XMLHTTPRequest 请求打开天气返回状态 0

Question

标题说的差不多了。我已经在我的地址栏中输入了 URL，它返回正常。我觉得它有些愚蠢，我只是看不到它，因为我已经盯着它看了一段时间了，但这是代码。

(function() {
  var httpRequest;
  document.getElementById("weatherButton").addEventListener('click', makeRequest);
  function makeRequest() {
    httpRequest = new XMLHttpRequest();
    if (!httpRequest) {
      alert('Cannot create XMLHTTP instance!');
      return false;
    }
    httpRequest.onreadystatechange = alertContents;
    httpRequest.open("GET", "api.openweathermap.org/data/2.5/weather?zip=94040&APPID=xxxxxxxxx&mode=json");
    httpRequest.send();
  }
  function alertContents() {
    if (httpRequest.readyState != 4) {
      return;
    }
    if (httpRequest.status == 200) {
      alert(httpRequest.responseXML);
    }
    if (httpRequest.status != 200) {
      alert(httpRequest.status + ": " + httpRequest.statusText_)
      alert(httpRequest.readyState)

    }
  }
})();

Answer 1

尝试指定完整路径：

httpRequest.open("GET", "https://api.openweathermap.org/data/2.5/weather?zip=94040&APPID=xxxxxxxxx&mode=json");

否则，它寻找[yourdomain]/api.openweathermap.org/...。

XMLHTTPRequest 请求打开天气返回状态 0

XMLHTTPRequest Request To Open Weather Returning Status 0

javascript

xmlhttprequest

openweathermap