从嵌套 json 中过滤项目

Question

有一个json：x = [{"name":"Peter", "list":[{"position":"high", "id":"ZZ"},                         {"position":"low", "id":"KJ"}]},{"name":"Alise", "list":[{"position":"high", "id":"TC"},                         {"position":"low", "id":"ZZ"}]}]我需要使用filter()删除“list”中包含“id”：“ZZ”的那些项目。我期望最终的结果如下：[{"name":"Peter", "list":[{"position":"low", "id":"KJ"}]},{"name":"Alise", "list":[{"position":"low", "id":"ZZ"}]}]我尝试实现这一目标：y = for (i=0;i<x.length;++i){        for (n=0;n<x[i]["list"].length;++n){             x[i]["list"][n].filter(id => "id" == "ZZ")        }    }浏览器控制台输出显示Uncaught TypeError: x[i]["list"][n].filter is not a function.你能告诉我如何过滤给定的 json 并摆脱 "id":"ZZ" 行吗？

Answer 1

map您可以为每个元素和filter每个list元素执行 a

我在这里使用展开运算符来保持其他属性相同，只需修改list

const x = [

  {

    name: "Peter",

    list: [

      { position: "high", id: "ZZ" },

      { position: "low", id: "KJ" },

    ],

  },

  {

    name: "Alise",

    list: [

      { position: "high", id: "TC" },

      { position: "low", id: "ZZ" },

    ],

  },

];

const res = x.map((el) => ({

  ...el,

  list: el.list.filter(({ id }) => id !== "ZZ"),

}));

console.log(res);

Answer 2

// const objectScan = require('object-scan');

const myData = [ { name: 'Peter', list: [{ position: 'high', id: 'ZZ' }, { position: 'low', id: 'KJ' }] }, { name: 'Alise', list: [{ position: 'high', id: 'TC' }, { position: 'low', id: 'ZZ' }] } ];

const prune = (data, id) => objectScan(['[*].list[*]'], {

  rtn: 'count',

  filterFn: ({ value, property, parent }) => {

    if (value.id === id) {

      parent.splice(property, 1);

      return true;

    }

    return false;

  }

})(data);

console.log(prune(myData, 'ZZ')); // returns the number of deletions

// => 2

console.log(myData);

// => [ { name: 'Peter', list: [ { position: 'low', id: 'KJ' } ] }, { name: 'Alise', list: [ { position: 'high', id: 'TC' } ] } ]

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

<script src="https://bundle.run/object-scan@13.8.0"></script>