Question

我有2个阵列

第一个数组有firstname和lastname 第二个数组只有firstname

我将索引一个名字并检查数组

function whereIsAlice(persons) {
  var index = 1;
  for (var i = 0; i < friends1.length; i++) {
    if (friends1[i].firstName === "Alice") {
      index = i;
      break;
    }
    if (friends2[i].firstName === "Alice") {
      index = i;
    }
  }
  return index
}
var friends1 = [{
    firstName: 'John',
    lastName: 'Gaudet'
  },
  {
    firstName: 'Lisa',
    lastName: 'Mcclanahan'
  },
  {
    firstName: 'Alice',
    lastName: 'Vore'
  }, // Alice is here, at index 2
  {
    firstName: 'Marine',
    lastName: 'Salsbury'
  },
];
var friends2 = [{
    firstName: 'Tim'
  },
  {
    firstName: 'Arthur'
  },
  {
    firstName: 'Juan'
  },
];
console.log(whereIsAlice(friends1)); //Should be 2
console.log(whereIsAlice(friends2)); // Should be -1

两者的输出均为2。我该如何解决？

Answer 1

如果你想编写自己的循环，你可以这样做。

var friends1 = [
  { firstName: 'John', lastName: 'Gaudet' },
  { firstName: 'Lisa', lastName: 'Mcclanahan' },
  { firstName: 'Alice', lastName: 'Vore' },
  { firstName: 'Marine', lastName: 'Salsbury' },
];

var friends2 = [
  { firstName: 'Tim' },
  { firstName: 'Arthur' },
  { firstName: 'Juan' },
];

const whereIsAlice = arr => {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i].firstName === 'Alice') {
      return i;
    }
  }
  return -1;
}

console.log(whereIsAlice(friends1)); //Should be 2
console.log(whereIsAlice(friends2)); // Should be -1

Answer 2

问题很容易。您正在比较whereIsAlice方法始终是第一个数组和第二个数组，然后该方法始终找到值并断开de for循环。

-Wno-deprecated-declarations

Answer 3

为了简化您的功能，您可以使用下一个代码

function whereIsAlice(persons) {
  return persons.map(function(e) { return e.firstName; }).indexOf('Alice');
  }

  var friends1 = [{
      firstName: 'John',
      lastName: 'Gaudet'
    },
    {
      firstName: 'Lisa',
      lastName: 'Mcclanahan'
    },
    {
      firstName: 'Alice',
      lastName: 'Vore'
    }, // Alice is here, at index 2
    {
      firstName: 'Marine',
      lastName: 'Salsbury'
    },
  ];
  var friends2 = [{
      firstName: 'Tim'
    },
    {
      firstName: 'Arthur'
    },
    {
      firstName: 'Juan'
    },
  ];
  console.log(whereIsAlice(friends1)); //Should be 2
  console.log(whereIsAlice(friends2)); // Should be -1

Answer 4

重写您的功能如下所示，这将适用于您的情况：

function whereIsAlice(persons) {
    var index= -1;
    for(var i=0;i<persons.length;i++)
        if(persons[i].firstName==="Alice"){index=i;break;}
    return index;
}

Answer 5

你们有些人从初始数组变异新数组（array.map）并执行indexOf。

其他人正在使用for循环，然后检查变量index。

那么为什么JS社区正在努力扩展语言结构，方法等？

我建议您更好地深入了解MDN并阅读findIndex？

＆＃13;

function whereIsAlice(persons) {
  return persons.findIndex(function(person) {
    return person.firstName === 'Alice';
  });
}

var friends1 = [
  {
    firstName: 'John',
    lastName: 'Gaudet'
  },
  {
    firstName: 'Lisa',
    lastName: 'Mcclanahan'
  },
  {
    firstName: 'Alice',
    lastName: 'Vore'
  }, // Alice is here, at index 2
  {
    firstName: 'Marine',
    lastName: 'Salsbury'
  },
];

var friends2 = [
  {
    firstName: 'Tim'
  },
  {
    firstName: 'Arthur'
  },
  {
    firstName: 'Juan'
  },
];
console.log(whereIsAlice(friends1));
console.log(whereIsAlice(friends2));

＆＃13;

使用ES6，它太短了，我没有看到创建方法的理由：

console.log(friends1.findIndex(friend => friend.firstName === 'Alice'));
console.log(friends2.findIndex(friend => friend.firstName === 'Alice'));

javascript在2个数组中查找对象的索引

5 个答案: