调整图像输出文件名的大小问题

Question

I'm having an issue with the code below, for some reason I get the following error:

**Warning: imagejpeg() expects parameter 1 to be resource**

Using the following code:

$image = "image1.png";
$output = "filename.jpg";

// Call function
$img = resize_image($image, $output);

function resize_image($file, $output) {
  imagejpeg($img, $output, 60);
}

I'm basically trying to pass in the output filename ($output = "filename.jpg";) into the imagejpeg() function but getting the above error.

Answer 1

It looks like your input parameter is the issue. Try this:

$image = "image1.png";
$output = "filename.jpg";

// Call function
$img = resize_image($image, $output);

function resize_image($file, $output) {
  imagejpeg(imagecreatefromjpeg($img), $output, 60);
}