artka54 Posted March 17, 2012 Share Posted March 17, 2012 Okey, I know that this can be searched, but I don't understand it anyway. I am totally beginner in jquery. So I have test.php and test.html I use ajax call (can't get jquery call working) and get results, such as tweet a href. And I have jquery function that renders tweet a href with image. The problem is, that I am using that jquery function in a wrong way, because it is not working, although it should be. test.php <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js" type="text/javascript"></script> <script src="http://platform.twitter.com/widgets.js" type="text/javascript"></script> <script> function clickButton() { if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("test").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","test.html",true); xmlhttp.send(''); } /// this is the function That I should use to get twitter icons $('.topic-term-link').ajaxComplete(function(){ //re-render twitter icons by calling the render() method on each $('#test').find('a.twitter-share-button').each(function(){ var tweet_button = new twttr.TweetButton( $( this ).get( 0 ) ); tweet_button.render(); }); }); </script> <p onclick="clickButton();">eeee</p> <div id="test"></div> test.html <a href="http://twitter.com/share" class="twitter-share-button">Tweet</a> BEEE Link to comment https://forums.phpfreaks.com/topic/259156-twitter-button-after-ajax-call/ Share on other sites More sharing options...
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