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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