PHP 抓取网页图片并且另存为的实现代码
下面是源代码，及其相关解释
<div class="codetitle"><a style="CURSOR: pointer" data="8866" class="copybut" id="copybut8866" onclick="doCopy('code8866')"> 代码如下:
<div class="codebody" id="code8866">
<?
PHP //URL是远程的完整
图片地址，不能为空,$filename 是另存为的
图片名字
//
默认把
图片放在以此脚本相同的目录里
function GrabImage($url,$filename=""){
//$url 为空则返回 false;
if($url == ""){return false;}
$ext = strrchr($url,".");//得到
图片的扩展名
if($ext != ".gif" && $ext != ".jpg" && $ext != ".bmp"){echo "格式
不支持！";return false;}
if($filename == ""){$filename = time()."$ext";}//以时间戳另起名
//开始捕捉
ob_start();
readfile($url);
$img = ob_get_contents();
ob_end_clean();
$size = strlen($img);
$fp2 = fopen($filename,"a");
fwrite($fp2,$img);
fclose($fp2);
return $filename;
}
//测试
GrabImage("//www.jb51.cc/images/
logo.gif","as.gif");
?>
ob_start : 打开
输出缓冲
This function will turn output buffering on. While output buffering is active no output is sent from the script (other than headers),instead the output is stored in an internal buffer. (
输出是在内部缓冲储存)
//
readfile : 读入
一个文件并写入到
输出缓冲
返回从
文件中读入的字节数。如果出错返回 FALSE 并且除非是以 @readfile() 形式
调用，否则会
显示错误信息。
// ob_get_contents : Return the contents of the output buffer(返回
输出缓冲的
内容)
This will return the contents of the output buffer without clearing it or FALSE,if output buffering isn't active. (如果
输出缓冲没有活动（打开），则返回 FALSE)
//
ob_end_clean() : Clean (erase) the output buffer and turn off output buffering(清除
输出缓冲)
This function
discards(丢弃) the contents of the topmost output buffer and turns off this output buffering.(丢弃并且关掉) If you want to further process the buffer's contents you have to call ob_get_contents() before ob_end_clean() as the buffer contents are
discarded when ob_end_clean() is called. （如果要用缓冲
内容，则在清理
输出缓冲之前要先
调用 ob_get_contents()）The function returns TRUE when it successfully
discarded one buffer and FALSE otherwise. Reasons for failure are f
irst that you called the function without an active buffer or that for some reason a buffer
Could not be deleted (possible for special buffer).
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
