PHP从url增加图像分辨率然后裁剪

问题描述：

I have images with resolution of 720x1280 and I would need to store them on server with resolution 800x1500. Since 720x1280 increased to height 1500 gives resolution of 844x1500 I would also need to crop image, remove 22 pixels from the left and right side.

For now I have this:

$img_url = file_get_contents($url);

$img = imagecreatefromstring($img_url);

$width = imagesx($img);
$height = imagesy($img);

$new_width = '800';
$new_height = '1500';

$thumb = imagecreatetruecolor($new_width, $new_height);

imagecopyresampled($thumb, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

imagejpeg($thumb, $name, 100);

imagedestroy($thumb);
imagedestroy($img);

But the problem is that image is not cropped, 22 pixels from the left and right side are not removed.

Is there a way to do this, to first increase image resolution from url and then crop?

我有分辨率为720x1280的图像，我需要将它们存储在分辨率为800x1500的服务器上。因为720x1280增加了高度1500给出844x1500的分辨率我还需要裁剪图像，从左侧和右侧删除22个像素。 p>

现在我有这个： p>

  $ img_url = file_get_contents（$ url）; 
 
 $ img = imagecreatefromstring（$ img_url）; 
 
 $ width = imagesx（$ img）; 
 $ height = imagesy（$ img）  ）; 
 
 $ new_width ='800'; 
 $ new_height ='1500'; 
 
 $ thumb = imagecreatetruecolor（$ new_width，$ new_height）; 
 
imagecopyresampled（$ thumb，$ img，0  ，0,0,0，$ new_width，$ new_height，$ width，$ height）; 
 
imagejpeg（$ thumb，$ name，100）; 
 
imagedestroy（$ thumb）; 
imagedestroy（$ img）;  
  code>  pre> 
 
 但问题是没有裁剪图像，左边和右边22像素都没有被删除。 p> 
 
 
 有没有办法做到这一点，首先从网址提高图像分辨率然后裁剪？ p> 
  div>

答

Googling php image crop reveals the secret:

$rect = [22, 0, 800, 1500]
$thumb = imagecrop($thumb, $rect)

答

In this line:

imagecopyresampled($thumb, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

you just scale your original image to new dimensions. As the documentation says:

If the source and destination coordinates and width and heights differ, appropriate stretching or shrinking of the image fragment will be performed.

You need to set:

$shiftX = 22*(1280/1500); // consider passing variables rather than constant values
$scaledWidth = 720-$shiftX*2;
imagecopyresampled($thumb, $img, 0, 0, $shiftX, 0, $new_width, $new_height, $scaledWidth, $height);

which cuts out the required (proportional) rectangle of the source image to paste it in you destination picture.

PHP从url增加图像分辨率然后裁剪

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