简单的例子
<?php
$reg = "#<img[^>]+src=(['\"])(.+)\\1#isU";
$url = "http://www.163.com";
$content = file_get_contents($url);
preg_match_all($reg, $content, $m);
foreach($m[2] as $src)
{
$data = file_get_contents($src);
file_put_contents('img/'.basename($src), $data);
}
?>
<?php
$reg = "#<img[^>]+src=(['\"])(.+)\\1#isU";
$url = "http://www.163.com";
$content = file_get_contents($url);
preg_match_all($reg, $content, $m);
foreach($m[2] as $src)
{
$data = file_get_contents($src);
file_put_contents('img/'.basename($src), $data);
}
?>
20061209093236509.jpg>hkjfghjfghjsdufauysdhgfyasdfguygyu";
preg_match("/<img[^>]+src=(['\"])(.+)\\1/isU",$str, $matches);
//print_r($matches);
$imgurl=$matches[1].".".$matches[2];$filename=basename($imgurl);
$dir="autodownpic";
if(!is_dir($dir)){
mkdir($dir);
}
$filename=$dir."/".$filename;
//####保存图片####
ob_start();
readfile($imgurl);
$img = ob_get_contents();
ob_end_clean();
$fp=@fopen($filename, "a");
fwrite($fp,$img);
fclose($fp);
$LPATH = "http://";
$LPATH .= $_SERVER["HTTP_HOST"];//主机名
$LPATH .= dirname($_SERVER["REQUEST_URI"]);//路径
$newimg = $LPATH."/".$filename;
echo $newimg;
//######执行文件名替换######
preg_replace($imgurl, $newimg, $str);
我要把图片的路径换成我服务器的地址,怎么上面的程序还是不行??
错在哪里?