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javascript - 如何检查特定对象键是否作为另一个对象键的值存在？

问题描述

要求：

我必须迭代多个对象，它包含属性。所以我必须迭代每个对象并检查与第一个对象键匹配的属性值（键名是映射）。

例如：

Node1, Node2, Node3

JSON: 
 {
  "Node1": {
    "type": "object",
    "properties": {
      "x": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "y": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node2": {
    "type": "object",
    "properties": {
      "body": {
        "type": "string",
        "mapping": "Node1.inputs.x",
        "checked": false,
        "defChecked": true
      },
      "subject": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "toemail": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "body_placeholders": {
        "type": "string",
        "mapping": "",
        "checked": true,
        "defChecked": true
      },
      "subject_placeholders": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node3": {
    "type": "object",
    "properties": {
      "email": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "Node2.inputs.toemail",
        "isSelected": true,
        "defChecked": true
      },
      "firstname": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      },
      "lastname": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "Node2.inputs.body_placeholders",
        "isSelected": true,
        "defChecked": true
      },
      "id": {
        "type": "uuid",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      }
    },
    "defChecked": true
  }
}

输出：

{
  "Node1": {
    "type": "object",
    "properties": {
      "y": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node2": {
    "type": "object",
    "properties": {
      "subject": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
     
      "subject_placeholders": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node3": {
    "type": "object",
    "properties": {
      "firstname": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      },
      
      "id": {
        "type": "uuid",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      }
    },
    "defChecked": true
  }
}

对于上面的 JSON - Node1 有一个属性'x' 并且这个属性映射或没有另一个对象属性（Node2 和 Node3）。如果您看到它 x 与 Node2 映射的 JSON，例如 (Node1.inputs.x)

如果匹配，那么我必须从 json 中删除对象 Node1.x。对其他人相似

标签： javascriptangulartypescript

解决方案

你可以这样做：

let input = {
  "Node1": {
    "type": "object",
    "properties": {
      "x": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "y": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node2": {
    "type": "object",
    "properties": {
      "body": {
        "type": "string",
        "mapping": "Node1.inputs.x",
        "checked": false,
        "defChecked": true
      },
      "subject": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "toemail": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      },
      "body_placeholders": {
        "type": "string",
        "mapping": "",
        "checked": true,
        "defChecked": true
      },
      "subject_placeholders": {
        "type": "string",
        "mapping": "",
        "checked": false,
        "defChecked": true
      }
    },
    "defChecked": true
  },
  "Node3": {
    "type": "object",
    "properties": {
      "email": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "Node2.inputs.toemail",
        "isSelected": true,
        "defChecked": true
      },
      "firstname": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      },
      "lastname": {
        "type": "text",
        "description": null,
        "default": null,
        "mapping": "Node2.inputs.body_placeholders",
        "isSelected": true,
        "defChecked": true
      },
      "id": {
        "type": "uuid",
        "description": null,
        "default": null,
        "mapping": "",
        "isSelected": true,
        "defChecked": true
      }
    },
    "defChecked": true
  }
};

for (let i = 1; i < 3; i++) {
   Object.entries(input["Node" + (i + 1)].properties).forEach(prop => {
      if (prop[1].mapping.includes("Node" + i)) {
        let attrToCancel = prop[1].mapping.split('.')[prop[1].mapping.split('.').length -1];
     
        delete input["Node" + i].properties[attrToCancel];
        delete input["Node" + (i + 1)].properties[prop[0]];
      }
   });
}

console.log(input)

获取Object.entries输入，然后拦截mapping属性，读取它并取消上一个属性Node。

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