PHP newbie CURL and array - arrays

Question

UPDATED: I've simplified the code (tried to)
I'm trying to download a series of images as set in an array, but something is clearly not right:
function savePhoto($remoteImage,$fname) {
$ch = curl_init();
curl_setopt ($ch, CURLOPT_NOBODY, true);
curl_setopt ($ch, CURLOPT_URL, $remoteImage);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, 0);
$fileContents = curl_exec($ch);
$retcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($retcode==200) {
$newImg = imagecreatefromstring($fileContents);
imagejpeg($newImg, ".{$fname}.jpg",100);
}
return $retcode;
}
$filesToGet = array('009');
$filesToPrint = array();
foreach ($filesToGet as $file) {
if(savePhoto('http://pimpin.dk/jpeg/'.$file.'.jpg',$file)==200) {
$size = getimagesize(".".$file.".jpg");
echo $size[0] . " * " . $size[1] . "<br />";
}
}
I get the following errors:
Warning: imagecreatefromstring()
[function.imagecreatefromstring]:
Empty string or invalid image in
C:\inetpub\vhosts\dehold.net\httpdocs\ripdw\index.php
on line 15
Warning: imagejpeg(): supplied
argument is not a valid Image resource
in
C:\inetpub\vhosts\dehold.net\httpdocs\ripdw\index.php
on line 16
Warning: getimagesize(.009.jpg)
[function.getimagesize]: failed to
open stream: No such file or directory
in
C:\inetpub\vhosts\dehold.net\httpdocs\ripdw\index.php
on line 26
*

Answer 1

try this instead:
function get_file1($file, $local_path, $newfilename)
{
$err_msg = '';
echo "<br>Attempting message download for $file<br>";
$out = fopen($newfilename, 'wb');
if ($out == FALSE){
print "File not opened<br>";
exit;
}
$ch = curl_init();
curl_setopt($ch, CURLOPT_FILE, $out);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_URL, $file);
curl_exec($ch);
echo "<br>Error is : ".curl_error ( $ch);
curl_close($ch);
//fclose($handle);
}//end function
// taken from: http://www.weberdev.com/get_example-4009.html
or file_get_contents

Answer 2

You should try file_get_contents in replacement of CURL (simpler but it does the job):
function savePhoto($remoteImage,$fname) {
$fileContents = file_get_contents($remoteImage);
try {
$newImg = imagecreatefromstring($fileContents);
imagejpeg($newImg, ".{$fname}.jpg",100);
} catch (Exception $e) {
//what to do if the url is invalid
}
}

Answer 3

I finally got it to work, with help from you all and a bit of snooping around :-)
I ended up using CURL:
function savePhoto($remoteImage,$fname) {
$ch = curl_init();
curl_setopt ($ch, CURLOPT_URL, $remoteImage);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, 0);
$fileContents = curl_exec($ch);
$retcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($retcode == 200) {
$newImg = imagecreatefromstring($fileContents);
imagejpeg($newImg, $fname.".jpg",100);
}
return $retcode;
}
$website = "http://www.pimpin.dk/jpeg";
$filesToGet = array('009');
$filesToPrint = array();
foreach ($filesToGet as $file) {
if(savePhoto("$website/$file.jpg",$file)==200) {
$size = getimagesize($file.".jpg");
echo $size[0] . " * " . $size[1] . "<br />";
} else {
echo "File wasn't found";
}
}