javascript - 比较两个 JSON 并创建一个具有共同值的 json

Question

我想通过比较两个 JSON 来创建两个基于公共(public)属性值的 JSON 数组。

this.linkedParticipants =[
  {
    "id": 3,
    "name": "Participant 2",
    "email": "participant2@fico.com"
  },
  {
    "id": 2,
    "name": "Participant 1",
    "email": "participant1@fico.com"
  },
  {
    "id": 1,
    "name": "Libin Varghese",
    "email": "LibinVarghese@fico.com"
  },
  {
    "id": 7,
    "name": "Participant 5",
    "email": "participant5@fico.com"
  }
]
this.appointmentList = [
  {
    "id": 32,
    "participant": {
      "id": 1,
      "name": "Libin Varghese",
      "email": "LibinVarghese@fico.com"
    }
  },
  {
    "id": 33,
    "participant": {
      "id": 7,
      "name": "Participant 5",
      "email": "participant5@fico.com"
    }
  }
]
this.invitedList = [];
this.confirmedList = [];
let temp = {}
this.linkedParticipants.forEach((participant, i) => {
  this.appointmentList.forEach((appointment, j) => {
    if (appointment.participant.name.indexOf(participant.name)) {
      temp = {
        'participant': participant
      }
      this.invitedList.push(temp);
    }
    else {
      this.confirmedList.push(appointment)
    }
  })
})

但是代码没有按预期工作，因为 this.invitedList 的两个值给出了重复值。我认为我的比较条件有一些问题。

Answer 1

您可以通过 filter() 的不同组合来实现此目的。 :

 var linkedParticipants =[
  {
    "id": 3,
    "name": "Participant 2",
    "email": "participant2@fico.com"
  },
  {
    "id": 2,
    "name": "Participant 1",
    "email": "participant1@fico.com"
  },
  {
    "id": 1,
    "name": "Libin Varghese",
    "email": "LibinVarghese@fico.com"
  },
  {
    "id": 7,
    "name": "Participant 5",
    "email": "participant5@fico.com"
  }
];
var appointmentList = [
  {
    "id": 32,
    "participant": {
      "id": 1,
      "name": "Libin Varghese",
      "email": "LibinVarghese@fico.com"
    }
  },
  {
    "id": 33,
    "participant": {
      "id": 7,
      "name": "Participant 5",
      "email": "participant5@fico.com"
    }
  }
];


var confirmedList = appointmentList.filter((a) => {
  return linkedParticipants.find((p) => p.id === a.participant.id);
});

var invitedList = linkedParticipants.filter((p) => {
  return !appointmentList.find((a) => p.id === a.participant.id);
});

console.log(confirmedList)
console.log(invitedList)