Javascript 如何从链接获取JSON数据并将其解析为HTML

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('

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('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?