xmlhttp js 请求

<html><script> var xhr = new XMLHttpRequest();xhr.open(‘GET‘, "http://ipinfo.io/json", true);xhr.send(); xhr.onreadystatechange = processRequest; function processRequest(e) { if (xhr.readyState == 4 && xhr.status == 200) { var response = JSON.parse(xhr.responseText); alert(response.ip); }}</script></html>

  

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