PHP - How To Get Rid Of .html File When Echo Out Images From Folder

I have a page using forms to help build listing templates for eBay. I have a folder where I have hundreds of logos stored. I know the logo names but not their extensions. . . . I have to test each potential (jpg, jpeg, gif, png, etc.) until I guess right. Here is an example code for the web form: <form action="extension_test2.php" method="post">
<p>Logo: <input name="e" value="" type="text" size="15" maxlength="30" /><p>
<input name="Submit" type="Submit"/>
</form>   and the form's result: <?

$e =$_POST['e'];

if(!empty($e)) {
echo '<img src="http://www.gbamedica...ebayimg/logos/'.$e.'">';
};

?>   Here is a link to the example: http://www.gbamedica...ension_test.php Use "olympus.jpg" for test. I am looking for code that can determine the file type and dynamically add the extension. Can it be done?

Hi, I want to be able to click on the photo and go to the next one in a folder. I have this code already, I just am not quite sure how to finish it.

-George
Code: [Select]
<?php
$count = $_GET['count'];
$dir = "images";
$names = array();
$handle = opendir($dir);
while ($name = readdir($handle)){
if(is_dir("$dir/$name")) {
if($name != '.' && $name != '..') {
echo "directory: $name\n";
}
} elseif ($name != '.DS_Store' ) {

$names[] = $name;
}

}
closedir($handle);
$numberofitems =  count($names);
$numberofitems--;


if ($count <= $numberofitems){


echo "<p>";
echo "<img src='images/".$names[$count]."'>";

} else {echo "end";}



?>

I made an upload image system, the images are stored in a folder, while the image name is stored in database. When I try to execute the image name, it works successfully, but when I try to disply the image from the folder by using the image name in the database, I fail each time.

The folder where the images are being stored is named: saveimage

Here is the query:

Code: [Select]
<?php
mysql_connect('localhost', 'root', '') or die ('Connection Failed');
mysql_select_db('imagedatabase');
$images = mysql_query("SELECT * FROM img WHERE email='$lemail'");
while($row = mysql_fetch_array($images))
{
echo "<img src='saveimage/'".$row['img_description'];
 }
?>

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!

I'm trying to write code that will let me pull 10 out of 15 images out of a folder and display them on my site. The images are all different, and I don't want dupes to show. So far, I have the following code figured out:

---------------

$s = array ("image.jpg", "image2.jpg"); // as many images as you want
$n = rand(1,len($s)); // randomly pick a number between 1 and the
length of the array
echo "<img src='". $s[$n] .'">"; // create an image tag for the randomly selected imagine (value of the randomly defined key)
array_pop($s, $n); // This piece isn't right, it needs to EXTRACT and delete the $n array element.

// Next random image
$n = rand(1,len($s));
echo "<img src='". $s[$n] .'">";

---------------
Any ideas what array_pop should be to work properly?

Thank you for the help!

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);

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!

<td><label for='images'> <b>File to upload:</b> </label></td>
            <td><input type='file'  name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></
                       </tr>
         
   <?php
            $target_path = "images/";
            $target_path = $target_path . basename( $_FILES['images']['name']);

            if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) {
                echo "The file ".  basename( $_FILES['images']['name']).
             " has been uploaded";
            } else{
                echo $row['drama_image'];
            }
         
     ?>

['drama_image'] is the name of the file

I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it.

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

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.

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.

The root directory: header.php stylesheet.css   In the following example I am trying to include the header.php file in a sub folder.   When I include the header.php like in the following example then the stylesheet.css file will not work anymo

<?php

include("../header.php");

?>

The stylesheet.css file is included in the head tags of the header.php file.   Is the above example the right way to do it? If yes, how can I do it so the stylesheet.css file will work too.