Re: [PHP] How do I return the error

[Tom Rogers]

Hi,

Wednesday, July 28, 2004, 11:56:16 PM, you wrote:
MO> Hi I have this function below - if it reurns false I want to
MO> also return the error so I can print it in my calling function.
MO> Can I do this?

MO> Cheers

MO> Matt

MO> function fileUpload($file) {
MO>   if ($file['type'] == "image/gif" || $file['type'] == "image/pjpeg"){
MO>     if (@copy ($file['tmp_name'], "images/" . $file['name']))
MO>       return true;
MO>   }else { 
MO>     return false;
MO>   }
MO> }

You can always return an array() something like this:

function fileUpload($file) {
        $r = array(false,'');
        switch($file['error']){
                case UPLOAD_ERR_INI_SIZE:
                        $r[1] = 'The uploaded file exceeds the upload_max_filesize 
directive in php.ini.';
                        break;
                case UPLOAD_ERR_FORM_SIZE:
                        $r[1] = 'The uploaded file exceeds the MAX_FILE_SIZE directive 
that was specified in the HTML form.';
                        break;
                case UPLOAD_ERR_PARTIAL:
                        $r[1] = 'The uploaded file was only partially uploaded.'; 
                        break;
                case UPLOAD_ERR_NO_FILE:
                        $r[1] = 'No file was uploaded';
                        break;
                default:
                        if ($file['type'] == "image/gif" || $file['type'] == 
"image/pjpeg"){ 
                                if (@copy ($file['tmp_name'], "images/" . 
$file['name']))
                                        $r[0] = true;
                                else
                                        $r[1] = 'Could not copy the file.';
                        }else { 
                                $r[1] = 'The uploaded file is not the right type.';
                        }
                        break;
        }
        return $r;
}

list ($status, $error) = fileUpload($file);
if(!$status) echo $error;


-- 
regards,
Tom

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