首页 > 解决方案 > 如何正确地将数据传递给映射方法

问题描述

我想在谷歌地图上添加一个标记。index.ts我将从根文件开始追溯:

import { rawData } from "./csvFileReader";
import { Address } from "./Address";

import { CustomMap } from "./CustomMap";

const ROW_DELIMITER = "\r\n";

// console.log(data);

const data: Address[] = [];

// const rawRow = data.toString();
// const address = new Address(rawData);

for (const rawRow of rawData.split(ROW_DELIMITER)) {
  data.push(new Address(rawRow));
}

console.log(data);

// const address = data;

// console.log(address);

// console.log(new Address(data.toString()));

const customMap = new CustomMap("map");
customMap.addAddressMarker();

这会导入,csvFileReader所以我不必到处重复:

import { readFileSync } from "fs";

export const rawData = readFileSync("src/Cales_trim_down.csv", {
  encoding: "utf-8",
});

它还导入了Address类:

import { rawData } from "./csvFileReader";

export class Address {
  Address: string;
  // General_Plan_Designation: string;
  Latitude: number;
  Longitude: number;

  static DELIMITER = ",";

  constructor(rawRow: string) {
    const data = rawRow.split(Address.DELIMITER);

    this.Address = data[0];
    // this.General_Plan_Designation = data[1];
    this.Latitude = parseFloat(data[2]);
    this.Longitude = parseFloat(data[3]);
  }
}

const ROW_DELIMITER = "\r\n";

const data: Address[] = [];

for (const rawRow of rawData.split(ROW_DELIMITER)) {
  data.push(new Address(rawRow));
}

这是我的CustomMap课:

import { Address } from "./Address";

export class CustomMap {
  private googleMap: google.maps.Map;

  constructor(divId: string) {
    this.googleMap = new google.maps.Map(document.getElementById(divId), {
      zoom: 1,
      center: {
        lat: 0,
        lng: 0,
      },
    });
  }

  addAddressMarker(address: Address): void {
    new google.maps.Marker({
      map: this.googleMap,
      position: {
        lat: address.Latitude,
        lng: address.Longitude,
      },
    });
  }
}

我成功获得了谷歌地图,但我需要帮助的是如何获取该data属性并将其传递customMap.addAddressMarker();

这些是我尝试过的不同方法:

// console.log(new Address(data.toString()));

标签: javascripttypescript

解决方案


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