php 使用 imagejpeg 保存和提供图像文件
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Using imagejpeg to save & serve image file
提问by V. Arora
I'm doing a bit of an experiment with PHP + Image manipulation. I'm trying to convert some images into black and white versions. I'm mostly figured it out but have one slight issue.
我正在做一些 PHP + 图像操作的实验。我正在尝试将一些图像转换为黑白版本。我主要是想通了，但有一个小问题。
In order to reduce the strain on the server, I wanted to save the B&W versions and only run the image filtering on images that haven't been run through the script before. So, I have something like this:
为了减少服务器的压力，我想保存黑白版本，只对之前没有通过脚本运行的图像运行图像过滤。所以，我有这样的事情：
<?php
header("Content-type: image/jpeg");
$file = $_GET['img'];
$name = md5($file).".jpg";
if(file_exists("/path/to/file" . $name)) {
ob_clean();
flush();
readfile("path/to/file" . $name);
exit;
}
else {
$image = imagecreatefromjpeg($file);
imagefilter($image, IMG_FILTER_GRAYSCALE);
imagejpeg($image, "/path/to/file" . $name);
imagedestroy($image);
};
?>
This does create the B&W versions of the file and save them to the server. The initial "if" statement is also working - it correctly serves the image if it already exists.
这会创建文件的黑白版本并将它们保存到服务器。最初的“if”语句也有效 - 如果图像已经存在，它会正确地提供图像。
The issue is that for new images that are run through, this saves them but doesn't output them to the browser. What can I use/change in order to do that?
问题在于，对于运行的新图像，这会保存它们但不会将它们输出到浏览器。我可以使用/更改什么来做到这一点？
Also, this is my first time doing anything like this. Any general tips you have about doing the above would be appreciated.
此外，这是我第一次做这样的事情。您对执行上述操作的任何一般提示将不胜感激。
回答by Stoic
A compact and correct form for above function can be:
上述函数的紧凑且正确的形式可以是：
<?php
header("Content-type: image/jpeg");
$file = $_GET['img'];
$name = md5($file).".jpg";
if(!file_exists("/path/to/file" . $name)) {
imagefilter($image, IMG_FILTER_GRAYSCALE);
imagejpeg($image, "/path/to/file" . $name);
} else {
$image = imagecreatefromjpeg("/path/to/file" . $name);
}
imagejpeg($image);
imagedestroy($image);
?>
回答by thedom
Because you are saving the image with imagejpeg()in the elsepart, your image will not be shown.
So you have to add
因为您imagejpeg()在else零件中保存图像，所以不会显示您的图像。所以你必须添加
readfile("/path/to/file". $name);
after imagedestroy();).
之后imagedestroy();)。
回答by Mike
You could wrap the image-output code in a function - something like this (untested):
您可以将图像输出代码包装在一个函数中 - 像这样（未经测试）：
function output_image ( $image_file ) {
header("Content-type: image/jpeg");
header('Content-Length: ' . filesize($image_file));
ob_clean();
flush();
readfile($image_file);
}
$file = $_GET['img'];
$name = md5( $file ) . ".jpg";
$image_file = "/path/to/file/" . $name;
if(!file_exists( $image_file )) {
$image = imagecreatefromjpeg( $file );
imagefilter( $image, IMG_FILTER_GRAYSCALE );
imagejpeg( $image, $image_file );
imagedestroy( $image );
}
output_image( $image_file );
