![]() |
|
首页 │ Apache │ Linux│ Java│ MySQL│ 注册│帮助 | |||
[php] $File = fsockopen($host, $port, $errno, $errstr, 60);
if (!$File){
echo "$errstr ($errno)
\\n";
return '';
}else{
$out = "GET $target HTTP/1.1\\r\\n";
$out .= "Host: $host\\r\\n";
$out .= "Referer: $referer\\r\\n";
$out .= "Connection: Close\\r\\n\\r\\n";
fputs($File, $out);
$makeFile = $buffer = "";
while ($buffer = fread($File,4096)){
$makeFile = $makeFile.$buffer;
}
fclose($File);[/php]
如上面一段代码,请看第6行:
$out = "GET $target HTTP/1.1\\r\\n";
在程序中target的值为: $target = "/image/all I have to give.jpg";
也就是说 $target 中间有空格,这样fsockopen就获取不到了,请问有什么办法解决吗?
(当然去掉空格的话地址就不对了)
[ 本帖最后由 sweety 于 2006-7-28 02:21 AM 编辑 ]

