无法让 php 以正确的名称保存文件名
由于某种原因，当使用此代码时，它输出的文件只是称为“.jpg”。
<?
$title = $GET['title'];
$url = $GET['url'];
$ourFileName = $title.jpg;
$ourFileHandle = fopen($ourFileName, 'w') or die("can't open file");
fclose($ourFileHandle);
$ch = curl_init ("http://www.address.com/url=$url");
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_BINARYTRANSFER,1);
$rawdata=curl_exec ($ch);
curl_close ($ch);
$fp = fopen("images/$ourFileName",'w');
fwrite($fp, $rawdata);
fclose($fp);
echo ("<img src=images/$ourFileName>");
?>
很确定它与声明 $ourFileName 有关，任何帮助将不胜感激。
For some reason, when using this code the file it outputs is just called ".jpg".
<?
$title = $GET['title'];
$url = $GET['url'];
$ourFileName = $title.jpg;
$ourFileHandle = fopen($ourFileName, 'w') or die("can't open file");
fclose($ourFileHandle);
$ch = curl_init ("http://www.address.com/url=$url");
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_BINARYTRANSFER,1);
$rawdata=curl_exec ($ch);
curl_close ($ch);
$fp = fopen("images/$ourFileName",'w');
fwrite($fp, $rawdata);
fclose($fp);
echo ("<img src=images/$ourFileName>");
?>
Pretty sure it has to do with declaring $ourFileName, any help would be appreciated.
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评论（5）
看起来应该是以下内容：
Looks like it should be the following:
当然应该是
看起来那里充满了安全问题！ 该脚本会用攻击者选择的文件覆盖其有权访问的任何文件！
Surely should be
Looks like a world of security problems there though! A script that overwrites any file it has permissions to with a file of an attackers choosing!
您确定
$GET['title']正确吗？ 用于访问 GET 参数值的超全局变量是$_GET，而不是$GET。Are you sure
$GET['title']is correct? The superglobal variable for accessing the GET parameter values is$_GET, not$GET.尝试这个
$ourFileName = $title . “.jpg”；
Try this
$ourFileName = $title . ".jpg";
看起来这行
应该是
只要您确定 $_GET['title']; 中有一个值就应该这样做。
Looks like this line
Should be
That should do it as long as you are sure there is a value in $_GET['title'];