So basically, I'm using PHP to read the contents of a directory (one that only contains html files) and then to create a list of links.
<?php
$dir="../zpress/pages/"; // Directory where files are stored
if ($dir_list = opendir($dir))
{
while(($filename = readdir($dir_list)) !== false)
{
?>
<p><a href="<?php echo $filename; ?>"><?php echo $filename;
?></a></p>
<?php
}
closedir($dir_list);
}
?>
It works perfectly on the pages where it lists them, but when i open the link it goes to www.mywebsite.com/zpress/g.html, whereas it's supposed to go to www.mywebsite.com/zpress/pages/g.html
Suggestions?
You'd need to output the correct path as well in your <a> area. The user's browser has absolutely NO way to tell that you're listing the contents of some OTHER area of the site, and will build urls based on the address of the current page. So you need to have:
<p><a href="../zpress/pages/<?php echo $filename; ?>"><?php echo $filename;
^^^^^^^^^^^^^^^^
Prefix the output so it goes to the right directory:
<p><?php echo $filename; ?></p>
Related
I've created an external php file (sort.php) that sorts the files in a folder on the server by time modified, and returns the most recent file.
<?php
function scanDir ($dir){
$fileTimeArray = array();
// Scan directory and get each file date
foreach (scandir($dir) as $fileTime){
$fileTimeArray[$fileTime] = filemtime($dir . '/' . $fileTime);
}
//Sort file times
var $latestFile = arsort($fileTimeArray);
return($latestFile[0]);
}
?>
I'm attempting to call this function inside of , in another php file, and set the src:
<img <?php echo 'src="'.scanDir("issues/preview").'"';?>/>
I've included sort.php at the top of the page in question.
The image src reads "(unknown)". What am I missing or doing wrong or both?
Thank you!
I would write
<?php $x = scanDir("issues/preview"); ?>
<img src="<?php echo $x; ?>"/>
Im trying to display images from backend of my app
<?php foreach ($img as $key=>$row): ?>
<div class="products_inside_wrapper intro_wrapper">
<div class="classes_inside_item bordered_wht_border">
<?php
foreach (explode(';',rtrim($row['images'],';')) as $key_img => $value_img)
{
?>
<?php echo Html::img('#backend/web'.'/'.$value_img);?>
<?php
}
?>
</div>
</div>
<?php endforeach; ?>
Tried with above code to display all images, but getting error Not allowed to load local resource when I open Google Chrome Inspect Element
i think you are using a local url instead of using this
<?php echo Html::img('#backend/web'.'/'.$value_img);?>
try using it like
<?= Html::img(Yii::getAlias('#web').'/images/'.$value_img]);?>
As stig-js answered you can't load local saved image directly, If you're really interested into loading resources from a local path, you can open image as a binary file with fopen and echo the content of it with a proper header to output. In general way, you can add a method to your model like this:
public function getImage($imageName)
{
$imagePath = '#backend/web' . '/' . $imageName;
$fileInfo = finfo_open(FILEINFO_MIME_TYPE);
$contentType = finfo_file($fileInfo, $imagePath);
finfo_close($fileInfo);
$fp = fopen($imagePath, 'r');
header("Content-Type: " . $contentType);
header("Content-Length: " . filesize($imagePath));
ob_end_clean();
fpassthru($fp);
}
P.S: Also you can use combination of this answer with showing image as base64 on HTML. See How to display Base64 images in HTML?
Images must be accesible by an url, like
yoursite.com/backend/imagedir/IMG'
If yoursite.com/backend points to your backend/web folder.
Backend alias points to your local path, so you need a custom alias to reach image folders.
Yii2 aliases: http://www.yiiframework.com/doc-2.0/guide-concept-aliases.html
I am aware that this kind of question has already been asked before, but I have a slightly different case.
foreach($dir as $file)
{
$file = '<li>'.basename($file).'</li>';
echo $file;
}
This is my script to display files in a folder and link to them. The way it is now, I use the $_GET['file'] on the other page to receive the information on the other page. The other page is supposed to display a photo/video with the file that has been linked, however I don't know how to use the $_POST or $_SESSION in this case, since it's a loop and I don't want the information about the file be in the link.
Also, I don't want any forms. I want to click the link with the name of the file and the other website to already have the information about the file and display the video or image.
Use like this by storing the media path in session and use the corresponding index to pass.
First file:
<?php
$i = 1;
session_start();
$dir = array('1.jpg', '2.jpg', '3.jpg');
echo '<ul>';
foreach($dir as $file)
{ //echo $i.'---'.$file."<br />";
echo '<li>'.$file.'</li>';
$_SESSION['media_'.$i] = $file;
$i++;
}
echo '</ul>';
?>
test.php
<?php
session_start();
echo $_SESSION['media_'.$_GET['file']];
?>
I have a directory tree like the below
root
index.php
/inc
1levelData.php
1level.php
/images
db.png
What I am trying to figure out is a way I can set an absolute path that can be used for both including files and images. In addition, a method that will work when loading both index.php as well as root\inc\1level.php I have put in some sample code below for what I am currently working with. This code seems to work well for include() but doesn't function for images. Is there a possible fix?
root/Index:
<?php
$fullPath = realpath($_SERVER['DOCUMENT_ROOT']);
echo "<img src = '".$fullPath."/inc/images/db.png'><BR>";
include $fullPath.'/inc/1levelData.php';
?>
root/inc/1levelData.php
<?php
echo "<img src = '".$fullPath."/inc/images/db.png'><BR>";
include ($fullPath.'/inc/images/2levelData.php');
?>
root/inc/1Level.php
<?php
$fullPath = realpath($_SERVER['DOCUMENT_ROOT']);
include ($fullPath.'/phpinfo.php');
?>
root/inc/2LevelData.php
<?php
echo "<img src = '".$fullPath."inc/images/db.png'><BR>";
?>
Normally if there is no htacces restriction is enabled it is possible to view the list of files under a folder hosted in web server using browsers. Except if there exist a index file like index.php it automatically go to the index page. (as far i know)
But is it possible to see the list of files though there exist an index file ?
thanks in advance
No, there is not. All web servers I'm aware of will only ever display a directory listing if there is no index page available (and, even then, only if directory listings are not disabled).
Build a file listing in PHP and display it in the index file.
Check out the info at http://php.net/manual/en/function.readdir.php . I used this for a client to display certain file types in a directory through the index.php file.
<?php
if ($handle = opendir('/path/to/files')) {
echo "Directory handle: $handle\n";
echo "Entries:\n";
while (false !== ($entry = readdir($handle))) {
echo "$entry\n";
}
closedir($handle);
}
?>
Put this in the web root directory a sindex.php
<?php
$pngFolder = <<< EOFILE
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
EOFILE;
if (isset($_GET['img']))
{
header("Content-type: image/png");
echo base64_decode($pngFolder);
exit();
}
$projectsListIgnore = array ('.','..');
$handle=opendir(".");
$projectContents = '';
while ($file = readdir($handle))
{
if (is_dir($file) && !in_array($file,$projectsListIgnore))
{
$projectContents .= '<li>'.$file.'</li>';
}
}
closedir($handle);
?>
<ul class="projects">
<?php $projectContents ?>
</ul>