Copy images from one folder to another - php

Question

My Web application stored in directory of XAMPP/htdocs/projectname/. And I have images(source) & img(destination) folders in above directory.I am writing following line of code to get the copy images from one folder to another. But I get the following Warnnigs: (Warning: copy(Resource id #3/image1.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs) and images are not copied into destination.
<?php
$src = opendir('../images/');
$dest = opendir('../img/');
while($readFile = readdir($src)){
if($readFile != '.' && $readFile != '..'){
if(!file_exists($readFile)){
if(copy($src.$readFile, $dest.$readFile)){
echo "Copy file";
}else{
echo "Canot Copy file";
}
}
}
}
?>

Answer 1

Just a guess (sorry) but I don't believe you can use $src = opendir(...) and $src.$readFile like that. Try doing this:
$srcPath = '../images/';
$destPath = '../img/';
$srcDir = opendir($srcPath);
while($readFile = readdir($srcDir))
{
if($readFile != '.' && $readFile != '..')
{
/* this check doesn't really make sense to me,
you might want !file_exists($destPath . $readFile) */
if (!file_exists($readFile))
{
if(copy($srcPath . $readFile, $destPath . $readFile))
{
echo "Copy file";
}
else
{
echo "Canot Copy file";
}
}
}
}
closedir($srcDir); // good idea to always close your handles

Answer 2

Replace this line in your code, this will work definitely.
if(copy("../images/".$readFile, "../img/".$readFile))

Answer 3

you are giving wrong path ,if path of your file say script.php is "XAMPP/htdocs/projectname/script.php" and images and img both are in "projectname" folder than you should use following values for $srcPath and $destPath,change their values to
$srcPath = 'images/';
$destPath = 'img/';

Answer 4

public function getImage()
{
$Path='image/'; //complete image directory path
$destPath = '/edit_image/';
// makes new folder, if not exists.
if(!file_exists($destPath) || file_exists($destPath))
{
rmdir($destPath);
mkdir($destPath, 0777);
}
$imageName='abc.jpg';
$Path=$Path.$imageName;
$dest=$destPath.$imageName;
if(copy($Path, $dest));
}