eZe616 Posted July 16, 2007 Share Posted July 16, 2007 Ok...I'm fairly new to ajax. I'm trying to build a simple ajax login form. I got so stuck that I'm just trying to understand getting POST values with the .open method. The Javascript I have so far is function ajaxfuntion() { var input1 = document.getElementById("text").value; xmlhttp.open("POST",t.php); xmlhttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); xmlhttp.onreadystatechange = function (){ if (xmlhttp.readyState == 4 && xmlhttp.status == 200 ) { document.getElementById("loginf").innerHTML = xmlhttp.responseText; } } xmlhttp.send(input1); } </script> The .php file to process the values being send <?php $txt = $_POST["text"]; echo $txt; ?> The HTML form code is as follow <div class="loginf"> <form action="" name="test" id="test" onSubmit="ajaxfunction(); return false;" method="post"> <fieldset> <label> <input type="text" name="text" id="text" maxlength="32" class="user" /> </label> <label> <input type="submit" value="test"/> </label> </fieldset> </form> </div> What I'm trying is to get the .php's echo statement to replace the form in the loginf div Is there anything wrong with my code? Quote Link to comment Share on other sites More sharing options...
eZe616 Posted July 16, 2007 Author Share Posted July 16, 2007 A head up.. I found a mistake in the code. where class="loginf" -> id="loginf" and in xmlhttp.open("POST",t.php); -> xmlhttp.open("POST","t.php"); but still not working thought ANybody have any idea as to why? Quote Link to comment Share on other sites More sharing options...
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