我正在尝试将外部图像保存在我的网站上
我有这个代码：
$file = $_GET['url'];
$newfile = $_SERVER['DOCUMENT_ROOT'] . '/images/name.jpg';
if ( copy($file, $newfile) ) {
echo "Copy saved";
}else{
echo "copy failed";
}
它运行良好并将图像保存在我的服务器上正常，但是当我尝试在 FOR 内部使用时，我有以下警告：
Warning: copy(): Filename cannot be empty in /home/animesad/public_html/bot/getimg.php on line 18
我的代码是：
<?php
require "../functions.php";
$p = $_GET['p'];
$url = "http://www.mysite.net/en/movie/page/".$p."/";
$source = curl_get_contents("$url");
preg_match_all('#<li title="(.*?)"><a href#',$source,$name); //outputs title post
preg_match_all('#<img src="(.*?)" width="140" height="200"/>#',$source,$img); //outputs img url
for($i=0;$i<10;$i++){
//print_r($name[1][$i]."<hr />"); //that code show the names OK. (Title posts);
$namefile = slugify($name[1][$i]); //slugify is my function that convert titles in slug.
$newfile = $_SERVER['DOCUMENT_ROOT'] .'/img/'.$namefile.'.jpg';
if ( copy($file, $newfile) ) {
echo "copy saved <hr />";
}else{
echo "copy failed <hr />";
}
}
?>
欢迎任何想法。
你忘了你的$file
$file = $_GET['url'];
for($i=0;$i<10;$i++){
$namefile = slugify($name[1][$i]); //slugify is my function that convert titles in slug.
$newfile = $_SERVER['DOCUMENT_ROOT'] .'/img/'.$namefile.'.jpg';
if ( copy($file, $newfile) ) {
echo "copy saved <hr />";
}else{
echo "copy failed <hr />";
}
}