Question

此问题的后续内容： How to get new width and height after an image has been rotated with imagerotate()?

我得到的答案是基于文件名的实际图像大小，但是如果我想从另一个宽度和高度开始。

我将如何实现这一目标？请参阅以下代码以了解我的尝试......

$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);

可能我错过了一些非常必要的东西......

更新

真实价值的一个例子......

image current width63.224619257754
image current height80.210337864315

//after calculation
image new width37.583523669887
image newt height21.716336015666

where angle is 41.10419020401479

Answer 1

查看我的评论

＆＃34;我实际上已经找到了解决方案＆＃34;

..那天我没有用我最好的英语技能......

我想你知道我的意思，这就是我解决问题的方法：

在我的方法中，我试图根据＆＃34;缩小图像＆＃34;的比率来计算新值。然后根据＆＃34;缩小图像＆＃34;的差异获得新的宽度和高度。和＆＃34;旋转后的图像＆＃34;。

这样的事情：

设置＆＃34;缩小图像＆＃34;之间的关系/比率和原始形象。
进行实际的图像旋转
从旋转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率因子相乘。
根据旋转角度获取新的宽度和高度

这确实无法正常工作（ step4 失败）。我一直在四处寻找，以寻找在旋转图像后如何计算宽度和高度的答案。但是这些宽度和高度没有返回与图像旋转后GD - 函数imagesx()和imagesy()返回的相同维度。我已使用sin()和cos()尝试了几项计算来检索宽度和高度，但仍未获得与imagesx()和{{1}完全相同的值}。

这让我思考......如果我将方法改为：

设置＆＃34;缩小图像＆＃34;之间的关系/比率和原始形象。
进行实际的图像旋转
根据比率乘以imagesy()和imagesx()来应用新尺寸 - 旋转后返回值

新代码：

imagesy()

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，那么比例就不会是准确（并会切断图像时间（取决于旋转角度））。

修改了代码，因此在创建调整大小的图像时，高度和宽度将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;

}

所以我的最终解决方案将包括以下步骤：

设置＆＃34;缩小图像＆＃34;之间的关系/比率和原始形象。
进行实际的图像旋转
根据比率乘以//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);和imagesx()来应用新尺寸 - 旋转后返回值
旋转图像时设置＆＃34;假＆＃34; 宽度和高度，因此调整大小将与原始图像成比例。

我希望这可以帮助那些在我遇到的同样问题上挣扎的人（几个小时到很多）！

在旋转后获得缩小图像的宽度和高度？

1 个答案: