检测对象某个属性的值为空

Question

I have a list of 100 objects. when a value of a property of the object is null I want to delete the whole object. Right now the if statement is not true, how can I fix this?

object:

{ id: 'blockstack-iou',
  name: 'Blockstack (IOU)',
  symbol: 'stx',
  image: 'missing_large.png',
  market_cap_usd: 174267825,
  market_cap_change_percentage: null,
  market_cap_relative_market_cap_percentage: 0 }

Code:

for (let i = 0; i < 100 ; i++) {

   // console.log(object[i]);
    console.log(Object.values(object[i]));
    //TODO: if one property value of the data is null delete object
    if (Object.values(object[i]) == null) {
        console.log("Null detected and object deleted");
        delete object[i];
    }
}

Output of console.log(Object.values(object[i]));

[ 'blockstack-iou',
  'Blockstack (IOU)',
  'stx',
  'missing_large.png',
  174267825,
  null,
  0 ]

Answer 1

You could use filter() on the original array to return items where Object.values(object[i]) doesn't include null .

 const object = [ {p1: 1, p2: 1}, {p1: 2, p2: null}, {p1: 3, p2: null}, {p1: 4, p2: 4} ]; const result = object.filter(o => !Object.values(o).includes(null)); console.log(result);

Answer 2

你可以做这样的事情。

const noNullPropertiesObjects = arrayWithObjects.filter(obj => !Object.values(obj).includes(null))