Jump to content

Result Set Limit and Navigation problem

saikiran

By saikiran
in PHP Coding Help

 Share

Recommended Posts

saikiran Newbie

saikiran

Posted
Posted
hi,

I am running a photo gallery in my site. in which , i am having a thumbnail page and a popup image page,

what i am doing is when user clicks the thumbnail, the bigger picture should open in a popup in a separate page.

I would like to add a navigation link in the popup page. i.e. with in the popup page, user can go and view the rest of the images in the gallery also.


[color=red]i am providing this as 1,2 ,3,4,5.....[Next][Previous] .  format.

i am not able to run the navigation part properly....[/color]


This is what the code for the 2 pages:

[color=red]Thumbnails page code [/color]

[code]<?php include_once('config.php');
$psql = "select * from bhakt_photo";
$result = mysql_query($psql);

if($result)
{
    while($row = mysql_fetch_array($result,MYSQL_BOTH))
    {
        echo "<table>";
        echo "<tr>";
        echo "<td>";?>
        <a  href='bhakti_popnew.php?id=<?php echo $row['id'];?>'><img src="http://www.seher.in/gall_upload/ <?php echo $row['mid_img'];?>" border="0" ></a>     
        <?php
        echo "</td>";
        echo "</table>";
    }
}

?>
[/code]


-----Popup page code-------------


[code]<?php include('config.php');
//new code

$id=$_GET['id']; //this the Id i am getting for the images

// how many rows to show per page

$rowsPerPage = 1;

// by default we show first page

$pageNum = 1;

// if $_GET['page'] defined, use it as page number
if(isset($_GET['page']))
{
    $pageNum = $_GET['page'];
}

// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;
//$offset =  $rowsPerPage;

$query = " SELECT * FROM bhakt_photo where id=$id " .
        " LIMIT $offset, $rowsPerPage";

echo $query."<br>"        ;

$result = mysql_query($query) or die('Error, query failed');

?>
<body leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">

        <?php
        while($row = mysql_fetch_array($result))
        {
            echo $row['big_img'];
        } ?>
      </table>
      <br>
<br>
<?php // how many rows we have in database

$query  = "SELECT COUNT(big_img) AS numrows FROM bhakt_photo ";

$result  = mysql_query($query) or die('Error, query failed');
$row    = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows'];

// how many pages we have when using paging?
$maxPage = ceil($numrows/$rowsPerPage);

// print the link to access each page
$self = $_SERVER['PHP_SELF'];
$nav  = '';

for($page = 1; $page <= $maxPage; $page++)
{
  if ($page == $pageNum)
  {
      $nav .= " $page "; // no need to create a link to current page
   
  }
  else
  { 
      $nav .= " <a href=\"$self?page=$page\">$page</a> ";   
   
  }
}
if ($pageNum > 1)
{
  $page  = $pageNum - 1;     
  $id  = $id-1;
 
  $prev  = " <a href=\"$self?page=$page&id=$id\">[Prev]</a> ";
  $first = " <a href=\"$self?page=1&id=$id\">[First Page]</a> ";
 
  //$prev  = " <a href=\"$self?page=$page\">[Prev]</a> ";
  //$first = " <a href=\"$self?page=1\">[First Page]</a> ";
}
else
{
  $prev  = '&nbsp;'; // we're on page one, don't print previous link
  $first = '&nbsp;'; // nor the first page link
}

if ($pageNum < $maxPage)
{
  $page = $pageNum + 1; 
  $id = $id +1;
 
  echo "<b>". $id."<br>";

      $next = " <a href=\"$self?page=$page&id=$id\">[Next]</a> ";
      $last = " <a href=\"$self?page=$maxPage&id=$id\">[Last Page]</a> ";


      //$next = " <a href=\"$self?page=$page\">[Next]</a> ";
      //$last = " <a href=\"$self?page=$maxPage\">[Last Page]</a> ";
}
else
{
  $next = '&nbsp;'; // we're on the last page, don't print next link
  $last = '&nbsp;'; // nor the last page link
}

// print the navigation link
echo $first . $prev . $nav . $next . $last;

// and close the database connection


// ... and we're done!
?>[/code]

the problem what i assume is - the 'id' value is not passing in the subsquent pages properly.

i have checked readymade photogallery packages...but i would like to have my own. I am not an expert in PHP too.

can anyone just tell me whats the problem in my code and help me out in completing this one.


cheers
saikiran
saisen76@hotmail.com
Link to comment
Share on other sites
saikiran Newbie

saikiran

Posted
  • Author
Posted
hi,

this is how i am calling it into the popup.

sorry for not putting the popup one

<?php include_once('config.php');
$psql = "select * from bhakt_photo";
$result = mysql_query($psql);

if($result)
{
    while($row = mysql_fetch_array($result,MYSQL_BOTH))
    {
        echo "<table>";
        echo "<tr>";
        echo "<td>";?>
[b]<a  href="javascript:;" onClick = "window.open('bhakti_popnew.php?id=<?php echo $row['id'];?>','Gallery','height=700,width=589,scrolling=yes')"><img src="http://www.seher.in/uploads/<?php echo $row['mid_img'];?>" border="0" ></a>[/b]
        <?php
        echo "</td>";
        echo "</table>";
    }
}

?>



cheers
saikiran
Link to comment
Share on other sites
This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
 Share
Go to topic listing
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.