Java 将文件上载到服务器会使用无效字符保存文件

我正在从java应用程序上传一个文件到php服务器，它可以很好地上传所有内容，但我注意到如果它有一个字符，比如“in it”，那么文件中的字符将是“not”

例如，如果我使用以下java代码上传文件：

public static void main(String[] args)
{
    FILE tmpFile = new FILE(EncryptionUtil.EncryptAndZip("tes't file.txt").getAbsoluteFile().toString());

    postData
    (
        UPLOAD_URL + "?filename=" + URLEncoder.encode(file.getFileName() + ".zip", "utf-8"),
        tmpFile.getFileName() + ".zip",
        tmpFile.getFileLoader().readAllBytes() // FILE.getFileLoader().readAllBytes() is just a wrapper and reads all bytes from a file to byte[]
    );
}

private static final String CrLf = "\r\n";
public static void postData(String url, String filename, byte[] byteData) throws IOException
{
    URLConnection conn = null;
    InputStream is = null;
    OutputStream os = null;

    try
    {
        URL obj = new URL(url);
        System.out.println("url:" + obj);
        conn = obj.openConnection();
        conn.setDoOutput(true);

        String message1 = "";
        message1 += "-----------------------------4664151417711" + CrLf;
        message1 += "Content-Disposition: form-data; name=\"uploadedfile\"; filename=\"" + filename + "\""
                + CrLf;
        message1 += "Content-Type: application/octet-stream" + CrLf;
        message1 += CrLf;

        String message2 = "";
        message2 += CrLf + "-----------------------------4664151417711--"
                + CrLf;

        conn.setRequestProperty("Content-Type",
                "multipart/form-data; boundary=---------------------------4664151417711");

        conn.setRequestProperty("Content-Length", String.valueOf((message1
                .length() + message2.length() + byteData.length)));

        os = conn.getOutputStream();

        os.write(message1.getBytes());

        int index = 0;
        int size = 1024;
        do
        {
            if ((index + size) > byteData.length)
            {
                size = byteData.length - index;
            }
            os.write(byteData, index, size);
            index += size;
        }
        while (index < byteData.length);

        os.write(message2.getBytes());
        os.flush();

        is = conn.getInputStream();

        char buff = 512;
        int len;
        byte[] data = new byte[buff];
        do
        {
            len = is.read(data);

            if (len > 0)
            {
                System.out.println(new String(data, 0, len));
            }
        }
        while (len > 0);

    }
    catch (Exception e)
    {
        e.printStackTrace();
    }
    finally
    {
        try
        {
            os.close();
        }
        catch (Exception e)
        {
        }
        try
        {
            is.close();
        }
        catch (Exception e)
        {
        }
        try
        {
        }
        catch (Exception e)
        {
        }
    }
}

下面是我的php文件upload.php

<?php 

$target_path = "" . $_GET['filename'];
echo '  name of file user wants to save as: '.$target_path.'<br/>';
if (!file_exists(dirname($target_path))) {
    mkdir(dirname($target_path), 0777, true);
}

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { 
    echo "actual save name: ".basename( $_FILES['uploadedfile']['name'])." <br/>"; 
} else{ 
    echo "There was an error uploading the file, please try again!<br />"; 
}

print_r($_FILES['uploadedfile']);
?> 

---

是什么导致了这个问题？我认为是php试图清理输入？我该如何阻止这种情况发生？

如OP所愿，结束这个问题

使用

它有助于取消引用带引号的字符串，这是原因

如果需要保存为tes't，并且如果需要，它会将tes't更改为tes't

从：

stripslashes的一个示例用法是，当PHP指令在PHP5.4之前默认为on时，您不会将此数据插入需要转义的数据库等位置。例如，如果您只是直接从HTML表单输出数据

---

我通常只想关闭魔术报价，如果它是开着的。顺便说一句，它在较新的PHP版本中已经被弃用了。

首先不允许使用这些字符。用于将“”替换为下划线，或不替换为任何内容。你也可以试试@Fred ii-stripslashes有效，因为某些原因我以为我已经关闭了自动卫生设备。太好了，很高兴听到它。@Fred ii-你能把答案贴出来让我接受吗？根据你的要求贴出来，干杯我以为我已经关掉了magic_quotes_gpc，这就是为什么我对逃跑的原因感到困惑。@user3045798我很高兴我们找到了问题的根源：

<?php 

$target_path = "" . $_GET['filename'];
echo '  name of file user wants to save as: '.$target_path.'<br/>';
if (!file_exists(dirname($target_path))) {
    mkdir(dirname($target_path), 0777, true);
}

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { 
    echo "actual save name: ".basename( $_FILES['uploadedfile']['name'])." <br/>"; 
} else{ 
    echo "There was an error uploading the file, please try again!<br />"; 
}

print_r($_FILES['uploadedfile']);
?>