PHP Copy content into a new file to a random directory..

[cristian77]

Salut,

Am nevoie de putin ajutor in codul de mai jos. Vreau sa citesc continutul unui fisier si sa il copiez intr-un fisier creat nou dintr-un random directory:

 

<?php
error_reporting(E_ALL);

$dirname = uniqid();
mkdir($dirname);

$newFileName = '/'.$dirname.'/newfile.txt';
$newFileContent = 'someinfofile.txt';

if (file_put_contents($newFileName, $newFileContent) !== false) {
    echo "File created (" . basename($newFileName) . ")";
} else {
    echo "Cannot create file (" . basename($newFileName) . ")";
}
?>

Primesc eroare ca nu exista fisierul in directorul respectiv..

[cristian77]

16 minutes ago, cristian77 said:

sier

Daca folosesc linia

$newFileName = $dirname.'/newfile.txt';

Primesc eroare: Notice: Undefined variable: dirname in C:\xampp\htdocs\test1.php on line 6
 

Am incercat deasemenea varianta de mai jos aceeasi poveste:

<? 
$dirname = uniqid();
mkdir($dirname);
copy('contentfile.txt', '/'.$dirname.'/newfile.txt');
?>

 

[cristian77]

Functioneaza, multumesc frumos! problema este ca orice as pune in continuare spre exemplu echo('done');

Imi da asa: Parse error: syntax error, unexpected 'echo' (T_ECHO) in C:\xampp\htdocs\test.php on line 6

Iar codul simplu pus in test.php vine cu eroarea: Notice: Use of undefined constant - assumed ' ' in C:\xampp\htdocs\test.php on line 6

Cu toate ca el lucreaza.

Edited January 21, 2019 by cristian77

[cristian77]

<? 
$dirname = uniqid();
mkdir($dirname);
copy('matrice.txt', $dirname.'/newfile.txt');
echo('done');

?>

Am rezolvat, functioneaza! Multumesc.

[Wav3]

Invata sa citesti erorile pentru ca iti spun mura-n gura unde este problema.

[BogdanWDK]

2 hours ago, cristian77 said:

<? 
$dirname = uniqid();
mkdir($dirname);
copy('matrice.txt', $dirname.'/newfile.txt');
echo('done');

?>

Am rezolvat, functioneaza! Multumesc.

echo "done";