Php 从任何文件类型的Url创建图像
我知道,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)?还是必须确定文件类型,然后使用适当的函数 当我说url时,我的意思是Php 从任何文件类型的Url创建图像,php,image,gd,Php,Image,Gd,我知道,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)?还是必须确定文件类型,然后使用适当的函数 当我说url时,我的意思是http://sample.com/image.png,而不是数据url您可以分析此代码 $url=$_SERVER['REQUEST_URI']; $url=explode('.',$url); $extension=$url[1]; switch($extension){ case'jpg': imagecreatefromjp
http://sample.com/image.png
,而不是数据url您可以分析此代码
$url=$_SERVER['REQUEST_URI'];
$url=explode('.',$url);
$extension=$url[1];
switch($extension){
case'jpg':
imagecreatefromjpeg();
break;
}
也许你想要这个:
$jpeg_image=imagecreatefromfile('photo.jpeg');
$gif_image=imagecreatefromfile('clipart.gif');
$png_image=imagecreatefromfile('transparent_checkboard.png');
$other_jpeg=imagecreatefromfile('picture.JPG');
//这要求您删除或重写文件。\u检查:
$jpeg_image=imagecreatefromfile('http://example.net/photo.jpeg' );
//当需要http://ARGS时,请参见以下操作:
$jpeg_image=imagecreatefromfile('http://example.net/photo.jpeg?foo=hello&bar=world' );
这是如何做到的:
函数imagecreatefromfile($filename){
如果(!file_存在($filename)){
抛出新的InvalidArgumentException('File'.$filename.'notfound');
}
开关(strtolower(路径信息($filename,路径信息扩展名))){
案例“jpeg”:
案件‘jpg’:
返回imagecreatefromjpeg($filename);
打破
案例“png”:
返回imagecreatefrompng($filename);
打破
案例“gif”:
返回imagecreatefromgif($filename);
打破
违约:
抛出新的InvalidArgumentException('File“.$filename.”不是有效的jpg、png或gif图像。“);
打破
}
}
通过对开关进行一些小的修改
即可为web url提供相同的功能:
/*如果(!file_存在($filename)){
抛出新的InvalidArgumentException('File'.$filename.'notfound');
}首先使用file\u get\u contents($url)
函数获取url,然后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。以下是示例代码:
$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);
谢谢。您可能还对最高级的版本感兴趣:
我使用这个函数。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;
}
}
用法(复杂示例):
最简单的方法是让php决定文件类型:
$image = imagecreatefromstring(file_get_contents($src));
这可能对你有帮助
$image=
imagecreatefromstring(文件获取内容(“您的图像路径在这里”)
示例:$image=imagecreatefromstring(文件获取内容('sample.jpg')代码>您需要先抓取图像,然后处理它,然后取消链接()。它是可能的(URL可以用作文件名),但这仍然需要获取它。存在问题,因为$extension
应该是数组的最后一部分,但在您的示例中,如果$URL=,它始终是第二部分http://www.example.org/images/picture.png“
然后,$扩展名将是“示例”
。获取最后一部分的正确方法(您的方法)是使用$extension=array_pop(explode('.',$url))代码>。这将保持$url
不变,并从字符串末尾正确提取$extension
。您不能依赖于扩展是否正确,来自@supersan的答案更可靠简单且完美。谢谢supersan!
/**
* 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;
}
}
$image = createImageFromFile("http://sample.com/image.png");
// 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
$image = imagecreatefromstring(file_get_contents($src));