ahmadkavand Posted February 28, 2010 Share Posted February 28, 2010 HiMake ajax in web.If I type my username in textbox, If there is my name in database,echo "user name is invalid"elseecho "username is valid";please help me. please compelet my php filepleasei will send html,js,php file in below.//////////////html file/////////////////////<html><head><script type="text/javascript" src="selectcustomer.js"></script></head><body><form>username:<input type="text" onkeyup="showCustomer(this.value)"></form><div id="txtHint"><b>look at.</b></div></body></html>//////////////////////////////////////////////////////////////////////////////////////javascript file/////////////////////var xmlhttpfunction showCustomer(str){xmlhttp=GetXmlHttpObject();if (xmlhttp==null) { alert ("Your browser does not support AJAX!"); return; }var url="getuser.php";url=url+"?q="+str;url=url+"&sid="+Math.random();xmlhttp.onreadystatechange=stateChanged;xmlhttp.open("GET",url,true);xmlhttp.send(null);}function stateChanged(){if (xmlhttp.readyState==4) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; }}function GetXmlHttpObject(){if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari return new XMLHttpRequest(); }if (window.ActiveXObject) { // code for IE6, IE5 return new ActiveXObject("Microsoft.XMLHTTP"); }return null;}////////////////////////////////////////////////////////////////////////////////////////php file/////////////////////////<?php$q=$_GET["q"];$con = mysql_connect("localhost", "root", "");if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("a1", $con);$sql="select username from a2 where firstname = $q;$result = mysql_query($sql);........mysql_close($con);?> please help me Link to comment Share on other sites More sharing options...
jeffman Posted February 28, 2010 Share Posted February 28, 2010 The first example here should give you some ideas. Link to comment Share on other sites More sharing options...
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