Android从Uri获取真实路径和文件名 [英] Android get the real path and file name from Uri

问题描述

public void fileSearching(){
    try{
        // find1, find2 are buttons that open the file explorer
        find1.setOnClickListener(new View.OnClickListener(){
            @Override
            public void onClick(View view){
                Intent i = new Intent();
                i.setAction(Intent.ACTION_GET_CONTENT);
                i.setType("file/*");
                check = 0;
                startActivityForResult(i, CODE);
            }
        });
        find2.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                Intent i = new Intent();
                i.setAction(Intent.ACTION_GET_CONTENT);
                i.setType("file/*");
                check = 1;
                startActivityForResult(i, CODE);
            }
        });
    }
    catch (Exception ex){}
}
@Override
public void onActivityResult(int request, int result, Intent data){
    if (request == CODE && result == RESULT_OK){
        Uri u = data.getData();
        String path = data.getData().getPath();
    }
}

重新运行 / external / file / 59 但我需要 / sdcard / downloads /... +文件名

我需要这个，因为我想得到一个sha-1，md5，... hash对于任何文件。

如果我使用/ external / file / xx，它会给我一个FileNotFoundException。

我尝试了一些已经提供的解决方案，但似乎没有为我工作或我不知道如何使用它$ / $>

Retrurns /external/file/59 but i need /sdcard/downloads/... + file name
I need this because i want to get a sha-1,md5,... hash for any file.
If i use the "/external/file/xx" it gives me a FileNotFoundException.
I've tried some solutions that were already presented but none seemed to work for me or i didn't know how to use it

推荐答案

Retrurns / external / file / 59

Retrurns /external/file/59

这是因为您没有看 Uri 。 getPath（） only 有意义，如果方案是 file 。您很少会使用文件方案获得 Uri 。通常，它会有一个内容方案，在这种情况下，路径对你来说毫无意义。

That is because you did not look at the scheme of the Uri. getPath() only has meaning if the scheme is file. You will rarely get a Uri with a file scheme. Usually, it will have a content scheme, in which case the path is meaningless to you.

我需要这个，因为我想为任何文件获得sha-1，md5，... hash。

I need this because i want to get a sha-1,md5,... hash for any file.

使用 ContentResolver 和 openInputStream（）来获取 InputStream 由 Uri 标识的内容。将 InputStream 传递给散列算法。

Use ContentResolver and openInputStream() to get an InputStream on the content identified by the Uri. Pass that InputStream to your hashing algorithm.

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