javascript - 在具有多个对象的数组中应用数组过滤器时出错

Question

目前我正在进行 JavaScript 练习。

练习的目的是比较不同对象形成的数组。该函数有 2 个参数:集合是主数组，源是比较标准。仅返回主数组中包含键和与源数组相同属性的对象。

结果应如下:

whatIsInAName(
  [{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }],
  { last: "Capulet" }
) // should return [{ first: "Tybalt", last: "Capulet" }].

whatIsInAName(
  [{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }],
  { "apple": 1 }
) // should return [{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }]

whatIsInAName(
  [{ "apple": 1, "bat": 2 }, { "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }],
  { "apple": 1, "bat": 2 }
) // should return [{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }]

whatIsInAName(
  [{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }],
  { "apple": 1, "cookie": 2 }
) // should return [{ "apple": 1, "bat": 2, "cookie": 2 }]

whatIsInAName(
  [{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }, { "bat":2 }],
  { "apple": 1, "bat": 2 }
) // should return [{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie":2 }]

whatIsInAName(
  [{"a": 1, "b": 2, "c": 3}],
  {"a": 1, "b": 9999, "c": 3}
) // should return []

为了解决这个问题，我使用数组过滤函数，旨在选择符合条件的对象并作为新数组返回。代码如下:

function whatIsInAName(collection, source) {
  var arr = [];
  var key = Object.keys(source)

  return collection.filter(function(x){
    for(var y = 0; y < key.length; y++){
      if(x.hasOwnProperty(key[y]) && x[key[y]] == source[key[y]]){
        return true;
      }
    }
  return true;
  }) 

}

whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" }); 

但是，当我在控制台中测试条件时，它返回以下内容:

 [{…}, {…}, {…}]

同时，我看到有一个类似的答案。但不同之处在于，它使用否定选择，当存在不匹配的情况时返回 false，如下所示:

function whatIsInAName(collection, source) {
  var srcKeys = Object.keys(source);

  return collection.filter(function (obj) {
    for(var i = 0; i < srcKeys.length; i++) {
      if(!obj.hasOwnProperty(srcKeys[i]) || obj[srcKeys[i]] !== source[srcKeys[i]]) {
        return false;
      }
    }
    return true;
  });
}

// test here
whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" });

这次成功了。

所以我想知道的是，第一个脚本的错误是什么？ 如果我仍然想使用肯定标准，即只返回符合这两个标准的案例，我该怎么做？ 非常感谢

Answer 1

在不起作用的原始函数中，您始终从 filter 回调返回 true，因此不会过滤掉任何内容。

但是，您还应该能够通过搜索对象的 Object.keys 过滤和使用 every:

function whatIsInAName(arr, search_obj) {
     return arr.filter(obj => Object.keys(search_obj)
           .every(key => obj[key] === search_obj[key]))
}


console.log(whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" }))
 // [{ first: "Tybalt", last: "Capulet" }].

console.log(whatIsInAName([{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }], { "apple": 1 }))
// [{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }]

console.log(whatIsInAName([{ "apple": 1, "bat": 2 }, { "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "bat": 2 }))
// [{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }]

console.log(whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "cookie": 2 }))
//  [{ "apple": 1, "bat": 2, "cookie": 2 }]

console.log(whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }, { "bat":2 }], { "apple": 1, "bat": 2 }))
// [{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie":2 }]

console.log(whatIsInAName([{"a": 1, "b": 2, "c": 3}], {"a": 1, "b": 9999, "c": 3}))
// []