javascript - 深入搜索属性值后返回对象

Question

我有一个函数可以在嵌套对象中成功搜索属性值中包含的关键字，它返回嵌套对象，但我希望它返回整个对象。

我正在搜索的数组如下所示:

var images = [
  {
    "node": {
      "comments_disabled": false,
      "edge_media_to_caption": {
        "edges": [
          {
            "node": {
              "text": "this text includes the hashtag #green"
            }
          }
        ]
      },
      "shortcode": "Byqehgyorxq"
    }
  },
  {
    "node": {
      "comments_disabled": false,
      "edge_media_to_caption": {
        "edges": [
          {
            "node": {
              "text": "this text includes the hashtag #red"
            }
          }
        ]
      },
      "shortcode": "Byqehgyorxq"
    }
  }
]

以下代码将返回 { "text": "this text includes the hashtag #red"}

function customFilter(object, result, key, value){
    if(object.hasOwnProperty(key) && object[key].includes(value))
        result.push(object);

    for(var i=0; i<Object.keys(object).length; i++){
        if(typeof object[Object.keys(object)[i]] == "object"){
            customFilter(object[Object.keys(object)[i]], result, key, value);
        }
    }
}
var result = []
customFilter(images, result, 'text', 'red');

我希望此函数返回整个节点对象，并具有所需的输出:

"node": { "comments_disabled": false, "edge_media_to_caption": { "edges": [ { "node": { "text": "this text includes the hashtag #red" } } ] }, "shortcode": "Byqehgyorxq"}

更新:修复无效对象

Answer 1

你可以通过查看对象的内部找到该对象。

function filter(array, key, value) {
    function find(object) {
        if (!object || typeof object !== 'object') return false;
        if (key in object && object[key].includes(value)) return true;
        return Object.values(object).some(find);
    }
    return array.find(find);
}

var images = [{ node: { comments_disabled: false, edge_media_to_caption: { edges: [{ node: { text: "this text includes the hashtag #red" } }] }, shortcode: "Byqehgyorxq" } }],
    result = filter(images, 'text', 'red');

console.log(result);

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