I want to create a path with dynamic variable.
echo $path='App\Widgets\mywidgetname';
I want to replace mywidgetname then I have to put a variable like this
$path='App\Widgets\'.$widgetname; // (but this not working)
Try this,
$path = 'App\Widgets\mywidgetname';
$path = str_replace('mywidgetname', $widgetname, $path);
OR
$path = 'App\Widgets\mywidgetname';
$path = substr($path, 0, strrpos($path, "\\") + 1) . $widgetname;
You can use:
$path = sprintf("App\Widgets\%s", $widgetname);
echo $path;
Related
SOLUTION
$path = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
$feedtitle = str_replace(" ", "", $feedtitle);
$feedtitle = strtolower($feedtitle);
$path = substr($path, 0, strrpos($path, "/"));
$feedlink = "http://" . "$path" . "/" . "$feedtitle" . '.xml';
PROBLEM
I am trying to generate a name for an rss feed from another field (feed title). In my query I have :
$path = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
$feedtitle = str_replace(" ", "", $feedtitle);
$feedtitle = strtolower($feedtitle);
$feedlink = "$path" . "$feedtitle" . '.xml';
If feedtitle is my rss feed, $feedlink will have the value myrssfeed.xml so that part of the code works but I'm having problems with the path. Instead of writing the current path the above writes the current path & the name of the current file! e.g. if the script I'm using for this is named feedlink.php the path that is stored in the $feedlink variable is :
mywebsite.com/mydir/feedlink.phpmyrssfeed.xml
How do I remove the name of the script?
With the str_replace() function
$feedlink = str_replace("feedlink.php", "", $feedlink);
Assuming that you requested the site as mywebsite.com/mydir/feedlink.php you should consider removing everything past the last / and then add your filename to that:
$path = substr($path, 0, strrpos($path, "/") + 1);
I have a script which displays the images present on a folder.
E.g. Script path (from HTDOCS):
/admin/global/thisscript.php
E.g. Pictures folder path (from HTDOCS):
/public/test/images/
Now I need to print under each image the path of the file starting from the HTDOCS directory.
E.g. if I have:
$picture1 = /WEB/mysite/htdocs/public/test/images/picture1.png
$picture1name = basename($picture1)
I'd like to print
print "/public/test/images/".$picture1name
having
/public/test/images
in a variable which is automatically update if picture1.png path changes.
If I use
$_SERVER['PHP_SELF']
it returns the path of the file I'm displaing. Also
__FILE__
__DIR__
basename
dirname
are not working for me.
How can I get the path?
Thank you
perhaps $_SERVER['DOCUMENT_ROOT']
edit:
Something like:
$path = "/WEB/mysite/htdocs/public/test/images/picture1.png";
$path = dirname(str_replace($_SERVER['DOCUMENT_ROOT'], '', $path));
Not sure if this could help but could you explode the path result you have and build it back from the HTDOCS entry? something like:
$fullpath = /WEB/mysite/htdocs/public/test/images/picture1.png
$pathbits = explode("/", $fullpath);
$newpath = "";
$foundstart = false;
foreach($pathbits as $key => $value){
if ($value == "htdocs" || $foundstart == true) {
$foundstart = true;
$newpath = $newpath + "/" + $value;
}
}
I'm trying to recursively iterate through a group of dirs that contain either files to upload or another dir to check for files to upload.
So far, I'm getting my script to go 2 levels deep into the filesystem, but I haven't figured out a way to keep my current full filepath in scope for my function:
function getPathsinFolder($basepath = null) {
$fullpath = 'www/doc_upload/test_batch_01/';
if(isset($basepath)):
$files = scandir($fullpath . $basepath . '/');
else:
$files = scandir($fullpath);
endif;
$one = array_shift($files); // to remove . & ..
$two = array_shift($files);
foreach($files as $file):
$type = filetype($fullpath . $file);
print $file . ' is a ' . $type . '<br/>';
if($type == 'dir'):
getPathsinFolder($file);
elseif(($type == 'file')):
//uploadDocsinFolder($file);
endif;
endforeach;
}
So, everytime I call getPathsinFolder I have the basepath I started with plus the current name of the directory I'm scandirring. But I'm missing the intermediate folders in between. How to keep the full current filepath in scope?
Very simple. If you want recursion, you need to pass the whole path as a parameter when you call your getPathsinFolder().
Scanning a large directory tree might be more efficient using a stack to save the intermediate paths (which would normally go on the heap), rather than use much more of the system stack (it has to save the path as well as a whole frame for the next level of the function call.
Thank you. Yes, I needed to build the full path inside the function. Here is the version that works:
function getPathsinFolder($path = null) {
if(isset($path)):
$files = scandir($path);
else: // Default path
$path = 'www/doc_upload/';
$files = scandir($path);
endif;
// Remove . & .. dirs
$remove_onedot = array_shift($files);
$remove_twodot = array_shift($files);
var_dump($files);
foreach($files as $file):
$type = filetype($path . '/' . $file);
print $file . ' is a ' . $type . '<br/>';
$fullpath = $path . $file . '/';
var_dump($fullpath);
if($type == 'dir'):
getPathsinFolder($fullpath);
elseif(($type == 'file')):
//uploadDocsinFolder($file);
endif;
endforeach;
}
Sorry, I'm bad English. I'm going to post my code now:
$image = 'http://example.com/thisisimage.gif';
$filename = substr($image, strrpos($image, '/') + 1);
echo '<br>';
echo $filename;
echo '<br>';
echo preg_replace('/^[^\/]+/', 'http://mydomain.com', $image);
echo '<br>';
$image is string;
$filename is image name (in example above, it returns 'thisisimage.gif')
Now i want replace all before $filename with 'http://mydomain.com', my code is above but it doesnt work.
Thanks!
$foo = explode($filename, $image);
echo $foo[0];
Explode "splits" one the given paramater ( in your case $filename ). It returns an array with where the keys are split on the string you gave.
And if you just want to change the url. you use a str_replace
$foo = str_replace("http://example.com", "http://localhost", $image);
//This will change "http://example.com" to "http://localhost", like a text replace in notepad.
In your case:
$image = 'http://example.com/thisisimage.gif';
$filename = substr($image, strrpos($image, '/') + 1);
$foo = explode($filename, $image);
echo '<br>';
echo $filename;
echo '<br>';
echo str_replace($foo[0], "http://yourdomain.com/", $url);
echo '<br>';
There's another approach in which you don't need a regular expression:
in Short:
$image = 'http://example.com/thisisimage.gif';
$url = "http://mydomain.com/".basename($image);
Explanation:
If you just want the file name without url's or directory path's, basename() is your friend;
$image = 'http://example.com/thisisimage.gif';
$filename = basename($image);
output: thisisimage.gif
Then you can add whatever domain you want:
$mydomain = "http://mydomain.com/";
$url = $mydomain.$filename;
Try this :
$image = 'http://example.com/thisisimage.gif';
echo preg_replace('/^http:\/\/.*\.com/', 'http://mydomain.com',$image);
This should simply work:
$image = 'http://example.com/thisisimage.gif';
$filename = substr($image, strrpos($image, '/') + 1);
echo '<br>';
echo $filename;
echo '<br>';
echo 'http://mydomain.com/'.$filename;
echo '<br>';
if you just like to add your own domain before the file name, try this:
$filename = array_pop(explode("/", $image));
echo "http://mydomain.com/" . $filename;
if you wanna only replace thedomain, try this:
echo preg_replace('/.*?[^\/]\/(?!\/)/', 'http://mydomain.com/', $image);
The other people here have given good answers about how to do it - regex has its advantages but also drawbacks - its slower, respectively requires more resources and for something simple as this, I would advice you to use the explode approach, but while speaking for regex functions you also may try this, instead your preg_replace:
echo preg_replace('#(?:.*?)/([^/]+)$#i', 'http://localhost/$1', $image);
It seems variable length positve lookbehind is not supported in PHP.
If I have filename.jpg, with PHP how would change it too filename123456789.jpg, where 123456789 is a timestamp, I currently do this,
$name = $filename;
$parts = explode(".", $name);
$filename = $parts[0].time().'.'.$parts[1];
however this just just leaves me with 123456789.
Your approach works fine too, but breaks if the filename has multiple dots. I'd rather use the pathinfo() function to accomplish this:
$info = pathinfo($filename);
$filename = $info['filename'].time().'.'.$info['extension'];
debug (output) your input and steps to find the error
function debug($var, $label = '') {
echo $label
. '<pre>'
. print_r($var, true)
. '</pre>';
}
$name = 'filename.bug.test.jpg';
debug($name, 'old filename');
$parts = explode('.', $name);
debug($parts, 'filenameparts');
$ext = array_pop($parts);
$prefix = implode('.', $parts);
$newName = $prefix . time() . '.' . $ext;
debug($newName, 'new filename');
as mention above use pathinfo instead of explode
i've used explode, couse i've used a dummy filename.
thats a no-brainer:
function getFileName($filename){
preg_match('#([^/]+)\.([\w]+)$#',$filename,$match);
return $match[1].time().'.'.$match[2];
}