Gulp Git 运行第一个提交任务然后推送任务

Question

我正在做项目，我想通过 gulp 提交和推送 git，但是当我运行 git task 时遇到了一些问题，所以推送然后任务不等待提交......

任何人都可以帮助我！我想让任务像首先运行提交然后自动推送，并且在完成提交任务之前不要运行推送任务......

Gulp Git 提交和推送的 Gulp 任务！

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

问题是你告诉 runSequence 插件并行运行任务。解决方法很简单：

// 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 仅在 gulp:commit 完成后运行，并且在推送完成后 gulp:done > 将运行。

另外，我建议你在 gulp:push 中返回流，并在 gulp:done 中使用 done 回调参数，如果你不这样做，gulp 不会'不知道这个任务什么时候结束：

// 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：您也必须在 gulp:commit 任务中返回流，我更改了答案以向您展示如何操作。

Answer 2

gulp-git commit 实际上并不返回流（按设计）。见Commit doesn't fire end #49。

要解决这个问题，您可以使用像 bluebird 这样的 promise 库。

这是 github 用户 bfricka 建议的对我有用的方法：

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.


        ;


    });

});