I have the following scenario in my Android project:
Project1 --> Build.gradle (1)
Project2--> Build.gradle (2)
Example: Define the following variable:
//first Gradle file
def getProductFlavor() {
//Logic here
Gradle gradle = getGradle()
String requestingTask = gradle.getStartParameter().getTaskRequests().toString()
return requestingTask
}
Instead of defining getCurrentTime() in the second Gradle file,
I call the getCurrentTime() from the first Gradle file.
Maybe my example is wrong and the default value needs to be implementing in another gradle file like the script gradle or somewhere else, but the intent of the example was to clarify what I'm trying to achieve.
The two projects are independents but both belong to the same android project. I want to use ONE def value in both of these gradle files.
I'm a gradle newbie by the way. Never mind if I'm asking this question the wrong way.
Feels like a scenario for external script:
main-project
....| sub-project1
........| src
........| build.gradle
....| sub-project2
........| src
........| build.gradle
....| common.gradle
build.gradle
apply from: '../common.gradle'
def flavor = getProductFlavor()
common.gradle
def getProductFlavor() {
Gradle gradle = getGradle()
String requestingTask = gradle.getStartParameter().getTaskRequests().toString()
return requestingTask
}
ext {
getProductFlavor = this.&getProductFlavor
}
Related
I need a property to (one) be available when a plugin is applied and (two) allow for a calculated override value in the settings.gradle file. A project property would be ideal as it can have a default set in gradle.properties:
# gradle.properties
myProp=originalValue
This is great because it can be overrode with a command line argument like -PmyProp=newValue, but I was not able to find a good way to override the property in the settings.gradle file before the build.gradle executes (i.e. before the plugins are applied).
For instance all of these leave rootProject.myProp unaltered at plugin application:
// settings.gradle
rootProject.getProperties.put("myProp", "overrideValue")
settings.ext.myProp = "overrideValue"
settings.extensions.myProp = "overrideValue"
gradle.startParameters.projectProperties.myProp = "overrideValue"
We cannot do any magic in the build.gradle either because no logic can exist before the plugins block:
// build.gradle
plugins {
id 'com.myCompany.myPlugin' version 1.0.0 // 'myProp' must be set by now
}
One workaround I can think of would be to use:
// settings.gradle
gradle.ext.myProp = "overrideValue"
... but there doesn't seem to be a good way to access gradle.ext properties in Java source code (for a plugin), or is there?
This seems to work for the gradle.ext.myProp use case, but it is surprising to me that the only workable approach is to cast the Gradle object to an ExtensionAware object:
// MyPlugin.java
String myProp = (String) project.getRootProject().getProperties().getOrDefault("myProp", null);
Gradle gradle = project.getRootProject().getGradle();
if ((myProp == null) && (gradle instanceof ExtensionAware)) {
ExtensionAware gradleExtensions = (ExtensionAware) gradle;
myProp = (String) gradleExtensions.getExtensions().getExtraProperties().get("myProp");
}
It seems like what I'm trying to do should be commonplace, so is there a better way like solely using project properties?
If so, then how do you change the values in the settings.gradle file?
This is probably not what you’re looking for but maybe it still helps: have you considered an initialization script? In such a script it is possible to override a project property.
Example:
$ ./gradlew -PmyProp=originalValue properties | grep myProp
myProp: originalValue
$ ./gradlew -PmyProp=originalValue -I init.gradle properties | grep myProp
myProp: overrideValue
… where init.gradle is the following:
allprojects {
project.ext.myProp = 'overrideValue'
}
Note that there are also other ways of specifying the init script.
Basically I want to do exactly this:
Use Gradle function from other gradle file
which is to say, call one function in one gradle script from another. The wrinkle is that my build.gradle is in Kotlin (build.gradle.kts) and the script my function is in is still in groovy.
I followed the above link for groovy-to-groovy, but I can't get this to work using the Kotlin DSL.
In my groovy file, functions.gradle, I have:
def buildVersionName() {
//Do some stuff
}
And
ext {
buildVersionName = this.&buildVersionName
}
Then, in my build.gradle.kts script, I have:
apply(from = "functions.gradle")
project.ext.buildVersionName()
When I sync, I get the error:
Unresolved reference: buildVersionName
In the build.gradle.kts do the following:
import groovy.lang.Closure
apply(from="functions.gradle")
val buildVersionName: Closure<Any> by ext
buildVersionName()
I have a KotlinJs only project which I use official kotlin2js gradle to build, and no problems there.
