Android开放文件 [英] Android open file

问题描述

我试图打开一个文件进行读取。

i was trying to open a file for reading..

使用扫描仪输入=新的扫描仪（文件名）; 文件找不到

但是当我用

InputStream in = openFileInput(filename);
Scanner input = new Scanner(in);

它的工作...为什么是code中的第一行错了吗？

it worked...why was the first line of code wrong?

推荐答案

文件存储在设备上的一个特定的，依赖于应用程序的位置，这是我想 openFileInput 加在文件名的开头。最后的结果（位置+文件名）的结构如下：

Files are stored on the device in a specific, application-dependent location, which is what I suppose openFileInput adds at the beginning of the file name. The final result (location + file name) is constructed as follows:

/data/data/<application-package>/files/<file-name>

还要注意文件指出 openFileInput 参数<一href="http://developer.android.com/reference/android/content/Context.html#openFileInput%28java.lang.String%29"相对=nofollow>不能包含路径分隔符。

要避免硬编码位置路径，这可能会在原则上，甚至是不同的，从设备到设备，您可以通过调用获取指向存储目录中的文件对象 getFilesDir ，并用它来阅读你的任何文件喜欢。例如：

To avoid hard-coding the location path, which could in principle even be different from device to device, you can obtain a File object pointing to the storage directory by calling getFilesDir, and use it to read whatever file you would like to. For example:

File filesDir = getFilesDir();
Scanner input = new Scanner(new File(filesDir, filename));

请注意，构建一个扫描仪通过传递字符串作为一个参数会导致扫描仪工作的内容串，即除preting它作为实际的内容进行扫描，而不是作为打开一个文件的名称。

Note that constructing a Scanner by passing a String as a parameter would result in the scanner working on the content of the string, i.e. interpreting it as the actual content to scan instead of as the name of a file to open.

这篇关于Android开放文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！