Git首先运行提交任务，然后推送任务

Question

我正在做项目，我想提交和推动git的使用，但我在运行git任务时遇到了一些问题，所以推送任务不等待提交.

任何人都可以帮我！我想让任务像第一次运行提交，然后自动推送和不运行推送任务，直到完成提交任务.

吞咽任务杯Git提交和推动！

​

var gulp              = require('gulp'),
    runSequence       = require('run-sequence'),
    gutil             = require('gulp-util'),
    git               = require('gulp-git'),
    prompt            = require('gulp-prompt');


/* task to commit and push all file on git
******************************************************************/
// git commit task with gulp prompt
gulp.task('gulp:commit', function(){
  // just source anything here - we just wan't to call the prompt for now
  gulp.src('./*')
  .pipe(prompt.prompt({
    type: 'input',
    name: 'commit',
    message: 'Please enter commit message...'
  }, function(res){
    // now add all files that should be committed
    // but make sure to exclude the .gitignored ones, since gulp-git tries to commit them, too
    return gulp.src([ '!node_modules/', './*' ], {buffer:false})
    .pipe(git.commit(res.commit));
   }));
});

// Run git push, remote is the remote repo, branch is the remote branch to push to
gulp.task('gulp:push', ['gulp:commit'], function(cb){
    git.push('origin', 'master', cb, function(err){
        if (err) throw err;
    });
});

// # task completed notification with green color!
gulp.task('gulp:done', ['gulp:push'], function(){
  console.log('');
  gutil.log(gutil.colors.green('************** Git push is done! **************'));
  console.log('');
});

// gulp task to commit and push data on git
gulp.task('git', function(){
  runSequence(['gulp:commit', 'gulp:push', 'gulp:done'])
});

​

Answer 1

问题是，您正在告诉runSequence插件并行运行这些任务。解决方案非常简单：

// gulp task to commit and push data on git
gulp.task('git', function(){
  return runSequence('gulp:commit', 'gulp:push', 'gulp:done');
});

通过这样做，您可以确保gulp: push 只在gulp:commit完成之后运行，而在完成push之后，gulp: done 才会运行。

此外，我建议您返回 gulp :push中的流，并在gulp: done 中使用已完成的回调参数，如果不执行，则该gulp不知道这些任务何时结束：

// git commit task with gulp prompt
gulp.task('gulp:commit', function(){
  // just source anything here - we just wan't to call the prompt for now
  return gulp.src('./*')
  .pipe(prompt.prompt({
    type: 'input',
    name: 'commit',
    message: 'Please enter commit message...'
  }, function(res){
    // now add all files that should be committed
    // but make sure to exclude the .gitignored ones, since gulp-git tries to commit them, too
    return gulp.src([ '!node_modules/', './*' ], {buffer:false})
    .pipe(git.commit(res.commit));
   }));
});

// Run git push, remote is the remote repo, branch is the remote branch to push to
gulp.task('gulp:push', ['gulp:commit'], function(cb){
  return git.push('origin', 'master', cb, function(err){
    if (err) throw err;
  });
});

// # task completed notification with green color!
gulp.task('gulp:done', ['gulp:push'], function(done){
  console.log('');
  gutil.log(gutil.colors.green('************** Git push is done! **************'));
  console.log('');
  done();
});

编辑:您必须返回gulp:commit任务中的流，我更改了我的答案，向您展示了如何实现。

Answer 2

实际上并不返回流(按设计)。见#49。

为了解决这个问题，您可以使用像蓝鸟这样的允诺库。

，这是对我有用的东西，如github用户bfricka :所建议的那样

gulp.task('commit-changes', function (cb) {


    var q = require('bluebird'); //commit needs a a promise
    return new q(function (resolve, reject) {



        gulp.src('../')
                .pipe(git.add())
                .pipe(stream = git.commit("my commit message")).on('error', function (e) {
            console.log('No changes to commit');
        })

                .on('error', resolve) //resolve to allow the commit to resume even when there are not changes.
                .on('end', resolve);



        stream.resume(); //needed since commmit doesnt return stream by design.


        ;


    });

});