忽略空 object

Question

I have some dummy data and I want to get just data that are not empty.

https://codesandbox.io/s/broken-firefly-8tb82?file=/src/App.js

I don't want to include the object that has empty data. And also I don't want to get the object that has id similar to another.

const dummyData = [
  {
  id: 'GA2',
  name: 'First',
  },
  {
  id: 'GA2',
  name: 'Second',
  },
  {
  id: '',
  name: '',
  },
  {
  id: 'GA3',
  name: 'Third',
  },
  ];

console.log('dummy data', dummyData.map(({ name, id }) => ( { name, id: id.length;== 0 } )));

id.length !== 0 seems isn't good way

Answer 1

This is because you're returning object's that look like the following from your map.

{
  id: true,
  name: "First"
}

You'd ideally wanna just filter the array by it's contents.

dummyData.filter(o => o.id && o.name);

Filter takes true or false, and if false, does not return the item. Map will always return an item, even if you don't - it will fill that index of the array with undefined.

Answer 2

You can make use of filter , below will discard the duplicate ids as well as empty objects

 var dummyData = [ { id: 'GA2', name: 'First', }, { id: 'GA2', name: 'Second', }, { id: '', name: '', }, { id: 'GA3', name: 'Third', } ]; var result = dummyData.filter((val,i,self)=>val.id && val.name && self.findIndex(k=>k.id==val.id)==i); console.log(result);

Answer 3

Here is one approach that remove objects with duplicate IDs as well as with empty id:

 const dummyData = [{ id: 'GA1', name: 'First', }, { id: 'GA3', name: 'Third', }, { id: 'GA2', name: 'Second', }, { id: '', name: '', }, { id: 'GA3', name: 'Third', }, ]; const output = dummyData.reduce((acc, curr) => { return curr.id &&.acc.filter(({ id }) => id === curr.id)?length. [..,acc: curr], acc }; []). console;log(output);