使用 PHP 将图像放置在另一个图像中

Question

我有一个图像，上面的文本是通过 php 表单动态生成的。我将徽标图像的位置保存到 mysql 数据库中的变量中。有没有办法拍摄这个图像并将其应用于图像中的固定位置？如果需要，必须将其缩小以适应此图像区域。

我已经有一个看起来像这样的脚本：

        $img = imagecreatefromjpeg('coupon/coupontemplate.jpg');
        $textColor = imagecolorallocate($img, 0, 0, 0); // black text

        //Write first the coupon title
        imagefttext($img, 16, 0, 20, 34, $textColor, 'coupon/arialbd.ttf', $title);

        //Then write the coupon description
        imagettftextbox($img, 13, 0, 20, 45, $textColor, 'coupon/arial.ttf', $description, 440);
        //If the checkbox to include logo is checked...
        if ($_POST['clogo'] == 'y') {

            $logo = $row['imagecompany'];
            $logo_file = "../directory/memberimages/$logo";
            $logo_file_type = getimagesize($logo_file);

            if ($logo_file_type['mime'] == 'image/jpg') {
                $logoImage = imagecreatefromjpeg($logo_file);
            } else if ($logo_file_type['mime'] == 'image/gif') {
                $logoImage = imagecreatefromgif($logo_file);
            } else if ($logo_file_type['mime'] == 'image/png') {
                $logoImage = imagecreatefrompng($logo_file);
            }

            }

        // Output image to the browser
        //header('Content-Type: image/jpeg');
        //imagejpeg($img);

        // Or save to file
        imagejpeg($img, 'my-text.jpg');
        imagedestroy($img);

    }
//}

任何人都知道如何做到这一点 - 从指定位置获取图像并将其放在另一个图像上？谢谢！

Answer 1

合并图像可以这样完成：参考

<?php
 # If you don't know the type of image you are using as your originals.
 $image = imagecreatefromstring(file_get_contents($your_original_image);
 $frame = imagecreatefromstring(file_get_contents($your_frame_image));

 # If you know your originals are of type PNG.
 $image = imagecreatefrompng($your_original_image);
 $frame = imagecreatefrompng($your_frame_image);

 imagecopymerge($image, $frame, 0, 0, 0, 0, 50, 50, 100);

 # Save the image to a file
 imagepng($image, '/path/to/save/image.png');

 # Output straight to the browser.
 imagepng($image);

Answer 2

我认为您正在寻找imagecopyresampled() 或imagecopyresized()。