Warning: file_get_contents(/data/phpspider/zhask/data//catemap/8/file/3.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 获得;加载页面时,与服务器的连接被重置";尝试将文件上载到tmp_name时_Php_File_Upload - Fatal编程技术网

Php 获得;加载页面时,与服务器的连接被重置";尝试将文件上载到tmp_name时

Php 获得;加载页面时,与服务器的连接被重置";尝试将文件上载到tmp_name时,php,file,upload,Php,File,Upload,我正在制作一个网站,在数据库中存储mp3文件。当我访问该站点时,它显示我成功连接到数据库,并且成功选择了数据库。但是,当我选择一个文件并按下上载按钮时,它会尝试连接很长一段时间,直到我的web浏览器显示“页面加载时服务器的连接被重置”注释掉的代码不起作用。编辑:这可能与我使用的托管服务有关 <!DOCTYPE html> <!-- HTML5 style --> <html lan="en"> <head> <title>Upload

我正在制作一个网站,在数据库中存储mp3文件。当我访问该站点时,它显示我成功连接到数据库,并且成功选择了数据库。但是,当我选择一个文件并按下上载按钮时,它会尝试连接很长一段时间,直到我的web浏览器显示“页面加载时服务器的连接被重置”注释掉的代码不起作用。编辑:这可能与我使用的托管服务有关

<!DOCTYPE html>
<!-- HTML5 style -->
<html lan="en">
<head>
<title>Upload Music</title>
<meta charset="utf-8" />
</head>
<body>
<form action="UploadMusic.php" method="POST" enctype="multipart/form-data">
    File:
    <input type="file" name="music"/> <input type="submit" value="Upload"/>
</form>

<?php
$DBConnect = mysql_connect("replaced", "for", "privacy");
     if ($DBConnect === FALSE) {
         echo "Did not successfully connect to the database server." . 
         "<p>Error code " . mysql_errno()
               . ": " . mysql_error() . "</p>";
     }
     else {
         echo "<p>Successfully connected to the database server</p>";
         if (mysql_select_db("replaced") === FALSE) {
             echo "Did not successfully select the database";
         }
         else {
             echo "Successfully selected the database";
             /*$file = $_FILES["music"]["tmp_name"];
             echo $file;
             if (!isset($file)) {
                 echo "Please select a mp3 file to upload";
             }
             else {
                 $music = file_get_contents($_FILES["music"]["tmp_name"]);
                 echo $music;
             }
             */

        }
    }
?>  

</body>
</html>

上传音乐
文件:

文件大小对于我的托管服务来说太大

<!DOCTYPE html>
<!-- HTML5 style -->
<html lan="en">
<head>
<title>Upload Music</title>
<meta charset="utf-8" />
</head>
<body>
<form action="UploadMusic.php" method="POST" enctype="multipart/form-data">
    File:
    <input type="file" name="music"/> <input type="submit" value="Upload"/>
</form>

<?php
$DBConnect = mysql_connect("replaced", "for", "privacy");
     if ($DBConnect === FALSE) {
         echo "Did not successfully connect to the database server." . 
         "<p>Error code " . mysql_errno()
               . ": " . mysql_error() . "</p>";
     }
     else {
         echo "<p>Successfully connected to the database server</p>";
         if (mysql_select_db("replaced") === FALSE) {
             echo "Did not successfully select the database";
         }
         else {
             echo "Successfully selected the database";
             /*$file = $_FILES["music"]["tmp_name"];
             echo $file;
             if (!isset($file)) {
                 echo "Please select a mp3 file to upload";
             }
             else {
                 $music = file_get_contents($_FILES["music"]["tmp_name"]);
                 echo $music;
             }
             */

        }
    }
?>  

</body>
</html>