cyber_alchemist Posted October 8, 2013 Share Posted October 8, 2013 I was writing a code which check whether the file upload was successful or not or the file/image is of supported type or not. The script though always returns a error of 'invalid file type' even if i upload the correct file type. what could be the the plausible causes for it ?? do i have a error in my code or something else ?? here is the code. if ( isset( $_POST[ 'submit' ] ) ) { $dir_1 = './images/'; $thumb_1 = '.images/thumb'; //making sure the uploaded file transfer was successful if ( $_FILES[ 'pic1' ][ 'error' ] != UPLOAD_ERR_OK ) { switch ( $_FILES[ 'pic1' ][ 'error' ] ) { case UPLOAD_ERR_INI_SIZE: die( 'The uploaded 1st image exceeds the upload_max_filesize directive ' . 'in php.ini.' ); break; case UPLOAD_ERR_FORM_SIZE: die( 'The uploaded 1st exceeds the MAX_FILE_SIZE directive that ' . 'was specified in the HTML form.' ); break; case UPLOAD_ERR_PARTIAL: die( 'The uploaded 1st image was only partially uploaded.' ); break; case UPLOAD_ERR_NO_FILE: die( 'No 1st image was uploaded.' ); break; case UPLOAD_ERR_NO_TMP_DIR: die( 'The server is missing a temporary folder.' ); break; case UPLOAD_ERR_CANT_WRITE: die( 'The server failed to write the uploaded the uploaded 1st image to disk.' ); break; case UPLOAD_ERR_EXTENSION: die( '1st image upload stopped by extension.' ); break; } //$_FILES[ 'pic1' ][ 'error' ] } //$_FILES[ 'pic1' ][ 'error' ] != UPLOAD_ERR_OK // making sure the file is being uploaded $error = 'The 1st image you have uploaded either was not of supported filetype.'; switch ( $type ) { case IMAGETYPE_GIF: $image_1 = imagecreatefromgif( $_FILES[ 'pic1' ][ 'tmp_name' ] ) or die( $error ); break; case IMAGETYPE_JPEG: $image_1 = imagecreatefromjpeg( $_FILES[ 'pic1' ][ 'tmp_name' ] ) or die( $error ); break; case IMAGETYPE_PNG: $image_1 = imagecreatefrompng( $_FILES[ 'pic1' ][ 'tmp_name' ] ) or die( $error ); break; default: die( $error ); } //$type $image_date_1 = @date( 'Y-m-d' ); list( $width, $height, $type, $attr ) = getimagesize( $_FILES[ 'pic1' ][ 'tmp_name' ] ); } Quote Link to comment Share on other sites More sharing options...
Solution Ch0cu3r Posted October 8, 2013 Solution Share Posted October 8, 2013 Where is the variable $type defined? Quote Link to comment Share on other sites More sharing options...
cyber_alchemist Posted October 8, 2013 Author Share Posted October 8, 2013 oh my bad Quote Link to comment Share on other sites More sharing options...
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