getResource（）在运行时到文件 [英] getResource() to a file at runtime

问题描述

我在 src 文件夹下放了一些 .txt 文件（在资源<中/ code>文件夹）。

I put some .txt files under the src folder (in the resources folder).

但我无法在运行时创建有效的文件资源。

But I can't create a valid File at runtime from this resource.

String path = this.getClass().getResource("/resources/file.txt").getFile();
File file = new File(path);

if (!file.exists()) {
}

我从eclipse运行我的程序。我没有在classpath中添加任何内容。

我想将我的文本文件嵌入到 .jar 文件中，当我运行我想要的应用程序时抓住这些文件并将其复制到某个位置。

I run my program from eclipse. I didn't put in classpath anything.
I want my text files to be embedded into the .jar file, when I run my app I want to grab those files and copy them into some location.

更新

如果我做 InputStream is = getClass（）。getResourceAsStream（/ resources / file.txt）;

I得到流!!

推荐答案

正如您在发布问题后不久发现的那样，这有效：

As you already discovered yourself soon after posting your question, this works:

InputStream is = getClass().getResourceAsStream("/resources/file.txt");

原因这是有效的，而原始代码没有，是因为zip文件中的文件（jar文件是zip文件）在提取之前不是真正的文件。但是提取文件是你想要做的，所以在你的程序中，它不是一个真正的文件。所以这个问题是 XY问题：你想创建一个File对象，但那是不可能的 - 你需要回顾一下你最初尝试做的事情，这是从zip条目中读取的。

The reason this works, while your original code doesn't, is because a "file" inside in a zip file (a jar file is a zip file) is not a real file until it has been extracted. But extracting the file is what you are trying to do, so at that point in your program, it's not a real file. So this question is an X-Y problem: you wanted to create a File object, but that wasn't possible - you needed to refer back to what you were originally trying to do, which was read from the zip entry.

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