I'm using gradle and gulp to build and launch my JEE web application that is based on JDK6+, and the web container is Jetty.
Gulp is used to process static resource, such as concat and minify javascript/css files. Also, there is a watch
task in my gulp script, it is used to watch static file changes and rebuild them automatically.
So to launch the application in Jetty, i need to do following things:
- Build static resource using gulp
[gulp task: build]
.
- Watch static file changes
[gulp task: watch]
, i don't call this gulp task directly, it is call by a gradle task via Exec class [gradle task: watch]
.
- Build my java code and launch Jetty
[gradle task: jettyRun]
.
Since i added the gradle task watch
as a dependency of jettyRun
task, so i though i only need to call gradle jettyRun
from command line, my application will be launched. But the result is not same as i expected.
Following is my script files:
build.gradle
apply plugin: 'war'
apply plugin: 'jetty'
repositories {
jcenter()
}
// omit the dependencies here
task watch(type: Exec){
workingDir "${projectDir}"
// Pass build type to npm and gulp.
commandLine "gulp", "watch"
}
jettyRun.dependsOn watch
gulpfile.js
var gulp = require('gulp');
gulp.task('build', function(callback) {
// removed the code to make this question as simple as possible.
if (callback != null) callback();
});
gulp.task('watch', function(callback) {
gulp.watch(['./src/static/**/*.js'], ['build']);
if (callback != null) callback();
});
The result:
Question:
Now, the process hungs when excutes the gradle watch
task, there is no change to execute jettyRun
task for gradle. I know this hang is caused by the watch process launched by gulp, because it is watching file changes. But i hope that gradle just only launch the gulp watch process and return immediately to execute the next jettyRun
task!
How to do this? Also, i want to see the output of watch task from stdout.
I know there is a ProcessBuilder in java, i have tried it, but it doesn't work. Maybe I have done something wrong. :(
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