PHP 中的图像裁剪产生空白结果

发布于 2024-08-27 00:48:03 字数 516 浏览 6 评论 0原文

我只是尝试使用 PHP 裁剪 JPEG 图像（无缩放）。这是我的函数以及输入。

function cropPicture($imageLoc, $width, $height, $x1, $y1) {
    $newImage = imagecreatetruecolor($width, $height);

    $source = imagecreatefromjpeg($imageLoc);
    imagecopyresampled($newImage,$source,0,0,$x1,$y1,$width,$height,$width,$height);
    imagejpeg($newImage,$imageLoc,90);
}

当我按如下方式调用它时 --cropPicture('image.jpg', 300, 300, 0, 0) -- 该函数正确完成，但我留下了一个 300x300 的黑色图像px（换句话说，一块空白画布）。我是否传递了错误的论点？

图像存在并且可写。

原文

I'm simply trying to crop a JPEG image (no scaling) using PHP. Here is my function, along with the inputs.

function cropPicture($imageLoc, $width, $height, $x1, $y1) {
    $newImage = imagecreatetruecolor($width, $height);

    $source = imagecreatefromjpeg($imageLoc);
    imagecopyresampled($newImage,$source,0,0,$x1,$y1,$width,$height,$width,$height);
    imagejpeg($newImage,$imageLoc,90);
}

When I call it as follows--cropPicture('image.jpg', 300, 300, 0, 0)--the function completes properly, but I'm left with a black image that is 300x300 px (in other words, a blank canvas). Am I passing in the wrong arguments?

The image exists and is writeable.

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梦归所梦 2024-09-03 00:48:03

作为 sobedai 答案的补充：您在 CropPicture() 中使用的任何函数都可能会失败。您必须测试每一个的返回值。如果出现错误，它们将返回 false 并且您的函数无法（正常）继续。

function cropPicture($imageLoc, $width, $height, $x1, $y1) {
  $newImage = imagecreatetruecolor($width, $height);
  if ( !$newImage ) {
    throw new Exception('imagecreatetruecolor failed');
  }

  $source = imagecreatefromjpeg($imageLoc);
  if ( !$source ) {
    throw new Exception('imagecreatefromjpeg');
  }

  $rc = imagecopyresampled($newImage,$source,0,0,$x1,$y1,$width,$height,$width,$height);
  if ( !$rc ) {
    throw new Exception('imagecopyresampled');
  }

  $rc = imagejpeg($newImage,$imageLoc,90);
  if ( !$rc ) {
    throw new Exception('imagejpeg');
  }
}

编辑：您可能还对 http://docs.php.net/error_get_last 感兴趣。示例脚本中的异常消息没有那么有用......

As an addition to sobedai's answer: Any of those functions you use in cropPicture() can fail. You have to test the return value of each and every one. In case of an error they return false and your function cannot continue (properly).

function cropPicture($imageLoc, $width, $height, $x1, $y1) {
  $newImage = imagecreatetruecolor($width, $height);
  if ( !$newImage ) {
    throw new Exception('imagecreatetruecolor failed');
  }

  $source = imagecreatefromjpeg($imageLoc);
  if ( !$source ) {
    throw new Exception('imagecreatefromjpeg');
  }

  $rc = imagecopyresampled($newImage,$source,0,0,$x1,$y1,$width,$height,$width,$height);
  if ( !$rc ) {
    throw new Exception('imagecopyresampled');
  }

  $rc = imagejpeg($newImage,$imageLoc,90);
  if ( !$rc ) {
    throw new Exception('imagejpeg');
  }
}

edit: You might also be interested in http://docs.php.net/error_get_last. The exception messages in the example script aren't that helpful...

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