javascript - 用户允许定位后刷新页面

Question

我正在使用 Nominatim API 从纬度和经度获取用户的地址(使用地理位置 HTML5 API 获取这些地址)。这是我的地理定位 API 代码块:

function geoFindMe() {
  function success(position) {
    localStorage.setItem('pos_lat', position.coords.latitude);
    localStorage.setItem('pos_lon', position.coords.longitude);
  }

  function error() {
    elLocation.textContent = 'Unable to retrieve your location';
  }

  if (!navigator.geolocation) {
    elLocation.textContent = 'Geolocation is not supported by your browser';
  } else {
    elLocation.textContent = 'Locating…';
    navigator.geolocation.getCurrentPosition(success, error);
  }
}

我正在将用户的纬度和经度保存在本地存储中，因为我没有找到返回这些值的方法。我需要这些，因为在其他函数中我使用 Nominatim API 来获取经纬度地址。

现在，我的问题是，用户第一次访问网站时，他的 localStorage 是空的。当用户不允许我知道他的位置时，无法获取地址。我的解决方案是在用户允许后刷新页面，但我不知道该怎么做，有人可以帮助我解决这个或其他解决方案吗？

编辑:我将只添加获取地址的代码块:

async function getLocationData() {
  geoFindMe();
  const latitude = localStorage.getItem('pos_lat');
  const longitude = localStorage.getItem('pos_lon');
  let url = `https://nominatim.openstreetmap.org/reverse?format=json&lat=${latitude}&lon=${longitude}`;
  const response = await fetch(url);
  if(!response.ok) throw response;
  const data = await response.json();

  let locationAddress = {};
  locationAddress.village = data.address.village;
  locationAddress.city = data.address.city;
  locationAddress.country = data.address.country;
  locationAddress.countryCode = data.address.country_code;

  return locationAddress;
}

Answer 1

尝试这样的事情......

  function getLocationData() {
        if (navigator.geolocation) {
            return navigator.geolocation.getCurrentPosition(success, fail);
        } else {
            return 'Sorry, your browser does not support geolocation services.';
        }
    }

    async function success(position) {
        let url = `https://nominatim.openstreetmap.org/reverse?format=json&lat=${position.coords.latitude}&lon=${position.coords.longitude}`;
        const response = await fetch(url);
        if (!response.ok) throw response;
        const data = await response.json();

        let locationAddress = {};
        locationAddress.village = data.address.village;
        locationAddress.city = data.address.city;
        locationAddress.country = data.address.country;
        locationAddress.countryCode = data.address.country_code;
        console.log(locationAddress);
        return locationAddress;
    }

    function fail() {
        return 'Fail...';
    }

    console.log(getLocationData())