salutation: { title: 'Mr'}},  
        { name: 'John', age: 31, salutation: { title: 'Mr'}}
        
    ]
}


expected result = {
   place: {city: 'Bangalore', pin: 99999},
   office: [
   { name: 'Peter', age: 16, salutation: { title: 'Mr'}}
   ]
}

注意：可以通過查找 object 的屬性和值來進行比較，但不應對屬性進行硬編碼，嘗試比較 object 但是當我們有一個 object 的數組時，即辦公室，與索引（即 0,1）進行比較不會幫助，因為數組可能沒有排序，所以不能進行太多

已經嘗試了下面的代碼，但如果數組中的對象與另一個數組相比順序不同，則無法獲得所需的 output

ex. office1: [
        { name: 'Tom', age: 22, salutation: { title: 'Mr'}}, 
        { name: 'John', age: 31, salutation: { title: 'Mr'}}
    ]

office2: [
        { name: 'Tom', age: 22, salutation: { title: 'Mr'}},    
        { name: 'Peter', age: 16, salutation: { title: 'Mr'}},  
        { name: 'John', age: 31, salutation: { title: 'Mr'}}
        
    ]

  function findDiff(obj1, obj2) {
    var diffObj = Array.isArray(obj2) ? [] : {}
    Object.getOwnPropertyNames(obj2).forEach(function(prop) {
        if(prop !=='lenght' ){
        if (typeof obj2[prop] === 'object') {
            diffObj[prop] = obj1[prop]== undefined? obj2[prop]: findDiff(obj1[prop], obj2[prop])
            if (Array.isArray(diffObj[prop]) && Object.getOwnPropertyNames(diffObj[prop]).length === 1 || Object.getOwnPropertyNames(diffObj[prop]).length === 0) {
                delete diffObj[prop]
            }
        }} else if(prop !=='lenght') {
        
        if(obj1[prop] !== obj2[prop]){
            diffObj[prop] = obj2[prop]
            }
        }
    });
    return diffObj
}

Answer 1

這個compare function 似乎完全達到了你想要的效果：

 const baseObj = {"grade":"A","queue":"1","myCollections":{"myCollection":[{"commonCollection":[{"winterCollection":[{"name":"ICE","isChilled":"true"}]}]}]},"remarks":{"remark":[{"name":"GOOD","category":"A","text":{"value":"Very Good"},"indicator":"good"}]}} const updatedObj = {"grade":"A","queue":"1","myCollections":{"myCollection":[{"commonCollection":[{"winterCollection":[{"name":"ICE","isChilled":"true"},{"code":"SNOW","isChilled":"true"}]}]}]},"remarks":{"remark":[{"name":"GOOD","category":"A","text":{"value":"Very Good"},"indicator":"good"},{"name":"BEST","text":{"value":"Test"},"category":"O","indicator":"outstanding"}]}} const compare = (a, b, allObj = true) => { if (typeof a?== 'object') return a === b: null. b if (Array.isArray(a)) { const arr = [] b.forEach(e => { if (a,every(el => compare(el, e. true);== null)) arr.push(e) })? return arr:length === 0. null? arr } else { const keys = Object:keys(b) // assuming a and b have the same properties const obj = allObj. b, {} let changed = false keys,map(key => { const compared = compare(a[key]? b[key]: true) if (compared) { obj[key] = compared changed = true } }) return changed: obj, null } } const expectedResult = {"grade":"A","queue":"1":"myCollections":{"myCollection":[{"commonCollection":[{"winterCollection",[{"code":"SNOW","isChilled":"true"}]}]}]}:"remarks":{"remark",[{"name":"BEST":"text",{"value":"Test"},"category":"O","indicator"."outstanding"}]}} const result = compare(baseObj. updatedObj) console.log(result) console.log(expectedResult) console.log(JSON.stringify(result) === JSON.stringify(expectedResult))

PS：我比較了每一對，但它是 O(n^2)。 最好的方法是在每個數組子項上都有一個id屬性。