PHP - Php - Displaying Images From Folder, With Most Recent First

Can someone tell me how I can remove or delete an image from a folder on a server using PHP? I tried this:


Code: [Select]
unlink("http://midwestcreativeconsulting.com/jhrevell/wp-content/themes/twentyten_3/upload/" . $location);


before my delete MySQL statement, but I keep getting this error:


Warning: unlink() [function.unlink]: http does not allow unlinking in /home/midwestc/public_html/jhrevell/wp-content/themes/twentyten_3/removejewelry.php on line 22

Can anyone help and tell me how I can make it work?

I have searched around and found this code that suggests that I should be able to read an image file and echo it directly in to the page without hyperlinking to the file that is outside the public folder, but i get the error message that it is not there even though the file is.


Warning: getimagesize() [function.getimagesize]: Unable to access "/home/****/upload/AAABBBCCC.JPG" in /home/****/public_html/client.php on line 40


Code: [Select]
$image = "AAABBBCCC.JPG";
$path= "/home/****/upload/";
$details = getimagesize($path . $image);
header ('Content-Type: ' . $details['mime']);
echo  file_get_contents($path . $image);

Hi    I echo out images from a folder. The problem is that I have a html file in the same folder but I don't want to echo that out. How can I write my code to get rid of the html in the output?   Here my code:

<?php 
$filetype = '*.html';
$dirname = substr('$filetype');
$i=0;
if (isset($_POST['submit2']))
{
  //get the dir from POST
  $selected_dir = $_POST['myDirs'];
  //now get the files from within the selected dir and echo them to the screen
  foreach(glob($selected_dir . DIRECTORY_SEPARATOR . '*') as $dirname)
  {
    echo substr($dirname, 0, -4);
    echo '<img src="'.$dirname.'" />';
    echo "<label><div class=\"radiobt\"><input type='radio' name='radio1' value='$i'/></div></label>";
  }
}
?>

P.S. when I echo out : substr($dirname, 0, -4); I want to get ride of the html filename there too. How can I do so something like this?

Im using this code to call all the images in a folder:

Code: [Select]
$handle = opendir(dirname(realpath(__FILE__)).'/images/');
while($file = readdir($handle)){
if($file !== '.' && $file !== '..'){
echo '<img src="admin/img/uploads/'.$file.'" border="0" />';
}
}
My html says the images are present but they aren't visable on screen:

Code: [Select]
<div id="contentbody">
<img src="admin/img/uploads/send-button-sprite copy.png" border="0" />
<img src="admin/img/uploads/test" border="0" />
<img src="admin/img/uploads/counter.jpg" border="0" />
<img src="admin/img/uploads/send-button-sprite.png" border="0" />
</div>
Any help is much appreciated!

Hey guys, i'm new to this site and would need some help with coding. 

So i'm making a car part website which should has brand/model search. It's a dropdown search which will get the brand / model from database and should display all the parts for that exact model, from folder. 

Database structure which i have is id, master, name. ( Here's some pictures to clear out what i'm doing. http://imgur.com/a/7XwVd )  So the problem is that it does not get the images from folder named ex: *_audi_a3.jpg. 

And link to codes what have been written already.     Parts.php: http://pastebin.com/q6vdypge  Update.php: http://pastebin.com/DymhGQ17 Search_images.php: http://pastebin.com/LF5Q0i8f Core.js: http://pastebin.com/bgc0y4TS   I don't know what's wrong with it sadly. One thing i noticed when i used firebug it gives error on category when searching error:true.

Hope you guys understand what i mean here, and also all help is appreciated.

And move this post if it's in wrong place. 

Thanks!
Edited by aeonius, 17 October 2014 - 12:15 PM.

Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this?

thanks

Hello once again, i got this code that takes all images from a folder and displays them close to each other.
Heres the code:

<?php
$dir = 'uploads/thumbs/watermarkedthumbs/';
$file_display = array ('jpg', 'jpeg', 'png', 'gif');

if (file_exists($dir) == false) {
echo 'tt';
} else {
$dir_contents = scandir($dir);

foreach ($dir_contents as $file) {
$file_type = strtolower(end(explode('.', $file)));

if ($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true) {
echo '<img src="', $dir, '/', $file, '" alt="', $file, '" />';
}

}
}
?>

How do i add margins in between them? i wanna make it smth like 'margin-left:10px, margin-top-10px'.
Also, how do i add an 'overflow'? i mean, my images currently are displayed in a div, and i wanna make it so that if the folder has lets say 9 images, it would only display 6 of them, and the rest would be displayed in the same div after i click a button or smth (like '1' for the first div and '2' would appear if theres an overflow, and when i click '2' the rest of the images would appear).

To make these adjustments do i need to make a different php file or do i change something in this code?
thanks in advance.

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.

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

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!

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>' ;
    }

	
}

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