php 从 Url 任何文件类型创建图像
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Create Image From Url Any File Type
提问by Steve Robbins
I know of imagecreatefromgif(), imagecreatefromjpeg(), and imagecreatefrompng()but is there a way to create an image resource (for png preferably) from a url of anytype of valid image? Or do you have to determine the file type and then use the appropriate function?
我知道imagecreatefromgif(), imagecreatefromjpeg(),imagecreatefrompng()但是有没有办法从任何类型的有效图像的 url 创建图像资源（最好是 png）？还是必须先确定文件类型，然后使用相应的函数？
When I say url I mean something like http://sample.com/image.png, not a data url
当我说 url 时，我的意思是类似http://sample.com/image.png，而不是数据 url
回答by supersan
The simplest way to do this is let php decide what is the file type:
最简单的方法是让 php 决定文件类型：
$image = imagecreatefromstring(file_get_contents($src));
回答by Sampo Sarrala - codidact.org
Maybe you want this:
也许你想要这个：
$jpeg_image = imagecreatefromfile( 'photo.jpeg' );
$gif_image = imagecreatefromfile( 'clipart.gif' );
$png_image = imagecreatefromfile( 'transparent_checkerboard.PnG' );
$another_jpeg = imagecreatefromfile( 'picture.JPG' );
// This requires you to remove or rewrite file_exists check:
$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg' );
// SEE BELOW HO TO DO IT WHEN http:// ARGS IS NEEDED:
$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg?foo=hello&bar=world' );
Here's how it's done:
这是它的完成方式：
function imagecreatefromfile( $filename ) {
if (!file_exists($filename)) {
throw new InvalidArgumentException('File "'.$filename.'" not found.');
}
switch ( strtolower( pathinfo( $filename, PATHINFO_EXTENSION ))) {
case 'jpeg':
case 'jpg':
return imagecreatefromjpeg($filename);
break;
case 'png':
return imagecreatefrompng($filename);
break;
case 'gif':
return imagecreatefromgif($filename);
break;
default:
throw new InvalidArgumentException('File "'.$filename.'" is not valid jpg, png or gif image.');
break;
}
}
With some small modifications to switchsame function is ready for web url's:
通过对switch相同功能的一些小的修改，可以为 web url 做好准备：
/* if (!file_exists($filename)) {
throw new InvalidArgumentException('File "'.$filename.'" not found.');
} <== This needs addiotional checks if using non local picture */
switch ( strtolower( array_pop( explode('.', substr($filename, 0, strpos($filename, '?'))))) ) {
case 'jpeg':
After that you can use it with http://www.tld/image.jpg:
之后，您可以将其用于http://www.tld/image.jpg：
$jpeg_image = imagecreatefromfile( 'http://example.net/photo.jpeg' );
$gif_image = imagecreatefromfile( 'http://www.example.com/art.gif?param=23&another=yes' );
Some proofs:
一些证明：
As you can read from official PHP manual function.imagecreatefromjpeg.phpGD allows loading images from URLs that is supported by function.fopen.php, so there is no need to fetch image firstand save it to file, and open that file.
正如您从 PHP 官方手册中看到的那样function.imagecreatefromjpeg.phpGD 允许从function.fopen.php支持的 URL 加载图像，因此无需先获取图像并将其保存到文件中，然后打开该文件。
回答by Radon8472
I use this function. It supports all types of urls and stream wrappers and all image types php can handle.
我用这个功能。它支持所有类型的 url 和流包装器以及 php 可以处理的所有图像类型。
/**
* creates a image ressource from file (or url)
*
* @version: 1.1 (2014-05-02)
*
* $param string: $filename url or local path to image file
* @param [bool: $use_include_path] As of PHP 5 the FILE_USE_INCLUDE_PATH constant can be used to trigger include path search.
