Question

让我们从一个简单的例子开始，以更好地理解我的问题，我有这个对象：

{
    "list_of_objects" : [
        {
            "name" : "0",
            "interesting_key" : "some_value",
        },
        {
            "name" : "1",
            "interesting_key" : "some_other_value",
            "list_of_objects" : [
                {
                    "name" : "10",
                    "interesting_key" : "something"
                },
                {
                    "name" : "11"
                },
            ]
        },
        {
            "name" : "2",
            "interesting_key" : "again_some_other_value",
        }
    ]
}

我想获取与interesting_key相对应的所有值，有没有一种方法可以在不循环访问每个对象列表的情况下？这样的结果：

["some_value", "some_other_value", "something", "again_some_other_value"]

我想这样做，所以我只能比较包含interesting_key的对象。

在此先感谢您的帮助，

Answer 1

您可以像这样递归搜索密钥：

const list_of_objects=[{"name":"0","interesting_key":"some_value",},{"name":"1","interesting_key":"some_other_value","list_of_objects":[{"name":"10","interesting_key":"something"},{"name":"11"},]},{"name":"2","interesting_key":"again_some_other_value",}]

function findNestedValues(array, key) {
  return array.reduce((r, obj) => {
    if(key in obj) 
      r.push(obj[key])
    
    if(obj.list_of_objects)
      r.push(...findNestedValues(obj.list_of_objects, key))
    
    return r
  }, [])
}

console.log(findNestedValues(list_of_objects, "interesting_key"))

Answer 2

这是一个字符串版本

const re = /"interesting_key":"(.*?)"/g

const str=JSON.stringify({"list_of_objects":[{"name":"0","interesting_key":"some_value",},{"name":"1","interesting_key":"some_other_value","list_of_objects":[{"name":"10","interesting_key":"something"},{"name":"11"},]},{"name":"2","interesting_key":"again_some_other_value",}]});
let result = [];

while ((match = re.exec(str)) != null) {
  result.push(match[1])
}
console.log(result)

更短

const re = /"interesting_key":"(.*?)"/g

const str=JSON.stringify({"list_of_objects":[{"name":"0","interesting_key":"some_value",},{"name":"1","interesting_key":"some_other_value","list_of_objects":[{"name":"10","interesting_key":"something"},{"name":"11"},]},{"name":"2","interesting_key":"again_some_other_value",}]});

let result = []
str.replace(re,(a,b) => result.push(b) )
console.log(result)

Answer 3

您可以创建一个递归函数，该函数遍历数据，如以下代码所示：

var list = [
  {
      "name" : "0",
      "interesting_key" : "some_value",
  },
  {
      "name" : "1",
      "interesting_key" : "some_other_value",
      "list_of_objects" : [
          {
              "name" : "10",
              "interesting_key" : "something"
          },
          {
              "name" : "11"
          },
      ]
  },
  {
      "name" : "2",
      "interesting_key" : "again_some_other_value",
  }
]
var interestingKey = [];
function getInterestingKey (list) {
  list.forEach(function(item) {
    if (item.interesting_key) {
      interestingKey.push(item.interesting_key);
    }
    if (item.list_of_objects) {
      getInterestingKey(item.list_of_objects);
    }
  })
  return interestingKey;
}

console.log(getInterestingKey(list))

根据对象嵌套列表中的键获取值

3 个答案: