我在弄清楚如何将Google地理编码器Api与另一个API结合使用时遇到了麻烦

Question

我尝试使用google api来获取经纬度坐标，然后将这些坐标传递给另一个根据位置查找医生的api。这两个APIS彼此独立工作，但我在使用promises将它们配置为协同工作时遇到了问题。

$(document).ready(function(){
  $("#submit").submit(function(){
    event.preventDefault();
    let newQuery = new ApiCall();
    const symptom = $("#user-input").val();
    const location = $("#user-location").val();
    let locationPromise = newQuery.locationCall(location);
    locationPromise.then(function(response){
    let locationBody = JSON.parse(response);
    },function(error){
      $('.showErrors').text(`There was an error processing your 
request: ${error.message}`);
    }).then(function(result){
    let promise = newQuery.newDataCall(symptom,locationPromise);
    promise.then(function(response){
      let body = JSON.parse(response);
      console.log(body);
      const output = parseData(body);
      const display = parseString(output);
      $(".output-field").html(display);
    }, function(error) {
      $('.showErrors').text(`There was an error processing your 
 request: ${error.message}`);
     });
   })
 });
});

我可以通过控制台记录API的两个结果，但当我尝试将位置传递给betterdoctor api时，它不会返回位置。

Answer 1

我不确定newQuery.newDataCall(symptom, locationPromise)是如何工作的，但我认为您误用了locationPromise。假设newDataCall的第二个参数接受位置坐标，那么您的代码应该如下所示：

$(document).ready(function(){
  $("#submit").submit(function(){
    event.preventDefault();
    let newQuery = new ApiCall();
    const symptom = $("#user-input").val();
    const location = $("#user-location").val();
    let locationPromise = newQuery.locationCall(location);
    locationPromise.then(function(response){
    let locationBody = JSON.parse(response);
    // return the result of the API call
      return locationBody;
    },function(error){
      $('.showErrors').text(`There was an error processing your 
request: ${error.message}`);
    }).then(function(result){
    // result === locationBody
    let promise = newQuery.newDataCall(symptom, result);
    promise.then(function(response){
      let body = JSON.parse(response);
      console.log(body);
      const output = parseData(body);
      const display = parseString(output);
      $(".output-field").html(display);
    }, function(error) {
      $('.showErrors').text(`There was an error processing your 
 request: ${error.message}`);
     });
   })
 });
});

希望这能有所帮助。