我有一个基于Sinatra的项目page,用户可以上传MP3文件。
<h2><%= I18n.t(:home_title) %></h2>
<%= I18n.t(:upload_body_text) %>
<form action="/<%= I18n.locale %>/upload" method="post" enctype="multipart/form-data">
<p>
<input type="file" name="song" size="40">
</p>
<div>
<input type="submit" value="<%= I18n.t(:home_submit) %>">
</div>
</form>
上传由此route处理:
post "/upload" do
File.open('uploads/' + params['song'][:filename], "w") do |f|
f.write(params['song'][:tempfile].read)
end
erb :main
end
文件上传后,它已损坏:
我该如何解决?
答案 0 :(得分:3)
您正在以文本模式打开文件(默认设置):
g E ? ? 8 º ? Ä ì ß ê ( ? ½ ^ ~ ? ? X ?
但你正在写二进制数据(MP3)。您需要在binary mode中打开目标文件:
import java.util.Scanner;
import java.util.Random;
public class Program4
{
public static void main(String[] args)
{
if(args.length < 2)
{
usage();
}
else if(args[0].equals("-e"))
{
encrypt(args);
}
else if(args[0].equals("-d"))
{
decrypt(args);
}
}
//Intro (Usage Method)
public static void usage()
{
System.out.println("Stream Encryption program by my name");
System.out.println("usage: java Encrypt [-e, -d] < inputFile > outputFile" );
}
//Encrypt Method
public static void encrypt(String[] args)
{ Scanner scan = new Scanner(System.in);
String key1 = args[1];
long key = Long.parseLong(key1);
Random rng = new Random(key);
int randomNum = rng.nextInt(256);
while (scan.hasNextLine())
{
String s = scan.nextLine();
for (int i = 0; i < s.length(); i++)
{
char allChars = s.charAt(i);
int cipherNums = allChars ^ randomNum;
System.out.print(cipherNums + " ");
}
}
}
//Decrypt Method
public static void decrypt(String[] args)
{ String key1 = args[1];
long key = Long.parseLong(key1);
Random rng = new Random(key);
Scanner scan = new Scanner(System.in);
while (scan.hasNextInt())
{
int next = scan.nextInt();
int randomNum = rng.nextInt(256);
int decipher = next ^ randomNum;
System.out.print((char)decipher + " ");
}
}
}
或IO库将尝试将EOL转换为Windows风格的CR-LF对:
File.open('uploads/' + params['song'][:filename], "w")
此外,您不应使用用户提供的名称("b" Binary file mode
Suppresses EOL <-> CRLF conversion on Windows. And
sets external encoding to ASCII-8BIT unless explicitly
specified.
)作为文件名而不彻底清除它;或者更好,不要使用他们的名字,将他们的名字存储在某个地方的数据库中,并使用表格的File.open('uploads/' + params['song'][:filename], "wb")
# --------------------------------------------------^
作为文件名。