Java – 从服务器下载文件到我的位置时,文件的名称始终为 ‘null’

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英文:

Java - The name of the file is always 'null' when downloading it from server to my location

问题

我有以下方法可以从服务器的某个文件夹下载文件。

public DefaultStreamedContent downloadFiles(String s) {
    try {
        File file = new File(s);
        
        InputStream is = new DefaultStreamedContent(new FileInputStream(file), Files.probeContentType(file.toPath())).getStream();
        byte[] array = IOUtils.toByteArray(is);
        
        return new DefaultStreamedContent(new ByteArrayInputStream(array));
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

这段代码运行良好,因为文件夹中的文件被正确下载(我已检查其内容)。问题是:当通过浏览器下载并在本地的“Downloads”文件夹中保存时,文件名始终为“null”。文件应该与服务器上存储的文件具有相同的名称。假设要下载的原始文件是server\folder1\folder2\file.extension。为什么在下载时,文件名会作为“null”保存在我的“Downloads”文件夹中?

我认为“文件路径”被视为文件的名称,因此浏览器/操作系统将其保存为“null”,因为斜杠“\”对于文件名无效。

我该如何解决这个问题?

英文:

I have the following method to download files from a certain folder in my server.

public DefaultStreamedContent downloadFiles(String s) {
	try {
		File file = new File(s);
		
		InputStream is = new DefaultStreamedContent(new FileInputStream(file), Files.probeContentType(file.toPath())).getStream();
   	 	byte[] array = IOUtils.toByteArray(is);
   	 	
        return new DefaultStreamedContent(new ByteArrayInputStream(array));
	} catch (Exception e) {
		e.printStackTrace();
		return null;
	}
}

The code works well, as the files in the folder are downloaded correctly (I checked it´s contents). The problem is: The file name, when downloaded by the browser and saved in my Downloads folder locally, is always null. The files should have the same name of those files stored in the server. Let´s assume that the original file to be downloaded is server\folder1\folder2\file.extension. Why, when downloading it, the file name goes as null to my Downloads folder?

Java – 从服务器下载文件到我的位置时,文件的名称始终为 ‘null’

I thought that the file path is assumed as the file´s name, and therefore the Browser/OS saves it as null because slashes \ are not valid for file´s names.

How can I solve this?

答案1

得分: 0

可以尝试类似这样的代码吗?不确定是否需要Files.probeContentType(file.toPath()

public DefaultStreamedContent downloadFiles(String s) {
    try {
        File file = new File(s);
        String filename = file.getName();
        
        InputStream is = new DefaultStreamedContent(new FileInputStream(file), Files.probeContentType(file.toPath())).getStream();
        byte[] array = IOUtils.toByteArray(is);
        
        return new DefaultStreamedContent(new ByteArrayInputStream(array), filename);
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}
英文:

Can you try something like this? Not sure if the Files.probeContentType(file.toPath() is required.

    public DefaultStreamedContent downloadFiles(String s) {
    try {
        File file = new File(s);
        String filename = file.getName();
        
        InputStream is = new DefaultStreamedContent(new FileInputStream(file), Files.probeContentType(file.toPath())).getStream();
        byte[] array = IOUtils.toByteArray(is);
        
        return new DefaultStreamedContent(new ByteArrayInputStream(array), filename);
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

答案2

得分: 0

我使用了这个答案来解决我的问题。
我基本上复制了代码并根据我的需求进行了调整。所以,它完美地运行了!

英文:

I solved my problem using this answer.
I basically copied the code and adapted to my needs. So, it works perfectly!

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  • 本文由 发表于 2023年7月12日 23:08:26
  • 转载请务必保留本文链接:https://go.coder-hub.com/76672060.html
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