-1
function getUsername()
{
var userName = document.form.screen_name.value;
document.getElementById("display").innerHTML = userName;
var apiName = "https://api.twitter.com/1/users/lookup.json?screen_name=" + userName + "&callback=?";
document.getElementById("display2").innerHTML = apiName;
$(document).ready(function(){
$.getJSON(apiName, function(twitter) {
alert(twitter.name);
$('#showdata').html("<p>item1="+twitter.follwers_count+" item2="+twitter.friends_count+"</p>");
});
});
Javascript代码。Twitter JSON API与jQuery
<form method="get" action="#" name="form">
Username: <input type="text" name="screen_name" id="username"/>
<input type="submit" value="submit" onclick="getUsername()" />
</form>
<p>Your username is <h2 id="display"></h2></p>
<p>Your api url is <h2 id="display2"></h2></p>
HTML代码
哪些错误与此代码?警报返回未定义。 谢谢
衷心感谢你,先生,就像一个魅力。 – user1137834 2012-01-09 02:42:11