获取图像旋转后缩小图像的宽度和高度

$ps['product_angle'] = 77; //Could be any angle
$filename = 'test.png'     //filename to the original product
list($source_width, $source_height) = getimagesize($filename);
$source_image = imagecreatefromjpeg($filename); 
//Example for clarification (this is parameters that is used to save new image)
$ps['product_width'] = 40;      //for example $source_width = 200
$ps['product_height'] = 80;     //and $source_height = 400

$angle = $ps['product_angle'];
if (intval($angle) <> 0) {
   //Actual dimensions of image from filename
    $current_source_x = imagesx($source_image); 
    $current_source_y = imagesy($source_image);
    //Get current ratio of "scaled down image"
    $ratio_x = $ps['product_width'] / $current_source_x;
    $ratio_y = $ps['product_height'] / $current_source_y;
    $source_image = imagerotate( 
        $source_image, 
        360-$angle, 
        imageColorAllocateAlpha($source_image, 255, 255, 255, 127)
    );
   //New dimensions from actual filename
   //This would be fine if I just wanted the new width and height
   //based on the filenames dimension, but I want new dimensions
   //for the "scaled downed version" of the image (40, 80)
   $new_image_source_x = imagesx($source_image);
   $new_image_source_y = imagesy($source_image);
   //I tried this, but obviously I'm doing something totally wrong        
   $new_width = ($new_image_source_x - $current_source_x)  * $ratio_x;
   $new_height = ($new_image_source_y - $current_source_y)  * $ratio_y;
   //Set new width after rotation                                                    
   $ps['product_width']  = $new_width;
   $ps['product_height'] = $new_height;

}
$ps['source_image'] = $source_image;
list($source_width, $source_height) =     getimagesize($filename);
$dest_width = (int)$ps['product_width']; 
$dest_height = (int)$ps['product_height'];
//Resize source-image to new width and height
//
imagecopyresized($dest_image, $ps['source_image'], 0, 0, 0, 0, $dest_width, $dest_height, $source_width, $source_height);    

image current width63.224619257754
image current height80.210337864315
//after calculation
image new width37.583523669887
image newt height21.716336015666
where angle is 41.10419020401479

查看我的评论

"我实际上已经找到了一个解决方案"

这不是我运用我最好的英语技能的日子。。。

我想你知道我的意思，下面是我解决问题的方法：

在我的方法中，我试图根据"缩小图像"的比率计算新值，然后根据"缩小的图像"answers"旋转后的图像"的差异获得新的宽度和高度。

类似这样的东西：

1. 设置"缩小图像"与原始图像之间的关系/比率
2. 进行实际的图像旋转
3. 从旋转后的图像中获取原始尺寸之间的差异，并将其与图像旋转前设置的比例因子相乘
4. 根据旋转角度获取新的宽度和高度

这确实不正确（在步骤4失败）。我一直在四处寻找如何在图像旋转后计算宽度和高度的答案。但这些宽度和高度并没有返回与GD-函数imagesx()和imagesy()在图像旋转后返回的尺寸相同的尺寸。我已经尝试了使用sin()和cos()来检索宽度和高度的几种计算，但仍然没有得到与imagesx()和imagesy()完全相同的值。

这让我思考。。。如果我将方法更改为：

1. 设置"缩小图像"与原始图像之间的关系/比率
2. 进行实际的图像旋转
3. 根据与imagesx()和imagesy()相乘的比率应用新尺寸标注-旋转后返回的值

新代码：

//Rotate resized image (if it should be)
$angle = $ps['product_angle'];
if (intval($angle) <> 0) {
    //Get current dimensions from file
    $current_source_x = imagesx($source_image);
    $current_source_y = imagesy($source_image);
    //Get current ratio of "scaled down image"
    $ratio_x = $ps['product_width'] / $current_source_x;
    $ratio_y = $ps['product_height'] / $current_source_y;
    //Rotate image
    $source_image = imagerotate($source_image, 360-$angle, imageColorAllocateAlpha($source_image, 255, 255, 255, 127));    
    //Now we get a new width from the imagerotate()-function, use those to set new_width from
    //ratio/propoprtions is used from origin width and height
    $ps['product_width']  = imagesx($source_image) * $ratio_x;
    $ps['product_height'] = imagesy($source_image) * $ratio_y;
}

这很好——几乎。。。现在的问题是，如果旋转后的图像的新宽度和/或高度比图像的原始尺寸大，那么比例就不准确（有时会切断图像（取决于旋转角度））。

修改了代码，以便在创建调整大小的图像时，高度和宽度与文件中的图像成比例。

//Rotate resized image (if it should be)
$angle = $ps['product_angle'];
if (intval($angle) <> 0) {
    //Get current dimensions from file
    $current_source_x = imagesx($source_image);
    $current_source_y = imagesy($source_image);
    //Get current ratio of "scaled down image"
    $ratio_x = $ps['product_width'] / $current_source_x;
    $ratio_y = $ps['product_height'] / $current_source_y;
    //Rotate image
    $source_image = imagerotate($source_image, 360-$angle, imageColorAllocateAlpha($source_image, 255, 255, 255, 127));    
    //Now we get a new width from the imagerotate()-function, use those to set new_width from
    //ratio/propoprtions is used from origin width and height
    $ps['product_width']  = imagesx($source_image) * $ratio_x;
    $ps['product_height'] = imagesy($source_image) * $ratio_y;
    //Set these so we can modifiy the width and height given from getimagesize()-function below
    $ps['source_width'] = imagesx($source_image) ;
    $ps['source_height'] =  imagesy($source_image);            
}

//If image is rotated, then width and height are adjusted with these values
if (isset($ps['source_width']) && isset($ps['source_height']) ) {
    $source_width = $ps['source_width'];
    $source_height = $ps['source_height'];                                    
}
//Set position where to place in the image to save
$dest_x = $ps['product_left'];
$dest_y = $ps['product_top'];
$dest_width = (int)$ps['product_width'];
$dest_height = (int)$ps['product_height'];
//Resize source-image to new width and height and then copy from source to destination point
imagecopyresized($dest_image, $ps['source_image'], $dest_x, $dest_y, 0, 0, $dest_width, $dest_height, $source_width, $source_height);  

因此，我的最终解决方案将包括以下步骤：

1. 设置"缩小图像"与原始图像之间的关系/比率
2. 进行实际的图像旋转
3. 根据与imagesx()和imagesy()相乘的比率应用新尺寸标注-旋转后返回的值
4. 当图像旋转时，设置"fake"宽度和高度，这样调整大小将与原始图像成比例

我希望这能帮助任何正在与我争论过的问题（几个小时到很多小时）作斗争的人！