gulp循环同步运行任务

Question

Now I have a problem below is my code

//code //config.json

[
    {
        "name": "1"
    },
    {
         "name": "2"
    },
    {
         "name": "3"
    }
]

//gulp

var gulp = require('gulp');
var fs = require('fs');
var gulpsync = require('gulp-sync')(gulp);

var config = JSON.parse(fs.readFileSync(paths.config, 'utf-8'));
var tasks = [];

warehouseConfig.forEach(function(item) {
    var task = item.name;
    tasks.push(task);
    template.dist = task;
    gulp.task('copy:' + task, function() {
        template.dist = task;
        console.log('start copy ' + task + 'dist:' + template.dist);
        gulp.start('copy');
    });
    gulp.task(task, ['copy:' + task], function() {
        return gulp.src(task)
            .pipe(gulp.dest(task + '/'));
    });
});

gulp.task('copy', ['copy:www', 'copy:plugins', 'copy:platforms']);

gulp.task( 'build:pre', gulpsync.sync(tasks));

gulp.task( 'build', ['clean'], function() {
    gulp.start('build:pre');
});

Now,I have the problem, the same task "copy" just run once,I want it run three times. Do you have any good ideas?

Answer 1

if you just want to run that gulp task run thrice you can actually run it in child process :

const exec = require('child_process').exec;
//replace gulp.start('copy') with 
exec('gulp copy');

this will make it async you can make it sync by replacing it with : child_process.execSync

here's link : https://nodejs.org/api/child_process.html#child_process_child_process_execsync_command_options

hope this works as gulp.start() is depreciated: https://github.com/gulpjs/gulp/issues/505