CODEWITHSUNDEEP
phpjqueryhtmlcssajax
Ian Taylor
Ian Taylor

Reputation: 179

jQuery page refresh not reading or executing php

sorry if the title is a little.. vague i couldnt pin it down.

So i am developing a friend request system which works i guess similar in concept to facebook. So you get a request and it lists them without a page reload.

However i get the div 'refreshing' or so i think i cant test the php which is where i have a problem, i will post the relevent code and files below.

It may look a little long winded but it shouldnt be too bad in reality. My php code should keep executing the query which is looking at the database in the updateFriendBox.php however it doesnt seem to be doing this. My code may be messy as well so I apologise.

myaccount.php

<script src="http://code.jquery.com/jquery-latest.js"></script>
<script language="javascript" type="text/javascript">
function refreshDiv()
{
    $.get('updateFriendBox.php', function(data){$('#refresh').html(data);});
}
$(function()
{
    refreshDiv();
    setInterval(refreshDiv, 5000);
});


function box(x){
    if($('#'+x).is(":hidden")){
        $('#'+x).slideDown(200);
    } else {
        $('#'+x).hide();
    }
}
</script>

<?php
$addBox = '<div style="display:inline; padding:5px;">
<a href="#" onclick="return false" onmouseup="javascript:box(\'fRequ\');">Show/Hide Friend Requests</a>
</div>';

// a bit further down in the code where its all called:

 <a href="#" class="tooltip"><? echo $addBox; ?></span></a>
          <div class="friendSlide" id="fRequ" style="height:240px; overflow:auto;">Your friend requests: <br />
          <div id="refresh"> <?php // this is where the refresh call is ?>
        </div>
        </center>
        </div>
</div>
</div>

updateFriendBox.php:

<script src="http://code.jquery.com/jquery-latest.js"></script>
<script language="javascript" type="text/javascript">
function acceptFriendRequest(x) {
    var url = "friendParse.php";
    $.post(url,{ request: "acceptFriend", reqID: x}, function(data){
        $("#req"+x).html(data).show().fadeOut(5000);
    });
}

function denyFriendRequest(x) {
    var url = "friendParse.php";
    $.post(url,{ request: "denyFriend", reqID: x}, function(data){
        $("#req"+x).html(data).show().fadeOut(5000);
    });
}
</script>
</head>
  <body>
        <?php
        include 'dbc.php';
        $sql = "SELECT * FROM friendRecu WHERE mem2='" . $_SESSION['user_id'] . "' ORDER BY id ASC LIMIT 10";
        $query = mysql_query($sql)or die(mysql_error());
        $num_rows = mysql_num_rows($query);
        if($num_rows < 1) { 
            echo "No friend requests";
        } else {
            while($row = mysql_fetch_array($query)){
                $requestID = $row['id'];
                $req = $row['mem1'];
                $sqlName = mysql_query("SELECT full_name FROM users WHERE id='$req'");
                while($row = mysql_fetch_array($sqlName)){
                    $requesterName = $row['full_name'];
                }
                echo '<hr /><table width="100%", cellpadding="5"><tr><td width="17%" align="left"><div style="overflow:hidden; height:50px; color:white;"></div></td> <td width="83%"><a href=viewmembers.php?uid=' . $req . '">' . $requesterName . '</a> added you as a friend
                <span id="req' . $requestID . '">
                <a href="#" onclick="return false" onmousedown="javascript:acceptFriendRequest(' . $requestID . ');">Accept</a>
                &nbsp; || &nbsp;
                <a href="#" onclick="return false" onmousedown="javascript:denyFriendRequest(' . $requestID . ');">Deny</a>
                </span></td></tr>
                </table>';
            }
        }
        ?>

Upvotes: 1

Views: 576

Answers (1)

AndrewR
AndrewR

Reputation: 6758

I think you are having a problem because your updateFriendBox.php is returning too much. Remove all that inline JS code, place it in an include file, and include it from myaccount.php. You also shouldn't have <head> and <body> sections in your updateFriendBox.php file.

The ajax call here doesn't create a whole new page, you're getting additional HTML to add to the original page.

So the only thing you should have there is your SQL query, the loop, and your HTML output for each data row.

Upvotes: 2

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