【发布时间】:2018-04-03 00:31:24
【问题描述】:
标题几乎说明了问题所在。我已将 URL 输入到地址栏中,并且返回正常。我觉得它有点愚蠢,我只是看不到它,因为我已经盯着它看了一段时间,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)
}
}
})();
【问题讨论】:
-
您还请求 JSON,但警告 responseXML?
标签: javascript xmlhttprequest openweathermap