javascript - 根据 Javascript 中的另一个对象过滤对象数组

Question

我有一个对象数组

var data = [
    {
      "body_focus": "upper",
      "difficulty": "three",
      "calories_min": "122",
      "calories_max": "250"
    },
    {
      "body_focus": "upper",
      "difficulty": "three",
      "calories_min": "150",
      "calories_max": "280"
    },
    {
      "body_focus": "lower",
      "difficulty": "two",
      "calories_min": "100",
      "calories_max": "180"
    },
    {
      "body_focus": "total",
      "difficulty": "four",
      "calories_min": "250",
      "calories_max": "350"
    }
]

我想根据另一个对象过滤该对象数组

var myObj = {
    "upper": true,
    "three": true
}

现在 myObj 有一个键 "upper" 和 "three" 并且它们的值为 true。因此，基于这些值，我想创建一个函数来获取 data 数组中的所有对象，其键 "body_focus" 的值为 "upper" 和 "difficulty" 键，其值为 "three"

所以函数应该只返回这些对象

[
    {
      "body_focus": "upper",
      "difficulty": "three",
      "calories_min": "122",
      "calories_max": "250"
    },
    {
      "body_focus": "upper",
      "difficulty": "three",
      "calories_min": "150",
      "calories_max": "280"
    }
]

这就是我试图解决问题的方式

var entry;
var found = [];
for (var i = 0; i < data.length; i++) {
    entry = data[i];
    for(var key in myObj) {
        if(entry.body_focus.indexOf(key) !== -1) {
            found.push(entry);
            break;  
        }
    }
}

我上面的代码只检查键 body_focus ，那么如何检查 body_focus 和 difficulty 呢？这可能看起来很愚蠢，但我已经被困了几个小时而且找不到解决方案

Answer 1

您可以使用一个对象进行搜索，它提供键和所需的值，例如

search = {
    body_focus: "upper",
    difficulty: "three"
}

然后遍历数组并使用实际对象的值检查 search 的所有属性。如果所有搜索条件都匹配，则返回 true。

var data = [{ body_focus: "upper", difficulty: "three", calories_min: "122", calories_max: "250" }, { body_focus: "upper", difficulty: "three", calories_min: "150", calories_max: "280" }, { body_focus: "lower", difficulty: "two", calories_min: "100", calories_max: "180" }, { body_focus: "total", difficulty: "four", calories_min: "250", calories_max: "350" }],
    search = { body_focus: "upper", difficulty: "three" },
    result = data.filter(function (o) {
        return Object.keys(search).every(function (k) {
            return o[k] === search[k];
        });
    });

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }