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@kjam
Created July 21, 2011 15:09
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Twitter Search
<html>
<head>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$.getJSON('http://search.twitter.com/search.json?q=earthquake&lang=en&callback=?', function(data) {
var data = data.results;
var html = "<ul>";
for(var i=0; i<data.length; i++) {
html += "<li><a href='http://twitter.com/" + data[i].from_user + "'>@"
+ data[i].from_user + "</a>: " + data[i].text + "</li>";
}
html += "</ul>"
$('.content').html(html);
});
});
</script>
</head>
<body>
<h2>Twitter</h2>
<div class="content">
</div>
</body>
</html>
@walid80
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walid80 commented Sep 10, 2014

<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $.getJSON('http://search.twitter.com/search.json?q=earthquake&lang=en&callback=?', function(data) { var data = data.results; var html = "
    "; for(var i=0; i@" + data[i].from_user + ": " + data[i].text + ""; } html += "
" $('.content').html(html); }); }); </script>

Twitter

@elserafi
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The Twitter REST API v1 is no longer active and should migrate to API v1.1. That's why the code mentioned above doesn't work.

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