使用 src/test/resources 作为带有 Gradle Java 的目录。

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英文:

Use src/test/resources as directory with gradle java

问题

我有两个工作目录,包含了 Java 代码和资源文件,一个用于测试,另一个用于生产。

src/main/java
src/main/resources

src/test/java
src/test/resources

当我运行代码并从 FileInputStream 中读取文件时,像这样:

new FileInputStream("./someFile.json");

只要文件位于项目的根目录,这样就可以正常工作。如果我将文件放在资源文件夹中,并直接指向它,就像这样:

new FileInputStream("src/test/resources/someFile.json");

是否有一种方法在 Gradle 中声明 src/test/resources 作为 FileInputStream 查找资源的主文件夹?

如果我将我的文件放在这样的位置,它就可以正常工作。

使用 src/test/resources 作为带有 Gradle Java 的目录。

但是,当我将文件放在 src/test/java/resources 中,就像这样:

使用 src/test/resources 作为带有 Gradle Java 的目录。

我会得到一个 FileNotFoundException。

我需要一种方法将这个目录添加到 Gradle 的扫描范围中。

将下面的内容添加到 build.gradle 中仍然会抛出 FileNotFoundException。

sourceSets {
    test {
        java {
            srcDirs = ['src/test/java']
        }
        resources {
            srcDirs = ['src/test/resources']
        }
    }
}
英文:

I have got two working directories containing java code and resources, one is for testing and the other one is production.

src/main/java <br>
src/main/resources

src/test/java <br>
src/test/resources

When running my code and reading a file from FileInputStream like this:

new FileInputStream(&quot;./someFile.json&quot;);

As long a the file is located in the root of the project this works fine. It also works if I have the file in the resource folder and point directly to it like this:

new FileInputStream(&quot;src/test/resources/someFile.json&quot;);

Is there a way declaring src/test/resources in gradle as the main folder where FileInputStream looks for resources?

If i put my file like this it works fine.

使用 src/test/resources 作为带有 Gradle Java 的目录。

But when I put the file in src/test/java/resources like this

使用 src/test/resources 作为带有 Gradle Java 的目录。

I get a fileNotFoundException. <br>

I need a way to add this directory to be scanned with gradle.

<br><br>

Adding this to build.gradle still throws a fileNotFoundException.

sourceSets {
test {
    java {
        srcDirs = [&#39;src/test/java&#39;]
    }
    resources {
        srcDirs = [&#39;src/test/resources&#39;]
        }
    }
}

答案1

得分: 4

如果您使用 java 或者,由于您似乎在构建命令行应用程序,使用 application 插件,一切都会为您预先设置好:

plugins {
    id 'application'
}

您可以移除所有的 sourceSet 配置,然后像这样加载文件:

new FileInputStream("/someFile.json");

(注意,我写的是 / 而不是 ./,这将把含义从“查找当前目录,项目根目录”更改为“查找类路径资源的根目录”)
根据Slaw在问题下的评论,最好是通过 Class#getResourceAsStream(String) 来定位文件 - 但这已经超出了您的问题范围。

英文:

If you use the java or, since you seem to be building a command line application, the application plugin, everything will be set up for you out of the box:

plugins {
    id &#39;application&#39;
}

You can remove all your sourceSet configurations and just load the file like so:

new FileInputStream(&quot;/someFile.json&quot;);

(Notice that I wrote / instead of ./ which changes the meaning from "look in the current directory, the project root" to "look in the root of a classpath resource".)
It would be even better to follow Slaw's comment under the question and locate the file via Class#getResourceAsStream(String) - but that's going beyond your question a bit.

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  • 本文由 发表于 2020年10月18日 16:27:19
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