PHP - Question About Loading Images From Folder And Adjusting How To Display Them

Hey guys - I have some code that pulls an image out of a folder and displays it on the page. The problem is, there won't always be an image to display in which case I'd rather the code not even display.

Here's my code so far: Code: [Select]
<tr><td><a href="images/uploads/<?php echo $row['loc_id']; ?>_1.jpg"><img id="morephoto" src="images/uploads/thumbs/<?php echo $row['loc_id']; ?>_1thb.jpg" /></a></td></tr>
The images are renamed on upload to have the id number for that row appended to the front of the file name and that is how I'm calling them back in.

I know I need to write an if statement that contains the code from the <tr> to the </tr> to display if the image exists, the problem is, since there isn't a field for this in the database, I don't know how to check it? I'm still a noob so I appreciate any help that is offered.

Thanks!

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?

Hi

I am trying to create a dynamic gallery in php with specific order of pictures on the page, but I can't find the function or piece of php code to do so.

Conditions:
1. The gallery must be dynamic, pictures will be coming from a folder via php, because when I'll add/upload more pictures in the folder they must be displayed on the page without adding manually the html tags in. ( this part is easy, problem is the condition 2 ).
2. The first row will have 5 pictures, the second - 4 pictures (important for the specific effect).

My code:

$files = glob("layout/gallery/*.jpg");

      rsort($files, SORT_NATURAL);

      for ($i=0; $i < count($files); $i++) {

        for( ; $i<5; $i++){
        $one = $files[$i];
        echo '<img src="'.$one.'">' . '<br><br>';
        }

echo "<br>";

        for( ; $i<9; $i++){
        $two = $files[$i];
        echo '<img src="'.$two.'">' . '<br><br>';
        }

      }

The code works well, but it just displays 9 pictures obviously. I was unable to make it dynamic displaying 5 pictures first, 4 pictures after and stay this way in a loop till displays all pictures from that folder.

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?

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

Basically I've built a CMS where by my clients can upload a number of images. On the success page I want to display the images they uploaded by file name.

The issue is the number of images can vary. They may upload 2 or 10 or 50 etc.

So far I've come up with this:

Code: [Select]
// number of files
$UN = 3; //I've set this to 3 for now, but this is passed from the upload page!

// server directories and directory names
$dir = '../properties';

$images = glob($dir.'/*.{jpg}', GLOB_BRACE); //formats to look for

$num_of_files = $UN; //number of images to display from number of uploaded files

foreach($images as $image)
{
$num_of_files--;

$newest_mtime = 0;
$image = 'BROKEN';
if ($handle = @opendir($dir)) {
while (false !== ($file = readdir($handle))) {
if (($file != '.') && ($file != '..')) {
$mtime = filemtime("$dir/$file");
if ($mtime > $newest_mtime) {
$newest_mtime = $mtime;
$image = "$file";
}
}
}
}

if($num_of_files > -1) //this made me laugh when I wrote it
echo $trimmed = ltrim($image, "../properties").'<br />'; //display images
else
break;
}
Without this piece of code:

Code: [Select]
$newest_mtime = 0;
$image = 'BROKEN';
if ($handle = @opendir($dir)) {
while (false !== ($file = readdir($handle))) {
if (($file != '.') && ($file != '..')) {
$mtime = filemtime("$dir/$file");
if ($mtime > $newest_mtime) {
$newest_mtime = $mtime;
$image = "$file";
}
}
}
}
It shows the first 3 files alphabetically. I want to view the last number of images added.

With the above code it simply shows the last image added 3 times!

So I need to get the time each image was added and then order by the newest added and limit to the number of images uploaded. Any suggestions please?

Kindest regards
Glynn

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

I am attempting to us glob to display contents of a users folder using a session variable.
Example: I have a session variable called department
Code: [Select]
$row_fullname['department'];
In department I have the name of the department the user belongs to such as: office, plant, maintenance, and groundskeeping
I created a folder called docs inside of docs there are 4 subfolders called office, plant, maintenance, and groundskeeping

I found this code which will display the contents of the folder:
Code: [Select]
<?php
$files = glob( './docs/office/*.*' );

foreach ( $files as $file )
{
    echo
        '<a href="./docs/office/' . basename( $file ) . '"target="_blank">'
        . basename( $file ) . '</a><br />';
}  
?>
The above code works fine, but I would like it to only display the contents of a departments folder only if the user is part on that department.

Here is an example that I know is completely wrong but it may help explain what I am trying to do.
Code: [Select]

<?php
]<?php
$files = glob( './docs/echo $row_fullname['department'];/*.*' );

foreach ( $files as $file )
{
    echo
        '<a href="./docs/echo $row_fullname['department'];/' . basename( $file ) . '"target="_blank">'
        . basename( $file ) . '</a><br />';
}  
?>Thanks for your time

I have this code below in a while loop, that gets looped through about 10 times and is simply resizing a slightly larger image to fit within a 50px max square. There is other code on my page, but I've narrowed the slowness down to this specific bit of code (i.e. if I take this snippet out, the page loads instantly).  With this snippet, the page takes between 5 and 10 seconds to load, which seems absurd for 10 images.  FYI, the original images are no larger than 200 px on either side, so it's not like it's looping through large image files. 

Anyone know why this might be taking so long?  Should I be doing it differently somehow?

Code: [Select]

list($width,$height) = getimagesize($myimage);
if ($width > $height) {
$datasofar .= "width=50 /></span>";
} elseif ($height > $width) {
$datasofar .= "height=50 /></span>";
} else { //height and width must be equal so just set width = 50, but could just as easily set height to 50
$datasofar .= "width=50 /></span>";
}

well I know the standard way of retrieving mysql data was through the following codes:

Code: [Select]
$query = "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'";
$result = mysql_query($query);
                $row = mysql_fetch_array($result);

This will return all properties inside a table row by an associative array indexed by column names. I am, however, wondering if there is an easier way to retrieve database info from more than one table. For now, what I am doing is:

Code: [Select]
                $result = mysql_query( "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'");
                $row = mysql_fetch_array($result);

                $result2 = mysql_query( "SELECT * FROM {$tablename2} WHERE columnmame2 = '{$var2}'");
                $row2 = mysql_fetch_array($result2);

which is a bit tedious and can cause problems when two or more coders work on the same project(it will be difficult to tell what is $row1, $row2 and $row3...). Is there away to write a simpler code than the one above? I mean, if it is possible to run mysql_fetch_array only once and retrieve database info from multiple tables?

I got this script:

But it give me error, file_get_contents cannot open stream.

I need to add the FTP connection with user/pass paramaters.

then look in set http url, to get the file contents(images) and transfer to ftp server location.

Can Anyone take alook and tell me if I am going down the right path and how to get there. Please

Code: [Select]
function postToHost($host, $port, $path, $postdata = array(), $filedata = array()) {
$data = "";
$boundary = "---------------------".substr(md5(rand(0,32000)),0,10);
$fp = fsockopen($host, $port);

fputs($fp, "POST $path HTTP/1.0\n");
fputs($fp, "Host: $host\n");
fputs($fp, "Content-type: multipart/form-data; boundary=".$boundary."\n");

// Ab dieser Stelle sammeln wir erstmal alle Daten in einem String
// Sammeln der POST Daten
foreach($postdata as $key => $val){
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$key."\"\n\n".$val."\n";
}

// Sammeln der FILE Daten
if($filedata) {
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$filedata['name']."\"; filename=\"".$filedata['name']."\"\n";
$data .= "Content-Type: ".$filedata['type']."\n";
$data .= "Content-Transfer-Encoding: binary\n\n";
$data .= $filedata['data']."\n";
$data .= "--$boundary--\n";
}

// Senden aller Informationen
fputs($fp, "Content-length: ".strlen($data)."\n\n");
fputs($fp, $data);

// Auslesen der Antwort
while(!feof($fp)) {
$res .= fread($fp, 1);
}
fclose($fp);

return $res;
}

$postdata = array('var1'=>'today', 'var2'=>'yesterday');
$filedata = array(
'type' => 'image/png',
'data' => file_get_contents('http://xxx/tdr-images/images/mapping/dynamic/deals/spot_map')
);

echo '<pre>'.postToHost ("localhost", 80, "/test3.php", $postdata, $filedata).'</pre>';