PHP - Images Called From Folder Won't Display On Screen

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!

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!

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!

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

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.

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

Ok, first off CSS is definitely not my strong suit so bear with me here.  This is my code:

<style media="screen" type="text/css">
	/* <!-- */

	.clear{
		clear:both;
	}
		 
	#grid{
		width: 100%;
	}
	 
	.grid-element{
		width:33%;
		float:left;
	}
	
	a.clickable-div {
		display: block;
		height: 100%;
		width: 100%;
		text-decoration: none;
	}
	
	.img {
		height:auto; 
		width:inherit;
		max-height:100%;
		max-width:100%;
	}
	
	.button {
		color: #900;
		font-weight: bold;
		font-size: 150%;
		text-transform: uppercase;
		display: block;
		width: 50%;
	}
	
	/* --> */
</style>
<div id="grid"><div class="grid-element" align="center">
				<img src="images/light_offline.png" class="img">
				<p><h1>Hue Lamp 1
			</h1></div><div class="grid-element" align="center">
				<a href="action.php?light=2&state=false">
					<img src="images/light_off.png" class="img"></a>
					<p><h1>Hue Lamp 2
			</h1></div><div class="grid-element" align="center">
				<a href="action.php?light=3&state=false">
					<img src="images/light_off.png" class="img"></a>
					<p><h1>Hue Lamp 3
			</h1></div></div><p>
<div align="center">
	<form method="POST" action="action.php">
		<input type="hidden" name="scene" value="02b12e930-off-0">
		<input type="submit" value="Scene 1" class="button">
	</form>
	<p><form method="POST" action="action.php">
		<input type="hidden" name="off">
		<input type="submit" value="All Off" class="button">
	</form>
</div>

This is the browser output.  A couple things are wrong:   1. Why aren't the images scaling so they are all the same size?  I assigned the same CSS class to each of the images but they are all appearing differently.  The original images are all the same size.   2. Why are my buttons in the far right corner?  I want to have them centered below the images so as to prevent the horizontal scroll bar, which I hate.     Attached Files  output.png   232.09KB   0 downloads

I'm have trouble getting my script to show images in a table across the screen. I want to show the picture and have the description shown below it. Right now it shows it in a column with the name next to it. Any help would be greatly appreciated.
Thanks in advance. Below is the script.

<?php
include 'config1.php';

// Connect to server and select database.
mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect");
mysql_select_db("vetman")or die("cannot select DB");


$result = mysql_query("SELECT * FROM $dbname WHERE year = '1954'") or die(mysql_error());
// store the record of the "" table into $row
//$current = '';

// keeps getting the next row until there are no more to get
if($result && mysql_num_rows($result) > 0)

{
    $i = 0;
    $max_columns = 3;
    echo "<table>";
echo "<br>";
    while($row = mysql_fetch_array($result))
   {
       // make the variables easy to deal with
       extract($row);

       // open row if counter is zero
       if($i == 0)

          echo "<tr>";

          echo "<td align=center>";
          ?>
          <div>
               <a><img src="<?php echo $tn; ?>"</a><a><?php echo $title; ?></a></div>



          <?php
          echo "</td>";

       // increment counter - if counter = max columns, reset counter and close row
       if(++$i == $max_columns)
       {
           echo "</tr>";
           $i=0;
       }  // end if
   } // end while
} // end if results

// clean up table - makes your code valid!
if($i > 0)
{
    for($j=$i; $j<$max_columns;$j++)
        echo "<td>&nbsp;</td>";
   echo '</tr>';
}
mysql_close();
 ?>
</table>

Hello to all.
I'm new here and i am very glad to be a part of this community.
I am new in PHP/mysql but i try to learn as much as i can in php & mysql.

And here is the problem:
Into  a php or html page i want to display 2 dropdown list to act as a filter for displaying a table.
I managed to display a table based on option select from first dropdown but i don'y know how to make  or activate the second dropdown list to act as a filter, and when i press "result" to display on the same page the results.

Thanks a lot.