Using AJAX, Javascript and PHP with MySQL to display search results -


my aim use ajax display php search results without having reload page. far have been able results, (i'm new ajax, , not know jquery), problem search results displaying in html table appear @ top page above everything, not in specified div. have used innerhtml try , display correctly.

here main code: 

    function searchresults(title) {         if (title == "") {             document.getelementbyid("response").innerhtml="";         }         var request= new xmlhttprequest();         request.onreadystatechange=function() {              if (request.readystate == 4 && request.status == 200) {                 var displaydiv= document.getelementbyid("response");                 displaydiv.innerhtml=request.responsetext;             }         }              request.open("get", "functions.php?titlea="+title, true);             request.send();             document.getelementsbyid("response").innerhtml="fuck";     }     </script>     <title>anime search</title> </head> <body>     <div class="main">          <div class= "header">         <h1>search database</h1>     </div> <!-- close header -->      <div class= "searcha">         <p>search anime database below</p>         <form onsubmit="searchresults(titlea)">             <label>title</label><input type="text" name="titlea" placeholder="enter title"/>     <input type="submit" value="submit"/>         </form>                  <div id="response">      </div> <!-- close response --> </div> <!-- close searcha -->

and here php:

if (isset($_get["titlea"])) {     $title= $_get["titlea"];     $connection= connect();      $username= $_session["username"];     $tablea= $username . "_anime";     $querya= "select * anime." . "`$tablea` title '%$title%'";     $resulta= mysqli_query($connection, $querya);      if ($resulta == false) {         die("no results found");     }      $numrows= mysqli_num_rows($resulta);      echo "<table class= 'tsearch'>             <thead>                 <th>title</th>                 <th>alt title</th>                 <th>seasons</th>                 <th>episodes</th>                 <th>ova's</th>                 <th>movies</th>                 <th>status</th>                 <th>downloaded</th>             </thead>             <tbody>";      while($row= mysqli_fetch_array($resulta, mysqli_both)) {                 echo "<tr>";                     echo "<td>" . $row["title"] . "</td>";                     echo "<td>" . $row["alt_title"] . "</td>";                     echo "<td>" . $row["seasons"] . "</td>";                     echo "<td>" . $row["total_episodes"] . "</td>";                     echo "<td>" . $row["ovas"] . "</td>";                     echo "<td>" . $row["movies"] . "</td>";                     echo "<td>" . $row["status"] . "</td>";                     echo "<td>" . $row["downloaded"] . "</td>";                  echo "</tr>";              }                  echo "</tbody>";             echo "</table>";              mysqli_close($connection);                  if ($resulta == false) {                     echo mysqli_error($connection);                 }             }

i have of course spend many hours trying find out whats wrong, plan on learning jquery, want work, please not tell me use jquery,

edit: link screenshot: http://i.imgur.com/ejkhati.png browser safari 7.0.4, tried firefox , got same problem thank in advance, rainy

here, let me give generic layout:

this may easiest way use ajax, use in every project. first link external ajax.js can use below script.

i don't know have done wrong base on description, works me locally. 1 reason may #response isn't loaded yet when execute script.

ajax.js

function ajaxobj( meth, url ) {     var x;     if (window.xmlhttprequest)         { //new browsers             x = new xmlhttprequest();         }     else         { //ie5, ie6         x = new activexobject("microsoft.xmlhttp");         }     x.open( meth, url, true );     x.setrequestheader("content-type", "application/x-www-form-urlencoded");     return x; } function ajaxreturn(x){     if(x.readystate == 4 && x.status == 200){        return true;      } }

script tag request ajax:

var ajax = ajaxobj("get", "functions.php");     ajax.onreadystatechange = function() {         if(ajaxreturn(ajax) == true) {             document.getelementbyid("response").innerhtml=ajax.responsetext;         }     } ajax.send("titlea="+title);

or if prefer jquery:

//you need load jquery first before using $(function() {  //this line means when document ready     var ajax = ajaxobj("get", "functions.php");         ajax.onreadystatechange = function() {             if(ajaxreturn(ajax) == true) {                 $("#response").html(ajax.responsetext);             }         }     ajax.send("titlea="+title); });

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