从服务返回的地理位置值未定义

Question

我已将地理定位调用置于我想在应用程序的不同区域使用的服务中。但是，未定义正在被返回。如何正确返回值以便检索？

服务代码

   getUserPosition() {
     this.geolocation.getCurrentPosition().then((position) => {
    this.latLng = new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
    this.currentLat = position.coords.latitude;
    this.currentLng = position.coords.longitude;
    return this.latLng;

    }, (err) => {
      console.log('There was an error');
    });
  }

从服务调用这个函数

  async ngOnInit() {
    // Since ngOnInit() is executed before `deviceready` event,
    // you have to wait the event.
    await this.platform.ready();
    await this.loadMap();

   this.val =  await this.location.getUserPosition(); 
   console.log("I should be here val "+ this.val);
  }

Answer 1

从函数返回详细信息

async getUserPosition() {
    await this.geolocation.getCurrentPosition().then((position) => {
      this.latLng = new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
      this.currentLat = position.coords.latitude;
      this.currentLng = position.coords.longitude;
      this.locationData = [this.latLng, this.currentLat, this.currentLng];
    }, (err) => {
      console.log('There was an error' + err.message);
    });
    // return all the details returned from the location
    return this.locationData;
  }

使用返回的详细信息

   this.geolocationData = await this.location.getUserPosition();
    this.latLng = this.geolocationData[0];
    this.currentLat = this.geolocationData[1];
    this.currentLng = this.geolocationData[2];

Answer 2

您似乎没有在 getUserPosition 函数中返回任何内容。你看，return this.latLng; 是then 的结果，而不是getUserPosition 的结果。为了解决这个问题，你可以等到 promise 完成然后返回值：

async getUserPosition() {
    try {
        const position = await this.geolocation.getCurrentPosition();
        this.latLng = new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
        this.currentLat = position.coords.latitude;
        this.currentLng = position.coords.longitude;
        return this.latLng;
    }, (err) => {
      console.log('There was an error');
    });
}

我强烈建议使用 typescript 来避免这种类型的错误——如果你正确地为函数添加类型，编译器会告诉你函数应该返回一个特定的类型，但会尝试返回 void。