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只应通过引用传递变量

// Other variables
$MAX_FILENAME_LENGTH = 260;
$file_name = $_FILES[$upload_name]['name'];
//echo "testing-".$file_name."<br>";
//$file_name = strtolower($file_name);
$file_extension = end(explode('.', $file_name)); //ERROR ON THIS LINE
$uploadErrors = array(
    0=>'There is no error, the file uploaded with success',
    1=>'The uploaded file exceeds the upload max filesize allowed.',
    2=>'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form',
    3=>'The uploaded file was only partially uploaded',
    4=>'No file was uploaded',
    6=>'Missing a temporary folder'
);
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有任何想法吗?2天后仍然卡住了.

php

220
推荐指数
9
解决办法
23万
查看次数

PHP获取完整的服务器名称,包括端口号和协议

在PHP中,是否有一种可靠而好的方法来获取这些东西:

协议:即http或https Servername:例如localhost Portnumber:例如8080

我可以使用获取服务器名称$_SERVER['SERVER_NAME'].

我可以得到协议,但我不认为它是完美的:

    if(strtolower(substr($_SERVER["SERVER_PROTOCOL"],0,5))=='https') {
        return "https";
    }
    else {
        return "http";
    }
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我不知道如何获得端口号.我使用的端口号不是80 ..它们是8080和8888.

谢谢.

php

17
推荐指数
4
解决办法
5万
查看次数

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