* @param [resource: $context] A valid context resource created with stream_context_create(). If you don't need to use a custom context, you can skip this parameter by NULL
* @param [&array: $info] Array with result info: $info["image"] = imageinformation from getimagesize, $info["http"] = http_response_headers (if array was populated)
*
* @see: http://php.net/manual/function.file-get-contents.php
* @see: http://php.net/manual/function.getimagesize.php
*
* @return bool|resource<gd> false, wenn aus Dateiinhalt keine gueltige PHP-Bildresource erstellt werden konnte (z.b. bei BMP-Datei)
* @throws InvalidArgumentException Wenn Datei kein gueltiges Bild ist, oder nicht gelesen werden kann
*
*/
function createImageFromFile($filename, $use_include_path = false, $context = null, &$info = null)
{
// try to detect image informations -> info is false if image was not readable or is no php supported image format (a check for "is_readable" or fileextension is no longer needed)
$info = array("image"=>getimagesize($filename));
$info["image"] = getimagesize($filename);
if($info["image"] === false) throw new InvalidArgumentException("\"".$filename."\" is not readable or no php supported format");
else
{
// fetches fileconten from url and creates an image ressource by string data
// if file is not readable or not supportet by imagecreate FALSE will be returnes as $imageRes
$imageRes = imagecreatefromstring(file_get_contents($filename, $use_include_path, $context));
// export $http_response_header to have this info outside of this function
if(isset($http_response_header)) $info["http"] = $http_response_header;
return $imageRes;
}
}
Usage (simple example):
用法（简单示例）：
$image = createImageFromFile("http://sample.com/image.png");
Usage (complex example):
用法（复杂的例子）：
// even sources with php extensions are supported and e.g. Proxy connections and other context Options
// see http://php.net/manual/function.stream-context-create.php for examples
$options = array("http"=>
array("proxy" => "tcp://myproxy:8080",
"request_fulluri" => true
)
);
$context = stream_context_create($options);
$image = createImageFromFile("http://de3.php.net/images/logo.php", null, $context,$info);
// ... your code to resize or modify the image
回答by edCoder
This may help you
这可能会帮助你
$image = imagecreatefromstring(file_get_contents('your_image_path_here'));
$image = imagecreatefromstring(file_get_contents('your_image_path_here'));
Example: $image = imagecreatefromstring(file_get_contents('sample.jpg'));
例子： $image = imagecreatefromstring(file_get_contents('sample.jpg'));
回答by Dev
First fetch the url using file_get_contents($url)function and the save the content to a file. After that you can use proper image manipulation functions to further changes. You can use following code to save image from url. Here is the sample code:
首先使用file_get_contents($url)函数获取 url并将内容保存到文件中。之后，您可以使用适当的图像处理功能进行进一步更改。您可以使用以下代码从 url 保存图像。这是示例代码：
$url = "http://sample.com/image.png";
$arr = explode("/",$url);
$img_file = dir(__FILE__).'/'.$arr[count($arr)-1];
$data = file_get_contents($url);
$fp = fopen($img_file,"w");
fwrite($fp,$data);
fclose($fp);
Thanks.
谢谢。
回答by test30
You might be also interested in most advanced version:
您可能还对最高级版本感兴趣：
http://salman-w.blogspot.com/2008/10/resize-images-using-phpgd-library.html
http://salman-w.blogspot.com/2008/10/resize-images-using-phpgd-library.html
<?php
/*
* PHP function to resize an image maintaining aspect ratio
* http://salman-w.blogspot.com/2008/10/resize-images-using-phpgd-library.html
*
* Creates a resized (e.g. thumbnail, small, medium, large)
* version of an image file and saves it as another file
*/
define('THUMBNAIL_IMAGE_MAX_WIDTH', 150);
define('THUMBNAIL_IMAGE_MAX_HEIGHT', 150);
function generate_image_thumbnail($source_image_path, $thumbnail_image_path)
{
list($source_image_width, $source_image_height, $source_image_type) = getimagesize($source_image_path);
switch ($source_image_type) {
case IMAGETYPE_GIF:
$source_gd_image = imagecreatefromgif($source_image_path);
break;
case IMAGETYPE_JPEG:
$source_gd_image = imagecreatefromjpeg($source_image_path);
break;
case IMAGETYPE_PNG:
$source_gd_image = imagecreatefrompng($source_image_path);
break;
}
if ($source_gd_image === false) {
return false;
}
$source_aspect_ratio = $source_image_width / $source_image_height;
$thumbnail_aspect_ratio = THUMBNAIL_IMAGE_MAX_WIDTH / THUMBNAIL_IMAGE_MAX_HEIGHT;
if ($source_image_width <= THUMBNAIL_IMAGE_MAX_WIDTH && $source_image_height <= THUMBNAIL_IMAGE_MAX_HEIGHT) {
$thumbnail_image_width = $source_image_width;
$thumbnail_image_height = $source_image_height;
} elseif ($thumbnail_aspect_ratio > $source_aspect_ratio) {
$thumbnail_image_width = (int) (THUMBNAIL_IMAGE_MAX_HEIGHT * $source_aspect_ratio);
$thumbnail_image_height = THUMBNAIL_IMAGE_MAX_HEIGHT;
} else {
$thumbnail_image_width = THUMBNAIL_IMAGE_MAX_WIDTH;
$thumbnail_image_height = (int) (THUMBNAIL_IMAGE_MAX_WIDTH / $source_aspect_ratio);
}
$thumbnail_gd_image = imagecreatetruecolor($thumbnail_image_width, $thumbnail_image_height);
imagecopyresampled($thumbnail_gd_image, $source_gd_image, 0, 0, 0, 0, $thumbnail_image_width, $thumbnail_image_height, $source_image_width, $source_image_height);
imagejpeg($thumbnail_gd_image, $thumbnail_image_path, 90);
imagedestroy($source_gd_image);
imagedestroy($thumbnail_gd_image);
return true;
}
?>
回答by ILker ?zcan
you analyse this code.
你分析这段代码。
$url=$_SERVER['REQUEST_URI'];
$url=explode('.',$url);
$extension=$url[1];
switch($extension){
case'jpg':
imagecreatefromjpeg();
break;
}
