Question

我有一个Web服务，它从db上的select返回一个json。问题是这种对象是层次结构的一部分，其中每个子节点都与父节点连接，并带有一个名为“idCdcParent”的字段。根元素是idCdcParent为null的位置。在这种情况下，罗马，米兰和那不勒斯是根本元素。这是json：

[{

  "idCdc": "2",
  "cdcName": "Roma",
  "order": "1",
  "isUsed": "false"
}, {

  "idCdc": "17",
  "idCdcParent": "5",
  "cdcName": "testGP",
  "order": "1",
  "isUsed": "false"
}, {

  "idCdc": "5",
  "idCdcParent": "2",
  "cdcName": "Progetti",
  "order": "2",
  "isUsed": "false"
}, {

  "idCdc": "18",
  "idCdcParent": "5",
  "cdcName": "testGPS",
  "order": "2",
  "isUsed": "false"
}, {

  "idCdc": "3",
  "cdcName": "Milano",
  "order": "4",
  "isUsed": "false"
}, {

  "idCdc": "7",
  "idCdcParent": "3",
  "cdcName": "l",
  "order": "4",
  "isUsed": "false"
}, {

  "idCdc": "4",
  "cdcName": "Napoli",
  "order": "5",
  "isUsed": "false"
}, {

  "idCdc": "9",
  "idCdcParent": "4",
  "cdcName": "cccc",
  "order": "6",
  "isUsed": "false"
}]

我希望它是一个具有层次结构的json，其中字段“idCdcParent”指的是父对象的id。像这样：

[{

    "idCdc": "2",
    "cdcName": "Roma",
    "order": "1",
    "isUsed": "false",
    "children": [{

        "idCdc": "5",
        "idCdcParent": "2",
        "cdcName": "Progetti",
        "order": "2",
        "isUsed": "false"
      },
      {

        "idCdc": "17",
        "idCdcParent": "5",
        "cdcName": "testGP",
        "order": "1",
        "isUsed": "false"
      }
    ]
  },
  {

    "idCdc": "5",
    "idCdcParent": "2",
    "cdcName": "Progetti",
    "order": "2",
    "isUsed": "false"
  },
  {

    "idCdc": "3",
    "cdcName": "Milano",
    "order": "4",
    "isUsed": "false",
    "children": [{
      "idCdc": "7",
      "idCdcParent": "3",
      "cdcName": "l",
      "order": "4",
      "isUsed": "false"
    }]
  },
  {
    "idCdc": "4",
    "cdcName": "Napoli",
    "order": "5",
    "isUsed": "false",
    "children": [{


      "idCdc": "9",
      "idCdcParent": "4",
      "cdcName": "cccc",
      "order": "6",
      "isUsed": "false"
    }]
  }
]

在javascript中可能吗？谢谢

Answer 1

确定有可能，只需遍历数组并为id建立一个查找表：

const byID = {};

for(const el of array) 
  byId[el.idCdc] = el;

现在您已经获得了查找表，它很容易遍历节点，并将它们转换为层次结构：

let root;

for(const el of array) {
  if(el.idCdcParent) {
     const parent = byID[el.idCdcParent];
     if(!parent.children) parent.children = [];
     parent.children.push(el);
  } else {
     root = el;
  }
}

Answer 2

您可以使用递归函数将实体数组转换为树。此方法创建新数组的层次结构实体，并且原始版本未被修改 - 可能是加号或减号，具体取决于您的特定用例：

function make_tree(treeable_entities) {
  function TreeNode(entity) {
    const children = get_children(entity.idCdc);
    const new_entity = Object.assign({}, entity);
    new_entity.children = children; 
    return new_entity;
  }

  function get_children(ofIdentifier) {
    return treeable_entities.filter(entity => entity.idCdcParent === ofIdentifier).map(TreeNode);
  }

  const is_root = (entity) => entity. idCdcParent === undefined;
  return treeable_entities.filter(is_root).map(TreeNode);
};

定义层次结构Json结构

2 个答案: