PHP - Images Displaying In Columns

Hi,

I am stuck. I have used dreamweaver to create a master details page with links to the details page successfully, but I would like to display the master details page results in 3 columns rather than the 1 column that dreamweaver defaults. As the site becomes richer the content will increase yielding many results, therfore the content displayed in a single column will require the users to scroll way down the page.

So, as it is now the results display as follows:

Plant1
Plant2
Plant3
Plant4
Plant5
Plant6
Plant7
Plant8
Plant9
etc...

I would like them to display as follows:
Plant1        Plant4        Plant7
Plant2        Plant5        Plant8
Plant3        Plant6        Plant9

Here is my code thus far:
<?php require_once('Connections/connGND.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "")
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;   
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

mysql_select_db($database_connGND, $connGND);
$query_rsBotanicalA = "SELECT * FROM plants WHERE CatIDB = 'A'";
$rsBotanicalA = mysql_query($query_rsBotanicalA, $connGND) or die(mysql_error());
$row_rsBotanicalA = mysql_fetch_assoc($rsBotanicalA);
$totalRows_rsBotanicalA = mysql_num_rows($rsBotanicalA);
?>
<table border="0" align="left">
  <?php do { ?>
    <tr>
      <td align="left"><a href="plantdetails.php?recordID=<?php echo $row_rsBotanicalA['PlantID']; ?>"> <?php echo $row_rsBotanicalA['BotanicalName']; ?>&nbsp; </a></td>
    </tr>
    <?php } while ($row_rsBotanicalA = mysql_fetch_assoc($rsBotanicalA)); ?>
</table>

Could someone help me to write this code, as I have been unsuccessful maintaining the links. I have been able get the results to display in the columsn as I would like, but I haven't been able to figure it out while maintaining the links!

Thanks.

I had this page working fine where it retrieves a set of images from the database depending on what thumbnail was click on the previous page. Now I am trying to add an edit and delete button to each of the pictures and it will not display any of it now!

This is the code I am using:
Code: [Select]
<?php
$id = $_GET[id];
?>


<?php
$max_items = 16; /* max number of news items to show */

require_once('config.php');
$db = mysql_connect (DB_HOST,DB_USER,DB_PASSWORD);
if(!$db) {
die('Failed to connect to server: ' . mysql_error());
}
mysql_select_db (DB_DATABASE,$db);

function displayNews($all = 0) {
    global $db, $max_items;
    
    if ($all2 == 0) {
        /* query for news up to $max_items */
        $query13 = "SELECT image,id,description,projectcode,thumbnail " . 
                 
                 "FROM project WHERE projectcode='$id' ORDER BY id DESC LIMIT $max_items";
    } else {
        /* query for all news */
        $query13 = "SELECT image,id,description,projectcode,thumbnail " . 
                 
                 "FROM project WHERE projectcode='$id' ORDER BY id DESC";
    }
    $result13 = mysql_query($query13) or print("<p>Error fetching entries from the database, error: " . "Statement: " . $query13 . "</p>" . mysql_error());


    while ($row13 = mysql_fetch_array($result13)) {
        
        $title = htmlentities ($row13['description']);
        $news = nl2br (strip_tags ($row13['projectcode'], '<a><b><i><u>'));
$image = $row13['image'];
$thumbnail = $row13['thumbnail'];
$id = $row13['id'];
        
        /* display the data */

        echo "<div class='portfolioproject'>";?>

        <div class='boxgrid captionfull'>
<a id="<?php echo $id ?>" class="example_group" href="images/gallery/<?php echo $image ?>" title="<?php echo $title ?>" rel="gallery"><img alt="" src="images/gallery/<?php echo $thumbnail ?>" width="122px" height="122px"</a>
               
          </a>}
               
        <?php

//Check whether the session variable SESS_MEMBER_ID is present or not

if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) {
echo "<form class='editbtn' action='editportfolio.php' method='POST'>";
echo "<input type='hidden' name='idf' value='$id' />";
echo "<input src='images/editbtn.png' type='image' value='Edit' />";
echo "</form>";
echo "<form class='editbtn' action='deleteportfolio.php' method='POST'>";
echo "<input type='hidden' name='ide' value='$id' />";
echo "<input src='images/delbtn.png' type='image' value='Delete' />";
echo "</form>";
} else {
echo "";
}


echo "</div>";

    }
    
    /* if not displaying all news, give a link to do so */
    if ($all == 0) {
        echo "<div class='show_all'><a class='show' href=\"{$_SERVER['PHP_SELF']}" .
             "?action=all\">Show all</a></div>\n";
    }
}

/* this is where the script decides what do do */

echo "\n";
switch($_GET['action']) {

    case 'all':
        displayNews(1);
        break;
    default:
        displayNews();
}
echo "\n";
?>

<?php
//Check whether the session variable SESS_MEMBER_ID is present or not
if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) {
echo "<span class='show_all'>";
echo "<a class='show' href='admincp.php'>&nbsp;&nbsp;Admin</a>";
echo "</span>";
echo "<span class='show_all'>";
echo "<a class='show' href='logout.php'>&nbsp;Logout</a>";
echo "</span>";
} else {
echo "";
}
?>

No error messages are displayed, only the 'show all' button, and the 'admin' and 'logout' buttons if signed in!

Any help on this would be much appreciated!

Thank you

Martin

Hello I am using this script right here
http://www.nearby.org.uk/sphinx/search-example5-withcomments.phps
and am trying to show images from my db but I can only get it to show the first image. Does anyone know how to make it show all images if images even exist for that particular search result?

I changed the query to $CONF['mysql_query'] = '
SELECT l.link_id AS id, l.title AS title, l.fulltxt AS body, l.url AS url, m.media_id AS im_id, m.title AS im_title, m.thumb_link AS im_t_link
FROM search1_links AS l LEFT JOIN search1_media AS m ON (m.link_id = l.link_id)
WHERE l.link_id IN ($ids)
';

and added

if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}
right here

            //Actully display the Results
            print "<ol class=\"results\" start=\"".($currentOffset+1)."\">";
            foreach ($ids as $c => $id) {
                $row = $rows[$id];
                
                $link = htmlentities(str_replace('$id',$row['id'],$CONF['link_format']));
                print "<li><a href=\"$link\">".htmlentities($row['title'])."</a><br/>";
                
                if ($CONF['body'] == 'excerpt' && !empty($reply[$c]))
                    print ($reply[$c])."</li>";
                else

           if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}

                    print htmlentities($row['body'])."</li>";
            }
            print "</ol>";
            
            if ($numberOfPages > 1) {
                print "<p class='pages'>Page $currentPage of $numberOfPages. ";
                printf("Result %d..%d of %d. ",($currentOffset)+1,min(($currentOffset)+$CONF['page_size'],$resultCount),$resultCount);
                print pagesString($currentPage,$numberOfPages)."</p>";
            }

Can anyone help me to be able to display all images if they exist? Thanks.

So I am trying to display an image from a database (I know it's not the best idea but it's what I was told to do). Anyway I'm having problems displaying it. Here is the code for displaying the image.
$cat = $_GET['cat'];

include 'includes/openDbConn.php';

$sql = 'SELECT * FROM CushionsCategories WHERE CushionCategory = "'.$cat.'"';

echo $sql;

$result = mysql_query($sql); 

echo '<table>';

while($val = mysql_fetch_assoc($result)){

$cush = 'SELECT * FROM Cushion WHERE SKU = "'.$val['CushionSKU'].'"';

echo $cush;

$cushres = mysql_query($cush);

$cushval = mysql_fetch_assoc;

$img = 'SELECT * FROM Images WHERE SKU = "'.$val['CushionSKU'].'"';

echo $img;

$imgres = mysql_query($img);

$imgval = mysql_fetch_assoc($imgres);

if( $i % 3 == 0 ) {

      echo '</tr><tr>';
  

}

echo '<td><img src="image.php?sku='.$val['CushionSKU'].'" name="'.$imgval['FileName'].'" description="'.$imgval['Description'].'" /></td>';

}

echo '</tr></table>';

and the page that gets the image:
$sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;
 $result = mysql_query($sql) or die("Invalid query: " . mysql_error());
 
        // set the header for the image
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0);

Thanks for any help in advance.

ok I got the images to upload to the directory on the server. now I got some code to try to display the image on the same page after you submit it to the directory. so it's like, submit form-> images goes in server directory->echoes back out on same page but just above the upload form. please any help greatly appreciated.
thanks.

here is the code so far.

Code: [Select]
<?php
// An Example about listing Images.
$dir = "uploads/";
$odir = opendir($dir);
while($file = readdir($odir)){
if(filetype($file) == "image/JPEG") {
echo $file;
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<p>UPLOAD YOUR PORTFOLIO AND PRICES</p>
<p>&nbsp;</p>
<p>&nbsp; <form enctype="multipart/form-data" action="uploader.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="100000" />
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form></p>
</body>

</html>

Hi All
take a look at http://www.buypromo.co.uk/index.php in IE 7 or 8

the images at the bottom of the page are not showing up, whereas therest are find, any ideas?

thanks

I'm having real problems with a rss box that I'm trying to incorporate onto the homepage of our website... The code I'm using as a basis I got from here -
http://www.dynamicdrive.com/dynamicindex18/rssdisplaybox/index.htm

It worked okay displaying title, date and content, however, we wanted it to display the first image of the blog entry as well. In the outputbody.php (the file that dictates what is shown in the rss box) I incorporated this code (shown in bold EDIT: with the bold tags around it! ) -


Code: [Select]
<?php
//Function for ouputting the body of each RSS item displayed (inside loop)- DynamicDrive.com
//For syntax pf bpdu, see: http://simplepie.org/docs/installation/from-scratch/ and http://simplepie.org/docs/reference/
//Function by default defines 3 different body outputs (templates). Modify or add additional templates as desired



    [b]  function get_image($text)
{
 
$imageurl="";
preg_match('/<\s*img [^\>]*src\s*=\s*[\""\']?([^\""\'>]*)/i' ,  $text, $matches);
$imageurl=$matches[1];
 
if (!$imageurl)
{
 preg_match("/([a-zA-Z0-9\-\_]+\.|)youtube\.com\/watch(\?v\=|\/v\/)([a-zA-Z0-9\-\_]{11})([^<\s]*)/", $text, $matches2);
 $youtubeurl = $matches2[0];
 if ($youtubeurl)
    $imageurl = "http://i.ytimg.com/vi/{$matches2[3]}/1.jpg"; 
 }
return $imageurl;
}[/b]

function shortencontent($content, $len=40){
if(strlen($content) > $len){
return substr($content, 0, $len - 6) . ' . . .';
} else {
return $content;
}
}

function outputbody($item, $template=""){
if ($template=="" || $template=="default"){ //DEFAULT TEMPLATE
?>
<div class="rsscontainer">
<div class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>"><?php echo $item->get_title(); ?></a></div>

<div class="rssdate"><?php echo $item->get_date('d M Y g:i a'); ?></div>     

<div class="rssdescription"><?php echo shortencontent($item->get_description(), 400); ?></div>

   
    [b]<div class="rssimage"><?php echo get_image(); ?></div>[/b]
</div>
<?php
} //end default template
else if ($template=="titles"){ //"TITLES" TEMPLATE
?>
<div class="rsscontainer">
<div class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>" target="_new"><?php echo $item->get_title(); ?></a></div>
<div>Category: <?php echo $item->get_category(); ?></div>
</div>
<?php
} //end titles template
else if ($template=="titlesdates"){ //"TITLESDATES" TEMPLATE
?>
<div class="rsscontainer">
<span class="rsstitle"><a href="<?php echo $item->get_permalink(); ?>"><?php echo $item->get_title(); ?></a></span>
<span class="rssdate"><?php echo $item->get_date('m/d/y g:i a'); ?></span>
</div>
<?php
} //end titlesdates template
else if ($template=="mytemplatename"){ //"mytemplatename" TEMPLATE
?>
//DEFINE ADDITIONAL CUSTOM TEMPLATE(s) USING SAME LOGIC STRUCTURE AS ABOVE
//For syntax of template body, see SimplePie docs: http://simplepie.org/docs/installation/from-scratch/ and http://simplepie.org/docs/reference/
<?php
}




else
die ("No template exists with such name!");
} //Closing function bracket
?>
While it creates an imagebox nothing is shown and when you open the empty image box in the browser it displays the following URL -

http://www.oururl.%20.%20.%20.</div>%20%20%20%20%20%20%20%20<div%20class=

I have little PHP knowledge but don't really know where to start with this... please can anyone help or point me in the right direction. I appreciate that it is frustrating dealing with someone with limited knowledge but please have a little patience with me...

Cheers people!

Hi!

I want the image of my site to be clickable so that it shows the next image contained in the result set of the sql query.

Code: [Select]
<?php
require_once('includes/connection.inc.php');
$conn = dbConnect();
$sql = 'SELECT filename FROM images ORDER BY image_id DESC LIMIT 1';
$msg = '';
if(!$sql) { echo "Something went wrong with the query. Please try again."; }
$result = $conn->query($sql) or die(mysqli_error());
if(mysqli_num_rows($result) == 0) {
header('Location: empty_blog.php');
$conn->close();
} else {
$row = $result->fetch_row();
$image = $row[0];
}
?>
<?php include('header.php'); ?>
<div class="content main">
<img id="image" src="images/<?php echo $image; ?>"/>
</div>
<?php include('includes/footer.inc.php'); ?>

All the filenames of the images are contained correctly in the $result variable, all I need to do now is to fetch the next image in the set. How would I go about this? Any help is greatly appreciated!

Hi, sincerely hope i can get some help here. im pulling some images from my database but i want to display them horizontally for example 3 or 4 images in a row then it moves to a new row. so far i have only being able to display the images vertically. I am not sure if i should be using css or there is a way to do it with php. would appreciate any assistance. I have been tryin to get this done for sometime now and i need to complete this project. I'm new at this

I am trying to display two images per row.  Will add in table tags later so it will look nicer.  Anyway, it is leaving off the second image.  So, I have 6 images and image number two is blank.

$dirname = "Files/$listingid/images/";

$images = glob($dirname."*");
   /*
   $html .= '<table width="100%">';

foreach($images as $image){
    
$html .= "<tr><td><center>";

$html .="<img src=\"$image\" width=\"300\">";

$html .= "</a></center></td>";

$html .= "</td></tr>";
}
  $html .= "</table>";
   */

$countImages = count($images) ;

$imagesPerRow = 2;


for ($i = 0 ; $i < $countImages; $i++) {
    //display image here
    $image = $images[$i] ;
    $html .= "<img width = \"200\" src='$image'>" ;

    
	if ($i % $imagesPerRow == 0) {
        //have displayed an entire row
        $html .= '<br>' ;
    }

	
}

Guys,
i need your help,i have one table employee with empno=number,image=blob, images uploaded sucessfully and insert into database but when i am trying to retrieve records image are not displaying instead of it some encrypted form shows please help me 

hi all
I am finishing off the final touches to my PHP postcard script and want to be able to display thumb file types only
I have found a script to resize the images but this creates larger ones so i want to be able to switch between the two.

<?php
// CHANGE PARAMETERS HERE BEGIN
$senderName  = " Holidays From Home "; // Eg.: John's Postcards
$senderEmail = "holidaysfromhome@voluntary.awardspace.co.uk";  // Eg.: john@postcard.com
// Change only if you have problems with urls
$postcardURL = "http://".$_SERVER["HTTP_HOST"].$_SERVER["SCRIPT_NAME"];
// CHANGE PARAMETERS HERE END
$result = 0;
$msg = "";
$msg1 = "";
$pic = "";

function displayPhotos() //edit photos and change image sizes 
{
 global $pic;
 $columns = 5;
 $act = 0;
 $act1 = 0;
 $sel = 0;
  // Open the actual directory
 if($handle = opendir("thumbs"))
 {
  // Read all file from the actual directory
  while($file = readdir($handle))
  {
   if(!is_dir($file))
   {
    if(isset($pic[1]))
    {
     if($pic[1] == $act1){$sel = "checked";}
     else{$sel = "unchecked";}
     }
    if($act == 0){echo "<tr>";}
    echo "<td align='center'><img src='thumbs/$file' alt='postcard' BORDER =1 /><br/><input type='radio' name='selimg' value='$file,$act1' $sel/></td>"; //displays the images

$act++;
    $act1++;
    if($act == $columns){$act = 0;echo "</tr>";}

       }
    }
  echo "</tr>";
  }    
 }
?>

my thumbnail images all end in (_th.jpg) and my large files all end in (.jpg). I am really stuck any help would be much appreciated 

The problem:

I'm trying to create a page which outputs images from a folder which I have been
able to do, but the problem I'm having is not being able to get the page to display
the most recent image according to file modification date at the top.

The first set of code below outputs the image timestamps in descending order, from newest to oldest which is
what I want, but as soon as I change/add a couple lines of code (Shown in the second lot of code) to get the image
file name along with the timestamp, the echoed list (timestamp and file names) gets muddled up in a random order.

In short;
As soon as the file names are retrieved with the timestamp, the list
goes from being organised descendingly, to not.

Show timestamp only code (1st lot of code):

<?php

//Open images directory
$ignore = array("..",".");

$dir = opendir("images1");

$images = array();

$sortedimages = array();

//List files in images directory
while (($file = readdir($dir)) !== false) 

if (!in_array($file, $ignore))

$images[] = $file;

foreach ($images as $image)

{

$filetime = filemtime("images1/$image");

$sortedimages[] = $filetime;

}

rsort($sortedimages);

foreach ($sortedimages as $sorted)

{

//foreach ($sorted as $key => $value)

//{

echo "$sorted<br/>";

}

//}

closedir($dir);

?>

The show timestamp and file name code (2nd lot of code):

<?php

//Open images directory
$ignore = array("..",".");

$dir = opendir("images1");

$images = array();

$sortedimages = array();

//List files in images directory
while (($file = readdir($dir)) !== false) 

if (!in_array($file, $ignore))

$images[] = $file;

foreach ($images as $image)

{

$filetime = filemtime("images1/$image");

$sortedimages[] = array($filetime => $image);

}

rsort($sortedimages);

foreach ($sortedimages as $sorted)

{

foreach ($sorted as $key => $value)

{

echo "$key and $value<br/>";

}

}

closedir($dir);

?>

The changes made in the 2nd script from the 1st:

//Changed:
$sortedimages[] = $filetime; ---> $sortedimages[] = array($filetime => $image);

//Included the previously commented out:
foreach ($sorted as $key => $value)
{

}

//Changed:
echo "$sorted<br/>"; ---> echo "$key and $value<br/>";

Thanks for any help!

this is the line in my script that I have to show the image:

Code: [Select]
$output .= "<img>{$row['disp_pic']}</img></br>\n";


As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean).

I am currently looking to insert images into a database and then display them in php.

I am thinking of storing the images as a VARCHAR data type but what would be a more suitable type?