Question

所以我上传的图片我尝试做的就是让用户旋转它，接收该值并旋转图像并用新图像替换原始图像。

所以我有一个按钮：

<a href="#" data-id="2" class="btn btn-primary">Rotate</a>

在我的js文件中：

$('.rotate').on('click', function(e){
   e.preventDefault();
   if(rotate < 360)
    rotate = rotate +90;
   else
    rotate = 0;

   var id = $(this).data('id');

   $('.img-container img').attr('style', '-webkit-transform: rotate(' + rotate + 'deg)');

   $.ajax({
       type: "POST",
       url: "functions/post",
       data: { rotate: rotate, id: id }
   });

}）;

and inside＆＃34; functions / post＆＃34;：

$degrees = $_POST['rotate'];
$postid = $_POST['postid'];

//user the postid to get the imagefilepath from db
$filename = '../img/uploads/' . $post->getimagepath();

// Content type
header('Content-type: image/jpeg');

// Load
$source = imagecreatefromjpeg($filename);

// Rotate
$rotate = imagerotate($source, $degrees, 0);

// Output
imagejpeg($rotate);

list($width, $height) = getimagesize($rotate);
$newImage = imagecreatetruecolor($width, $height);
imagecopyresampled($newImage, $source, 0, 0, 0, 0, $width, $height, imagesx($source), imagesy($source));
imagejpeg($newImage, realpath('../img/uploads/' . $post->getimagepath()));
// Free the memory
imagedestroy($source);
imagedestroy($rotate);
imagedestroy($newImage);

但它没有保存旋转的文件.. :( 这个代码我是通过搜索和谷歌搜索得到的，但似乎无法让它对我有用。这些代码都不是一成不变的，所以任何需要改变的东西都会让它变得有效，然后它就会变得非常好并且非常感激。

Answer 1

您没有保存旋转的图像，而是重新保存原始图像。

试试这个：

$degrees = $_POST['rotate'];
$postid = $_POST['postid'];

//user the postid to get the imagefilepath from db
$filename = '../img/uploads/' . $post->getimagepath();

// Content type
header('Content-type: image/jpeg');

// Load
$source = imagecreatefromjpeg($filename);

// Rotate
$rotate = imagerotate($source, $degrees, 0);

imagejpeg($rotate, realpath('../img/uploads/' . $post->getimagepath()));
// Free the memory
imagedestroy($source);
imagedestroy($rotate);

（未经测试）但你明白了。

php旋转图像并重新保存

1 个答案: