Considering you already have a gulpfile.js
, the following should work, although not tested.
const gulp = require('gulp');
const { spawn } = require('child_process');
// Task to build a project
function buildProject(projectPath) {
return new Promise((resolve, reject) => {
const buildProcess = spawn('dotnet', ['build', projectPath]);
buildProcess.stdout.on('data', (data) => {
console.log(data.toString());
});
buildProcess.stderr.on('data', (data) => {
console.error(data.toString());
});
buildProcess.on('close', (code) => {
if (code === 0) {
resolve();
} else {
reject(new Error(`Build process exited with code ${code}`));
}
});
});
}
// Task to build all projects under the Feature folder
gulp.task('build-feature-projects', async function () {
try {
await buildProject('./FeatureFolder/Project1');
await buildProject('./FeatureFolder/Project2');
await buildProject('./FeatureFolder/Project2/Project3');
console.log('All feature projects built successfully.');
} catch (error) {
console.error('Error building feature projects:', error);
}
});
// Default task
gulp.task('default', gulp.series('build-feature-projects'));
If you want to build all projects inside the folder you can do the following:
const fs = require('fs');
const path = require('path');
// Task to build all projects under the Feature folder
gulp.task('build-feature-projects', async function () {
try {
const featureFolder = './FeatureFolder';
const projects = fs.readdirSync(featureFolder)
.map(fileName => path.join(featureFolder, fileName))
.filter(file => fs.statSync(file).isDirectory());
for (const project of projects) {
await buildProject(project);
}
console.log('All feature projects built successfully.');
} catch (error) {
console.error('Error building feature projects:', error);
}
});