CODEWITHSUNDEEP
php
josh
josh

Reputation: 173

Problems in Moving a file

i am trying to move a file with the following code:

$f = 'image.jpeg';
$source ='/Waiting/$f';
$destination = '/Excepted/$f';
copy($source, $destination) or die("Error 1");

However I get this error every time:

Warning: copy(Waiting/$f) [function.copy]: failed to open stream: No such file or directory in C:\xampp\htdocs\gallery\v2\excepted.php on line 13

I realize it is because there is no file but it is not picking up the variables.

Upvotes: 1

Views: 170

Answers (4)

Harsh
Harsh

Reputation: 2096

try this

$f = 'image.jpeg';
$source = "/Waiting/$f"; // both values are in double quotes
$destination = "/Excepted/$f"; // both values are in double quotes
copy($source, $destination) or die("Error 1");

hope it works

Upvotes: 1

M.V.
M.V.

Reputation: 1672

Your code should be:

$f = 'image.jpeg';
$source ='/Waiting/'.$f;
$destination = '/Excepted/'.$f;
copy($source, $destination) or die("Error 1");

This means that you combine string Waiting and variable $f... This should work...

Hope it helps!

Try using this (since PHP 5.0 Linux or PHP 5.3.1 on Windows): 

$f = 'image.jpeg';
$source ='/Waiting/'.$f;
$destination = '/Excepted/'.$f;
rename($source, $destination);

This function should to the job better... It will do all at once (make file on new destination and delete old one)...

Upvotes: 1

saturngod
saturngod

Reputation: 24967

because you are using ' instead of "

your code should be

$f = "image.jpeg";
$source ="/Waiting/$f";
$destination = "/Excepted/$f";
copy($source, $destination) or die("Error 1");

' is not working for $variable.

example:

<?php
$f="file";
echo 'myfile: $f';
echo "<br/>";
echo "myfile: $f";
?>

you will see following result

myfile: $f
myfile: file

Upvotes: 2

JohnP
JohnP

Reputation: 50009

You need to use double quotes if you want variable parsing to work

$source = "/Waiting/$f";

Upvotes: 1

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