[ Skuffman @ 08.07.2011. 11:33 ] @
Imam problem koji nikako da riješim Naime imam neki link na koji kad kliknem ucita mi tablicu sa podacima, a na dnu tablice imam input field za dodati nesto u tablicu. I tu se javlja problem, jer ako ne koristim ajax nego samo imam tablicu onda insert radi, ali nakon ajax-a on ne radi.. u nastavku je kompletan kôd. Unaprijed zahvaljujem na pomoći index.php Code: <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("#insert").live("click",function() { var boxval = $("#content").val(); var dataString = 'content='+ boxval; if(boxval==''){ alert("Please Enter Some Text"); } else{ $.ajax({ type: "POST", url: "demo.php", data: dataString, cache: false, success: function(html){ $("table#update tbody").prepend(html); $("table#update tbody tr").slideDown("slow"); document.getElementById('content').value=''; } }); } return false; }); $(".load").live("click",function() { $.ajax({ type: "POST", url: "test.php", success: function(msg){ $("#container").ajaxComplete(function(event, request, settings){ $("#container").html(msg); }); } }); }); }); </script> </head> <body> <a href="#" class="load">load</a> <div id="container"> </div> </body> </html> demo.php (vrši insert u bazu i vraća podatke) Code: <?php $sql_check = mysql_query("SELECT * FROM xerox_evidencija order by id desc"); if(isset($_POST['content'])){ $content=$_POST['content']; mysql_query("insert into xerox_app_evidencija(kopirka_id) values ('$content')"); $sql_in= mysql_query("SELECT * FROM xerox_app_evidencija order by id desc"); $r=mysql_fetch_array($sql_in); $msg=$r['kopirka_id']; $msg_id=$r['id']; ?> <tr class="prva"> <td class="<?php echo $msg_id; ?>"><?php echo $msg; ?></td> </tr> <?php } ?> test.php (to se zove klikom na link, da se učita cijela tablica) Code: <table id="update" class="timeline" width="300" border="1"> <thead> <tr> <th>Naziv</th> <th>Akcija</th> </tr> </thead> <tbody> <?php $sql = "SELECT * FROM xerox_app_evidencija order by id desc"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { //print data ?> <tr> <td><?php echo $row['kopirka_id']; ?></td> </tr> <?php } ?> </tbody> </table> <form method="post" name="form" action=""> <input type="text" name="content" id="content" maxlength="145" ></input> <input type="submit" value="Update" name="submit" id="insert"></input> </form> zanimljivo je da nakon učitavanja tablice radi provjera da li je prazan input Svaka ideja zašto ne radi je dobrodošla ;) |