How to determine whether a gif image is a dynamic image in PHP_PHP Tutorial

How does PHP determine whether a gif image is an animated image?

Example

The code is as follows

代码如下
/*  * 判断图片是否为动态图片（动画）  */ function isAnimatedGif($filename) {  $fp=fopen($filename,'rb');  $filecontent=fread($fp,filesize($filename));  fclose($fp);  return strpos($filecontent,chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0')===FALSE?0:1; }

/*

* Determine whether the picture is a dynamic picture (animation)
*/

function isAnimatedGif($filename) {

代码如下
function IsAnimatedGif($filename) { $fp = fopen($filename, 'rb'); $filecontent = fread($fp, filesize($filename)); fclose($fp); return strpos($filecontent,chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0') === FALSE?0:1; } echo IsAnimatedGif("51windows.gif"); ?>

$fp=fopen($filename,'rb');

$filecontent=fread($fp,filesize($filename));

fclose($fp);

return strpos($filecontent,chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0')===FALSE?0:1;
}

Or do this Use PHP to determine whether a gif image is animated (multi-frame)

代码如下
function check($image){ $content= file_get_contents($image); if(preg_match("/".chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0'."/",$content)){ return true; }else{ return false; } } if(check('/home/lyy/luoyinyou/2.gif')){ echo'真是动画'; }else{ echo'不是动画'; } ?>

The code is as follows
function IsAnimatedGif($filename) { $fp = fopen($filename, 'rb'); $filecontent = fread($fp, filesize($filename)); fclose($fp); return strpos($filecontent,chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0') === FALSE?0:1; } echo IsAnimatedGif("51windows.gif"); ?>

Example 2 The gif animation is in gif89 format, and it was found that the beginning of the file is gif89. But many transparent pictures also use the gif89 format, GOOGLE: You can check whether the file contains: chr(0×21).chr(0xff).chr(0×0b).'NETSCAPE2.0' chr(0×21).chr(0xff) is the header of the extended function segment in the gif image, 'NETSCAPE2.0' is the program name for the extended function execution The program code is as follows:

The code is as follows
function check($image){ $content= file_get_contents($image); if(preg_match("/".chr(0x21).chr(0xff).chr(0x0b).'NETSCAPE2.0'."/",$content)){ return true; }else{ return false;<🎜> }<🎜> }<🎜> if(check('/home/lyy/luoyinyou/2.gif')){<🎜> echo'It's really animated';<🎜> }else{<🎜> echo'Not an animation';<🎜> }<🎜> ?>

The test found that reading 1024 bytes is enough, because the data stream read at this time exactly contains chr(0×21).chr(0xff).chr(0×0b).'NETSCAPE2.0' http://www.bkjia.com/PHPjc/895096.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/895096.htmlTechArticleHow does PHP determine whether a gif image is a dynamic image? The example code is as follows /* * Determine whether the image is a dynamic image (animation) ) */ function isAnimatedGif($filename) { $fp=fopen($filenam...