conman90 Posted February 5, 2009 Share Posted February 5, 2009 It's a script to take an uploaded image and resize it to a thumbnail. Here's the code: <?php function resizeImage($originalImage,$toWidth,$toHeight){ list($width,$height)=getimagesize($originalImage); $xscale = $width/$toWidth; $yscale = $height/$toHeight; if ($yscale>$xscale){ $new_width = round($width * (1/$yscale)); $new_height = round($height * (1/$yscale)); } else { $new_width = round($width * (1/$xscale)); $new_height = round($height * (1/$xscale)); } $imageResized = imagecreatetruecolor($new_width, $new_height); $imageTmp = imagecreatefromjpeg($originalImage); imagecopyresampled($imageResized, $imageTmp, 0, 0, 0, 0, $new_width, $new_height, $width, $height); return $imageResized; } $oldimage = $_FILES['uploadfile']['tmp_name']; resizeImage($oldimage, 100, 100); $filename = "pics/thumbnails/".$_FILES['uploadfile']['name']; imagejpeg($imageResized,$filename,100); ?> And it continually gives me this error: Warning: imagejpeg(): supplied argument is not a valid Image resource in _____ on line 26 I checked all the arguments and the one that's causing problems is $imageResized but I can't figure out why. Link to comment https://forums.phpfreaks.com/topic/143891-help-i-keep-getting-an-invalid-argument-error/ Share on other sites More sharing options...
Philip Posted February 5, 2009 Share Posted February 5, 2009 Because you're not utilizing the return correctly. Try: $imageResized = resizeImage($oldimage, 100, 100); $filename = "pics/thumbnails/".$_FILES['uploadfile']['name']; imagejpeg($imageResized,$filename,100); Link to comment https://forums.phpfreaks.com/topic/143891-help-i-keep-getting-an-invalid-argument-error/#findComment-755024 Share on other sites More sharing options...
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