Javascript 如何从链接获取JSON数据并将其解析为HTML
这是我的,我必须向它发出请求,检索数据,将其放入JavaScript并显示在页面上,但我的请求会产生一个空响应。可能有什么问题,我怎样才能做得更好Javascript 如何从链接获取JSON数据并将其解析为HTML,javascript,html,ajax,client-server,Javascript,Html,Ajax,Client Server,这是我的,我必须向它发出请求,检索数据,将其放入JavaScript并显示在页面上,但我的请求会产生一个空响应。可能有什么问题,我怎样才能做得更好 // Get the modal element var modal = document.getElementById('simpleModal'); var modalBtn = document.getElementById('modalBtn'); var closeBtn = document.getElementById('
// Get the modal element
var modal = document.getElementById('simpleModal');
var modalBtn = document.getElementById('modalBtn');
var closeBtn = document.getElementById('closeBtn');
modalBtn.addEventListener('click', openModal);
var requestResultData;
// Listen for a close click
closeBtn.addEventListener('click', closeModal);
//Outside click
window.addEventListener('click', clickOutside);
// Function
function openModal() {
modal.style.display = 'block';
dataRequest();
}
function dataRequest () {
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState === 4){
console.log(xhr);
}
};
xhr.open('GET','http://www.omdbapi.com/?i=tt5687270&apikey=480344f1');
xhr.send();
}
// function to close modal
function closeModal() {
modal.style.display = 'none';
}
// Function for an outside click
function clickOutside(e) {
if(e.target == modal) {
modal.style.display = 'none';
}
}
您可以使用fetchapi
fetch("https://yourwebsite.com/data.json")
.then(res=>res.json())
.then(res=>{
//do something with the data here
console.log(res);
});
关于获取api的更多信息错误“已达到请求限制!”,但这是暂时的,您想将响应放在哪里?在模态中?