Gradle 项目和资源文件夹以及用于创建文件的逻辑根

Question

I have a problem understanding how Intellij is working with a Gradle project and the resources folder.

I have created a default Gradle project its created a module 'group', and a module when looking in the module group the src/main/resources folder shows as a 'resource folder' (however it doesn't in the stand-alone module, where groovy/java/resources are all grey).

So that sort out seems to work when compiling code generally.

I tried however to create a file in Groovy script like this

File newFile = new File ("resources/temp.txt")
def fpath = newFile.toURL()
if (!newFile.exists()) {
    println "creating new $fpath file "
    newFile.createNewFile()
}

However run you run this it fails at bit like this

creating new file:/D:/Intellij - Azure/quickstart-java/graph/src/main/groovy/playpen/resources/temp.txt file 
Caught: java.io.IOException: The system cannot find the path specified
java.io.IOException: The system cannot find the path specified
    at java_io_File$createNewFile$1.call(Unknown Source)
    at playpen.TinkerPop-Example.run(TinkerPop-Example.groovy:47)

The File seems to have relative root .../src/main/groovy/playpen which is where my script is. there is no src/main/groovy/playpen/resources/ so it fails

if a use File("/resources/temp.txt") and look at the URL it shows as D:\\resources\\temp.txt` defaulting to same drive as where the script is defined.

If you remove the resources prefix - the file gets created in playpen - again assumed root is same as the source program script.

What I want is to read a file from the 'resources' folder but unless I go to absolute file paths it just ignores the 'resources' folder and only looks in the Groovy source folders.

So for example if I copy the temp.txt into the resources folder and run this

File newFile = new File ("temp.txt")
def fpath = newFile.toURL()
if (!newFile.exists()) {
    println "creating new $fpath file "
    newFile.createNewFile()
} else {
    println "reading file at $fpath"
}

it just creates a new temp.txt in the playpen package where the script runs and doesn't see a copy from 'resources' folder.

So what format of 'file name' do I use so that the 'resources' folder is naturally used to resolve file names - without having to use absolute file names?

Equally if want to create a File programmatically and save that in the 'resources' folder where the script runs from src/main/groovy/playpen , what's the path name that puts it in the correct location.

I'm missing something basic here and can't figure out how to read/or write from the resources folder.

Answer 1

ended up with brute force and ignorance on this one - someone may have a 'smarter'answer, but this one appears to be working. Slightly tweaked some code i got working.

I'm using groovy here rather than java (just less boilerplate noise), and nice File groovy methods

steps - (1)first locate root of your IDE/project using System.getProperty("user.dir") (2) get the current resource file for 'this' - the class you run from ide (3) see if the resource is in $root/out/test/.. dir - it so its a test else its your normal code. (4) set resourcePath now to correct 'location' either in src/main/resources or src/test/resources/ - then build rest of file from this stub (5) check if file exists - if so delete and rewrite this (you can make this cleverer) (6) create file and fill its contents

job done this file should now appear where you expect it. Happy to take cleverer ways to get do this - but for anyone else stuck this seems to do the trick

void exportToFile (dir=null, fileName=null) {

    boolean isTest = false
    String root = System.getProperty("user.dir")
    URL url = this.getClass().getResource("/")
    File loc = new File(url.toURI())
    String canPath = loc.getCanonicalPath()
    String stem = "$root${File.separatorChar}out${File.separatorChar}test"
    if (canPath.contains(stem))
        isTest = true

    String resourcesPath
    if (isTest)
        resourcesPath = "$root${File.separatorChar}src${File.separatorChar}test${File.separatorChar}resources"
    else
        resourcesPath = "$root${File.separatorChar}src${File.separatorChar}main${File.separatorChar}resources"

    String procDir = dir ?: "some-dir-string"
    if (procDir.endsWith("$File.separatorChar"))
        procDir = procDir - "$File.separatorChar"
    String procFileName = fileName ?: "somedefaultname"
    String completeFileName
    File exportFile = "$resourcesPath${File.separatorChar}$procDir${File.separatorChar}${procFileName}"
    exportFile = new File(completeFileName)
    println "path: $procDir, file:$procFileName, full: $completeFileName"

    exportFile.with {
        if (exists())
            delete()
        createNewFile()
        text = toString()
    }

}