How to setup the output folder, currently, the building of subproject will result in a build which locates inside the subproject folder, how to set it to somewhere else? I tried:
sourceSets {
main {
kotlin.outputDir = new File(‘./out/‘)
}
}
and
sourceSets {
main.kotlin.outputDir = new File(‘./out/’)
}
No luck.
What I want is to no matter how many subprojects are there, the output folder should be in some path like ./build/projectA and ./build/projectB, rather than all in their own folder. How to do this?
Currently, it's done through the task configuration, namely setting its kotlinOptions.outputFile:
compileKotlin2Js.kotlinOptions.outputFile = "out/output.js"
It's briefly mentioned in the tutorial: Getting Started with Kotlin and JavaScript with Gradle
I have a multi-project Gradle build, which is currently configured through a single build.gradle file.
There are over 70 modules in this project, and the single (gigantic) build.gradle file has become cumbersome to use, so I'd like to split it into small per-module buildscript files.
Now, I don't want to have 70 small build.gradle files (one in each module), as that would make navigating to a specific build.gradle a pain in the IDE (the only difference between the files is their path).
What I want is my per-module buildscript files to be named after the module name.
Instead of this:
root
|--foo\
|--| build.gradle
|--bar\
|--| build.gradle
I want this:
root
|--foo\
|--| foo.gradle
|--bar\
|--| bar.gradle
Since this doesn't seem to be officially supported, I tried hacking around the root build.gradle a bit, but it seems that applying a .gradle file happens before the projects are configured, so this gives an error for projects that depend on other projects:
in root build.gradle:
subprojects { subProject ->
rootProject.apply from: "${subProject.name}/${subProject.name}.gradle"
}
foo.gradle, which is not a standard build.gradle file:
project('foo') {
dependencies {
compile project(':bar')
}
}
Is there any way of making it work like this?
A web search for "gradle rename build.gradle" rendered the below example settings.gradle file:
rootProject.buildFileName = 'epub-organizer.gradle'
rootProject.children.each { project ->
String fileBaseName = project.name.replaceAll("\p{Upper}") { "-${it.toLowerCase()}" }
project.buildFileName = "${fileBaseName}.gradle"
}
Note that the author is here also renaming the root project's build script, which you may or may not want.
One of the authors of Gradle, Hans Dockter, has said somewhere (I believe it was in his "Rocking the Gradle" demo from 2012), that he felt one of their biggest mistakes was using build.gradle as the default file name.
You can customize name of your build scripts in settings.gradle file. Check recent presentation from Ben Muschko about multi-project builds or look at Gradle sources where similar customization is done.
rootProject.children.each {
it.buildFileName = it.name + '.gradle'
}
You can find this content in Gradle in action, manning
I ran across the following configuration in a Gradle project's buildfile (build.gradle):
codenarcMain {
configFile = rootProject.file("gradle/codenarc/CodeNarcMain.groovy")
}
When I Google "Gradle rootProject" I find this link which makes it look like rootProject is a ProjectDescriptor instance.
But looking at ProjectDescriptor, I don't see any property called file. I see a buildFile and projectDir, but no file property.
Ultimately, I am trying to get Gradle to load the CodeNarc config file from outside the build directory. On my system I have an env var called $CODENARC_HOME with the following directory structure:
CODENARC_HOME/ (say this is /home/myuser/tools/codenarc/)
CodeNarcMain.groovy
CodeNarcTest.groovy
README.md
Now I would like to change the CodeNarc config in Gradle to look something like this:
codenarcMain {
configFile = rootProject.file("CODENARC_HOME/CodeNarcMain.groovy")
}
And then, no matter where CODENARC_HOME is defined, the Gradle build will still be able to locate the config file.
So my questions:
What is the file property on rootProject, and why don't I see it in the API docs (linked above)?; and
How to get rootProject.file(...) to honor system/env vars inside its file path string argument?
rootProject in settings.gradle is-a ProjectDescriptor. rootProject in build.gradle is-a Project. Environment variables and system properties can be accessed in the standard Java way:
codenarcMain {
configFile = rootProject.file("${System.getenv("CODENARC_HOME")}/CodeNarcMain.groovy")
// or: System.getProperty("codenarc.home")
}