Question

我正在尝试将一系列图像叠加在一起，并将结果保存到blob字段中的mysql数据库中。我正在使用codeigniter的'active-record'语法来执行此操作。我遇到了以下错误：

Error Number: 1064</p><p>You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'id #48 WHERE `id` = '5'' at line 1
UPDATE `users_thumbnails` SET `thumbnail` = Resource id #48 WHERE `id` = '5'

有人可以解释我做错了什么吗？谢谢！

$base = imagecreatefromjpeg('application/assets/images/vel1_bg.jpg');
    foreach($array as $key => $value){
                $item = imagecreatefrompng('application/assets/images/items/item' . $value[0] . '.png');
                list($width, $height, $type, $attr) = getimagesize('application/assets/images/items/item'. $value[0] . '.png');
                imagecopymerge($base, 
                            $item,
                            $value[1],
                            $value[2],
                            0,
                            0,
                            $width,
                            $height,
                            100);
                //imagedestroy($item);
            }

    $data = array('thumbnail' => $base);
    $this->db->where('fbid', $this->session->userdata('id'));
    $this->db->update('users_thumbnails', $data);

Answer 1

使用

  ...
 ob_start();
 imagepng($base);
 $baseimg = ob_get_clean();
 $data = array('thumbnail' => $baseimg);
  ...

其他$ base在您将其传递给查询时不会自动转换为图片，只会将其转换为Resource id #48

编辑：捕获内容

如何将PHP imagecopymerge函数创建的映像存储在mysql数据库中

1 个答案: