PHP remove part of a url in an array element [duplicate] - php

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find the filename from a string with php
(3 answers)
Closed 7 years ago.
I have a .csv file which I have converted into an array $outputData. The 6th $data[6] element of each array contains an image file path, e.g.
http://images.pleaserusa.com/pic/unique.image.jpg
I need to remove the the filepath http://images.pleaserusa.com/pic/ of each image and leave just the image name itself, e.g.
unique.image.jpg
I have looked on here and Google and tried to use preg_match() and str* functions, but I'm not really getting anywhere with it.
Rizer123 I have placed the "basename" function but can't get it to replace inside the element . i can only get it to echo/print outside the array / i have placed the code below and an excerpt array out put below .
domina-456-b.jpgArray
(
[0] => DOM456/B
[1] => 6" Lace-Up Pump W/ D-Ring&Ribbon Lace
[2] => Devious
[3] => Single Soles
[4] => DOMINA-456
[5] => Blk Pat
[6] => http://images.pleaserusa.com/pic/domina-456-b.jpg
[7] => 5-15
)
domina-460-b.jpgArray
(
[0] => DOM460/B
[1] => 6" Oxford Lace Up Pump
[2] => Devious
[3] => Single Soles
[4] => DOMINA-460
[5] => Blk Pat
[6] => http://images.pleaserusa.com/pic/domina-460-b.jpg
[7] => 5-16
)
<?php
$input = 'Parser/Inv_item.csv';
$output = 'hd_inv_items.csv';
if (false !== ($ih = fopen($input, 'r'))) {
$oh = fopen($output, 'w');
header('Content-Type: text/plain');
while (false !== ($data = fgetcsv($ih))) {
// this is where you build your new row
$outputData = array($data[0], $data[1], $data[2], $data[3], $data[5],$data[6], $data[9], $data[8]);
$path = $data[9];
$name = basename($path); // $name == '1.jpg'
print_r($name);
print_r($outputData);
//fputcsv($oh, $outputData);
}
fclose($ih);
fclose($oh);
}
?>
As you can see the image file path is still in the array and the basename result is printing outside of the array . How do i fix this issue
Regards Biwwabong

Answer 1

No need for explosions or string manipulations for this, a simple basename() should suffice:
$path = 'http://images.pleaserusa.com/pic/unique.image.jpg';
echo basename($path);
output:
unique.image.jpg
pathinfo() would work as well:
echo pathinfo($path, PATHINFO_BASENAME);

Answer 2

I always do this with explode()
$url = "http://images.pleaserusa.com/pic/unique.image.jpg";
$parts = explode("/",$url);
$pic = $parts[count($parts)-1];
You are welcome ;-)