I have a string with a path like so:
C:/myfolder/mysubfolder/myfile.doc
I need to truncate the string to become:
myfile.doc
Needless to say, I have a list of such paths with different lengths and different folder depths. What I need is something like trancating the rest of the string beginning from the last of the string till the first / is encountered.
How can I achieve this in PHP.
Many thanks
$path = 'C:/myfolder/mysubfolder/myfile.doc':
$filename = basename($path);
echo $filename; // "myfile.doc"
See Manual "basename()". For more detailed information about a path see pathinfo()
You can use the basename() function for this.
echo basename("C:/myfolder/mysubfolder/myfile.doc");
Output:
myfile.doc
Note that this doesn't exactly do what you described, because for these inputs:
/etc/
.
it would return
etc
.
So an (untested) alternative could be:
$fullname = "C:/myfolder/mysubfolder/myfile.doc";
$parts = explode("/", $fullname);
$filename = $parts[count($parts)-1];
echo $filename;
If you mean you want the filename segment, you can use basename()
http://php.net/manual/en/function.basename.php
Related
I have a path "../uploads/e2c_name_icon/" and I need to extract e2c_name_icon from the path.
What I tried is using str_replace function
$msg = str_replace("../uploads/","","../uploads/e2c_name_icon/");
This result in an output "e2c_name_icon/"
$msg=str_replace("/","","e2c_name_icon/")
There is a better way to do this. I am searching alternative method to use regex expression.
Try this. Outputs: e2c_name_icon
<?php
$path = "../uploads/e2c_name_icon/";
// Outputs: 'e2c_name_icon'
echo explode('/', $path)[2];
However, this is technically the third component of the path, the ../ being the first. If you always need to get the third index, then this should work. Otherwise, you'll need to resolve the relative path first.
Use basename function provided by PHP.
$var = "../uploads/e2c_name_icon/";
echo basename( $var ); // prints e2c_name_icon
If you are strictly want to get the last part of the url after '../uploads'
Then you could use this :
$url = '../uploads/e2c_name_icon/';
$regex = '/\.\.\/uploads\/(\w+)/';
preg_match($regex, $url, $m)
print_r ($m); // $m[1] would output your url if possible
You can trim after the str_replace.
echo $msg = trim(str_replace("../uploads/","","../uploads/e2c_name_icon/"), "/");
I don't think you need to use regex for this. Simple string functions are usually faster
You could also use strrpos to find the second last /, then trim off both /.
$path = "../uploads/e2c_name_icon/";
echo $msg = trim(substr($path, strrpos($path, "/",-2)),"/");
I added -2 in strrpos to skip the last /. That means it returns the positon of the / after uploads.
So substr will return /e2c_name_icon/ and trim will remove both /.
You'd be much better off using the native PHP path functions vs trying to parse it yourself.
For example:
$path = "../uploads/e2c_name_icon/";
$msg = basename(dirname(realpath($path))); // e2c_name_icon
This question already has answers here:
How do I get a file name from a full path with PHP?
(14 answers)
Closed 8 years ago.
My input value is just like this: Oppa/upload/default.jpeg
I want to slice the value of an input according by / cause i want to get the image file name. Does anyone know some tricks to do this?
example: i want to get default.png
<input type="text" value="Oppa/upload/default.png" id="fileLink" name="fileLink" />
Use basename():
$path = "Oppa/upload/default.jpeg";
echo basename($path); //will output "default.jpeg"
echo basename($path, '.jpeg'); //will output "default"
The first parameter is the path of which the trailing component will be removed. If the first parameter ends in the optional second parameter, the second parameter will also be cut off.
On Windows, both slash (/) and backslash (\) are used as directory
separator character. In other environments, it is the forward slash
(/).
- PHP manual
You should use basename() PHP function.
This will work for you
$path = "Oppa/upload/default.jpeg";
echo basename($path);
Use pathinfo() php function
$path = "http://domain.tld/Oppa/upload/default.png";
$info = pathinfo ( $path, PATHINFO_BASENAME ); // returns default.png
Yet another solution, using preg_match()
<?php
$path = "http://domain.tld/Oppa/upload/default.png"; // or "C:\\domain.tld\Oppa\upload\default.jpg";
$pattern = '/[\/|\\\\]((?:.(?!\/|\\\\))+)$/';
if(preg_match($pattern, $path, $matches)){
echo $matches[1]; // default.png or default.jpg
}
?>
Note: People claim to have problems using basename() with asian characters
I am supposing that if your image path will not be change from Oppa/upload/ than this Should work using explode :
$str = "Oppa/upload/default.jpeg";
$s= explode("Oppa/upload/",$str);
echo $s[1];
Another Best thing you can do with defining the relative path as a constant, so :
const path = "Oppa/upload/";
$str = "Oppa/upload/default.jpeg";
$s= explode(path,$str);
echo $s[1];
will also work.
I have a script to upload files with PHP.
I already do some cleaning to remove ugly characters.
I would also like to to remove dots in the filename, EXCEPT for the last one, which indicates the file extension.
Anyone has an idea how I could do that.?
For example, how would you get
$filename = "water.fall_blue.sky.jpg";
$filename2 = "water.fall_blue.sky.jpeg";
to return this in both cases..?
water.fall_blue.sky
Use pathinfo() to extract the file name (the "filename" array element is available since PHP 5.2); str_replace() all the dots out of it; and re-glue the file extension.
Here's an example of how this can be done:
<?php
$string = "really.long.file.name.txt";
$lastDot = strrpos($string, ".");
$string = str_replace(".", "", substr($string, 0, $lastDot)) . substr($string, $lastDot);
?>
It converts filenames like so:
really.long.file.name.txt -> reallylongfilename.txt
Check here: example
[Edit] Updated script, dot position is cached now
FILENAME = this/is(your,file.name.JPG
$basename=basename($_FILES['Filedata']['name']);
$filename=pathinfo($basename,PATHINFO_FILENAME);
$ext=pathinfo($basename,PATHINFO_EXTENSION);
//replace all these characters with an hyphen
$repar=array(".",","," ",";","'","\\","\"","/","(",")","?");
$repairedfilename=str_replace($repar,"-",$filename);
$cleanfilename=$repairedfilename.".".strtolower($ext);
RESULT = this-is-your-file-name.jpg
I have two example filename strings:
jquery.ui.min.js
jquery.ui.min.css
What regex can I use to only match the LAST dot? I don't need anything else, just the final dot.
A little more on what I'm doing. I'm using PHP's preg_split() function to split the filename into an array. The function deletes any matches and gives you an array with the elements between splits. I'm trying to get it to split jquery.ui.min.js into an array that looks like this:
array[0] = jquery.ui.min
array[1] = js
If you're looking to extract the last part of the string, you'd need:
\.([^.]*)$
if you don't want the . or
(\.[^.]*)$
if you do.
I think you'll have a hard time using preg_split, preg_match should be the better choice.
preg_match('/(.*)\.([^.]*)$/', $filename, $matches);
Alternatively, have a look at pathinfo.
Or, do it very simply in two lines:
$filename = substr($file, 0, strrpos($file, '.'));
$extension = substr($file, strrpos($file, '.') + 1);
At face value there is no reason to use regex for this. Here are 2 different methods that use functions optimized for static string parsing:
Option 1:
$ext = "jquery.ui.min.css";
$ext = array_pop(explode('.',$ext));
echo $ext;
Option 2:
$ext = "jquery.ui.min.css";
$ext = pathinfo($ext);
echo $ext['extension'];
\.[^.]*$
Here's what I needed to exclude the last dot when matching for just the last "part":
[^\.]([^.]*)$
Using a positive lookahead, I managed this answer:
\.(?=\w+$)
This answer matches specifically the last dot in the string.
I used this - give it a try:
m/\.([^.\\]+)$/
I have a following string and I want to extract image123.jpg.
..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length
image123 can be any length (newimage123456 etc) and with extension of jpg, jpeg, gif or png.
I assume I need to use preg_match, but I am not really sure and like to know how to code it or if there are any other ways or function I can use.
You can use:
if(preg_match('#".*?\/(.*?)"#',$str,$matches)) {
$filename = $matches[1];
}
Alternatively you can extract the entire path between the double quotes using preg_match and then extract the filename from the path using the function basename:
if(preg_match('#"(.*?)"#',$str,$matches)) {
$path = $matches[1]; // extract the entire path.
$filename = basename ($path); // extract file name from path.
}
What about something like this :
$str = '..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length';
$m = array();
if (preg_match('#".*?/([^\.]+\.(jpg|jpeg|gif|png))"#', $str, $m)) {
var_dump($m[1]);
}
Which, here, will give you :
string(12) "image123.jpg"
I suppose the pattern could be a bit simpler -- you could not check the extension, for instance, and accept any kind of file ; but not sure it would suit your needs.
Basically, here, the pattern :
starts with a "
takes any number of characters until a / : .*?/
then takes any number of characters that are not a . : [^\.]+
then checks for a dot : \.
then comes the extension -- one of those you decided to allow : (jpg|jpeg|gif|png)
and, finally, the end of pattern, another "
And the whole portion of the pattern that corresponds to the filename is surrounded by (), so it's captured -- returned in $m
$string = '..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length';
$data = explode('"',$string);
$basename = basename($data[1]);