比较两个忽略一个键的对象

Question

I have two objects as below:

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

I need to compare both objects ignoring the third key - value pair 'c'. Basically the comparison should return true ignoring the 'c'

Can anyone help me with an optimistic solution in angular or javascript

Thanks in advance

Answer 1

Simply

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

Or if an a and b are going to be dynamic, then store keyToIgnore

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

Demo

 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 ) ) 

Answer 2

Following is one way to achieve this:

 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); 

The above first filters out the non-required property and then compares the values of remaining properties in the two objects.

Answer 3

You could check explcit every property of the objects.

 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));