如何添加一个.on
侦听器来阻止gulp崩溃,并在下次检测到文件重新加载时再试一次?
这就是我目前所尝试的
'use strict';
const gulp = require('gulp');
const jspm = require('gulp-jspm');
gulp.task('default', () => {
return gulp.watch('./src/**/*.js', ['bundle'])
.on('error', function (error) {
console.log('error');
this.emit('end');
});
});
gulp.task('bundle', () => {
return gulp.src('./src/main.js')
.pipe(jspm({ selfExecutingBundle : true })
.on('error', function (error) {
console.log('error');
this.emit('end');
})
)
.on('error', function (error) {
console.log('error');
this.emit('end');
})
.pipe(gulp.dest('./dist/'));
});
其中没有一个捕获错误,或阻止gulp崩溃。
答案 0 :(得分:0)
您可以在此处查看我的gulpfile.js
。
const gulp = require('gulp');
const gulp_jspm = require('gulp-jspm');
const paths = {
//Your main.js
main: 'public/js/main.js',
//Your all js files
scripts: 'public/js/**',
dest: 'public/dest/'
};
gulp.task('scripts', () => {
gulp.src(paths.main)
.pipe(gulp_jspm({
selfExecutingBundle: true
}))
.pipe(gulp.dest(paths.dest));
});
function reportChange(event){
console.log('File ' + event.path + ' was ' + event.type + ', running tasks...');
}
gulp.task('watch', () => {
gulp.watch([paths.scripts], ['scripts']).on('change', reportChange);
});
gulp.task('default', ['watch', 'scripts']);