删除具有空数组值的对象

Question

所以我在json 文件中有一个对象，如下所示：

{
    "0": {
        "damage_type": "Scratch",
        "regions": []
    },
    "1": {
        "damage_type": "Dent",
        "regions": []
    },
    "2": {
        "damage_type": "Dent",
        "regions": [
            "front side",
            "front window"
        ]
    }
}

我想要完成的是删除具有空regions 的对象。像这样：

{
    "2": {
        "damage_type": "Dent",
        "regions": [
            "front side",
            "front window"
        ]
    }
}

我正在使用 for 循环，但仍然不成功：

jsonfile.readFile(theJsonFile, function (err, obj) {
    if (err) console.error(err)

    for (var i = 0; i <= Object.keys(obj).length - 1; i++) {
      if (obj[i].damage_type.length < 1) {
        delete obj[i]
      } 
    }
}

有什么想法吗？

Answer 1

你可以通过key删除一个json节点

var json = { ... };
var key = "foo";
delete json[key]; // Removes json.foo from the dictionary.

Answer 2

为什么您的代码不起作用？

您从每个节点错误地检查damage_type.length 的条件。您需要删除带有空regions 的节点。所以你必须检查regions.length。

您的有效解决方案

const obj = { 
  "0": { "damage_type": "Scratch", "regions": [] },
  "1": { "damage_type": "Dent", "regions": [] },
  "2": { "damage_type": "Dent", "regions": [ "front side", "front window" ] }
};

for (var i = 0; i < Object.keys(obj).length; i++) {
  if (!obj[i].regions || obj[i].regions.length === 0) {
    delete obj[i]
  } 
}
console.log(obj);

此解决方案之所以有效，是因为对象的键与索引“0”、“1”和“2”相同。如果密钥是其他东西，这将不起作用。

更正的解决方案

使用Object.keys 获取密钥并删除正确的节点。

工作小提琴

const obj = { 
  "0": { "damage_type": "Scratch", "regions": [] },
  "1": { "damage_type": "Dent", "regions": [] },
  "2": { "damage_type": "Dent", "regions": [ "front side", "front window" ] }
};
const keys = Object.keys(obj);
for (var i = 0; i <= keys.length - 1; i++) {
  if (!obj[keys[i]].regions || obj[keys[i]].regions.length === 0) {
    delete obj[keys[i]]
  } 
}
console.log(obj);

通用解决方案

使用Object.entries获取键值组合

const obj = { 
  "0": { "damage_type": "Scratch", "regions": [] },
  "1": { "damage_type": "Dent", "regions": [] },
  "2": { "damage_type": "Dent", "regions": [ "front side", "front window" ] }
};
const entries = Object.entries(obj);
entries.forEach(([key, value]) => {
  if (!value.regions || value.regions.length === 0) {
    delete obj[key]
  }
})
console.log(obj);

Answer 3

做出改变，试试这个

let json = {
    "0": {
        "damage_type": "Scratch",
        "regions": []
    },
    "1": {
        "damage_type": "Dent",
        "regions": []
    },
    "2": {
        "damage_type": "Dent",
        "regions": [
            "front side",
            "front window"
        ]
    }
};


function walk(obj) {
  for (var key in obj) {
     let length = Object.keys(obj[key].regions).length; // get the length
     if(length==0){
       delete obj[key];
       
     }
  }
 console.log(json);
}
walk(json);

输出

{ '2':
   { damage_type: 'Dent',
     regions: [ 'front side', 'front window' ]
 } 
}

Answer 4

调用解析后的值damages 而不是obj，您可以将所有条目作为键/值对抓取，对其进行迭代并通过其ID 删除无区域“损坏”：

Object.entries(damages).forEach(([id, damage]) => {
  if (!damage.regions.length) {
    delete damages[id];
  }
});

或者，您可以在不触及原始结构的情况下创建一个全新的结构：

const entries = Object.entries(damages);
const filteredEntries = entries.filter(([_, damage]) => damage.regions.length);
const filteredDamages = Object.fromEntries(filteredEntries);

Answer 5

在这种情况下，首先你需要一个数组。要将对象转换为数组，请使用条目entry

let obj = {
    "0": {
        "damage_type": "Scratch",
        "regions": []
    },
    "1": {
        "damage_type": "Dent",
        "regions": []
    },
    "2": {
        "damage_type": "Dent",
        "regions": [
            "front side",
            "front window"
        ]
    }
}

转换数组：

const myentry= Object.entries(obj);

现在消除或过滤数组的最好方法是使用过滤器方法，然后获取另一个具有过滤器值的对象。

Filter

const myFiletrObj = myentry.filter(([_, obj]) => obj.regions.length).reduce(withObjectAssign, {})

function withObjectAssign(object, [key, value]) {
  return Object.assign(object, {[key]: value})
}

这将为您提供过滤后的输出。