Question

我有两个对象如下：

obj1 = {"a":"firstValue", "b":"secondValue", "c":"thirdValue157das15714daa4"} 
obj2 = {"a":"firstValue", "b":"secondValue", "c":"thirdValue1348sdfsf114sfs45fsd"}

我需要比较两个对象，忽略第三个键 - 值对'c'。基本上，比较应该返回true忽略'c'

任何人都可以帮助我使用angular或javascript中的乐观解决方案

提前致谢

Answer 1

简单地

obj1.a == obj2.a && obj1.b == obj2.b

如果a和b是动态的，那么请存储keyToIgnore

var keyToIgnore = [ "c" ];
var isEqual = Object.keys( obj1 ).every( s => keyToIgnore.includes(s) || obj1[ s ] == obj2[ s ] );

<强>演示

＆＃13;

var obj1 = {
  "a": "firstValue",
  "b": "secondValue",
  "c": "thirdValue157das15714daa4"
}
var obj2 = {
  "a": "firstValue",
  "b": "secondValue",
  "c": "thirdValue1348sdfsf114sfs45fsd"
}
var keyToIgnore = [ "c" ];
var fnCompare = ( obj1, obj2 ) => Object.keys(obj1).every(s => keyToIgnore.includes(s) || obj1[s] == obj2[s]);

console.log( fnCompare( obj1, obj2 ) )

＆＃13;

Answer 2

以下是达到此目的的一种方法：

＆＃13;

const obj1 = {
  "a": "firstValue",
  "b": "secondValue",
  "c": "thirdValue157das15714daa4"
};

const obj2 = {
  "a": "firstValue",
  "b": "secondValue",
  "c": "thirdValue1348sdfsf114sfs45fsd"
};

const result = Object
  .keys(obj1)
  .filter(k => k !== 'c')
  .every(k => obj1[k] === obj2[k]);

console.log(result);

＆＃13;

上面首先筛选出非必需属性，然后比较两个对象中剩余属性的值。

Answer 3

您可以检查explcit对象的每个属性。

＆＃13;

function check(o1, o2) {
    const filter = v => v !== 'c' ;

    var keys1 = Object.keys(o1).filter(filter),
        keys2 = Object.keys(o2).filter(filter);

    return keys1.length === keys2.length && keys1.every(k => o1[k] === o2[k]);
}

var obj1 = { a: "firstValue", b: "secondValue", c:"thirdValue157das15714daa4" },
    obj2 = { a: "firstValue", b: "secondValue", c: "thirdValue1348sdfsf114sfs45fsd" };

console.log(check(obj1, obj2));

＆＃13;

比较忽略一个键的两个对象

3 个答案: