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];
}
Related
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;
I have a site where with jQuery/ajax I want to upload my image.
The problem is when I have strange filename for my image. Like with dots or other.
I have tried with this mode but doesn't work fine, it replace the dot in file extension for example if I have
image.test.png
become
imagetestpng
but I want this:
imagetest.png
This is my code:
$name = $_FILES['upload']['name'];
$size = $_FILES['upload']['size'];
$name = preg_replace("/[^a-zA-Z0-9_\-]+/", "", $name);
echo($name);
How to solve this?
Thanks
First, you need to decompose the file name:
$info = pathinfo($name);
Then apply your filter on both parts:
$name = preg_replace("/[^\w-]+/", "", $info['filename']);
// check if we have an extension
if (isset($info['extension'])) {
$name .= '.' . preg_replace('/[^\w]/', '', $info['extension']);
}
Demo
You can use this to replace the characters in the filename while preserving the file extension.
$name = preg_replace('/[^a-zA-Z0-9_-]+/',
"",
pathinfo($name, PATHINFO_FILENAME)
) . (pathinfo($name, PATHINFO_EXTENSION)?"." . pathinfo($name, PATHINFO_EXTENSION):"");
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.
I have a filename with a date in it, the date is always at the end of the filename.
And there is no extension (because of the basename function i use).
What i have:
$file = '../file_2012-01-02.txt';
$file = basename('$file', '.txt');
$date = preg_replace('PATTERN', '', $file);
Im really not good at regex, so could someone help me with getting the date out of the filename.
Thanks
I suggest to use preg_match instead of preg_replace:
$file = '../file_2012-01-02';
preg_match("/.*([0-9]{4}-[0-9]{2}-[0-9]{2}).*/", $file, $matches);
echo $matches[1]; // contains '2012-01-02'
If there is always an underscore before the date:
ltrim(strrchr($file, '_'), '_');
^^^^^^^ get the last underscore of the string and the rest of the string after it
^^^^^ remove the underscore
I suggest you to try:
$exploded = explode("_", $filename);
echo $exploded[1] . '<br />'; //prints out 2012-01-02.txt
$exploded_again = explode(".", $exploded[1]);
echo $exploded_again[0]; //prints out 2012-01-02
Shorten it:
$exploded = explode( "_" , str_replace( ".txt", "", $filename ) );
echo $exploded[1];
With this, use regexp when you really need to :
current(explode('.', end(explode('_', $filename))));
This should help i think:
<?php
$file = '../file_2012-01-02.txt';
$file = basename("$file", '.txt');
$date = preg_replace('/(\d{4})-(\d{2})-(\d{2})$/', '', $file);
echo $date; // will output: file_
?>
I've been trying to create a directory following a specific structure, yet nothing appears to be happening. I've approached this by defining multiple variables as follows:
$rid = '/appicons/';
$sid = '$artistid';
$ssid = '$appid';
$s = '/';
and the function I've been using runs thusly:
$directory = $appid;
if (!is_dir ($directory))
{
mkdir($directory);
}
That works. However, I want to have the following structure in created directories: /appicons/$artistid/$appid/
yet nothing really seems to work. I understand that if I were to add more variables to $directory then I'd have to use quotes around them and concatenate them (which gets confusing).
Does anyone have any solutions?
$directory = "/appicons/$artistid/$appid/";
if (!is_dir ($directory))
{
//file mode
$mode = 0777;
//the third parameter set to true allows the creation of
//nested directories specified in the pathname.
mkdir($directory, $mode, true);
}
This should do what you want:
$rid = '/appicons/';
$sid = $artistid;
$ssid = $appid;
$s = '/';
$directory = $rid . $artistid . '/' . $appid . $s;
if (!is_dir ($directory)) {
mkdir($directory);
}
The reason your current code doesn't work is due to the fact you're trying to use a variable inside a string literal. A string literal in PHP is a string enclosed in single quotes ('). Every character in this string is treated as just a character, so any variables will just be parsed as text. Unquoting the variables so your declarations look like the following fixes your issue:
$rid = '/appicons/';
$sid = $artistid;
$ssid = $appid;
$s = '/';
This next line concatenates (joins) your variables together into a path:
$directory = $rid . $artistid . '/' . $appid . $s;
Concatenation works like this
$directory = $rid.$artistid."/".$appid."/"
When you're assigning one variable to another, you don't need the quotes around it, so the following should be what you're looking for.
$rid = 'appicons';
$sid = $artistid;
$ssid = $appid;
and then...
$dir = '/' . $rid . '/' . $sid . '/' . $ssid . '/';
if (!is_dir($dir)) {
mkdir($dir);
}