仅在gulp文件中调用了最后一个回调

Question

在gulpfile中，我有3个任务，当我运行以下代码时，它仅执行最后一个任务的回调。我希望如果我运行gulp命令，在完成[[clean任务后，它应该执行copy：db和default任务的回调。

Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
})

Gulp.task('copy:db', Gulp.series("clean"), function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
})

Gulp.task('default', Gulp.series("copy:db"), function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
});

当我运行命令gulp时，它显示以下日志。
[12:46:37] Starting 'default'...
[12:46:37] Starting 'copy:db'...
[12:46:37] Starting 'clean'...
Clean
[12:46:37] Finished 'clean' after 26 ms
[12:46:37] Finished 'copy:db' after 28 ms
[12:46:37] Finished 'default' after 31 ms

谁能告诉我我要去哪里错了？

Answer 1

为了使代码按您所描述的那样工作，回调函数需要作为参数传递给.series()。例如：

Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
})

Gulp.task('copy:db', Gulp.series(clean, function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
}))

Gulp.task('default', Gulp.series(copy:db, function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
}));

恕我直言，拥有三个完全独立的任务会更简单：
Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
});

Gulp.task('copy:db', function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
});

Gulp.task('default', function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
});

然后用以下任一方式呼叫他们：
Gulp.task('default', gulp.series(clean, copy:db, js));

或
Gulp.task('default', gulp.series(clean, gulp.parallel(copy:db, js)));

希望有帮助：）
附加说明：
gulp / vars的命名约定通常是camelCase，例如：gulp和typeScriptProject而不是Gulp或TypeScriptProject。
您完全可以不用使用gulp.]]来写var {gulp, task, src, dest, watch, series, parallel} = require('gulp');
而不是直接定义任务，您可以通过使用CommonJS exports模块表示法声明任务来使代码更易于阅读。
如果您在使用引号时保持一致，则生活会更轻松一些，而不是混合使用单打和双打。两者都允许遍历
遵循Gulp自己的文档，也许是开始的地方，their sample code on github提供了一些建立基本gulpfile的好例子。
如果把所有的东西都包起来，你会得到这个：
/*
 * Example of requires with gulp methods also requiring gulp.
 */
var { 
    gulp,
    dest,
    series,
    parallel,
    src,
    task,
    watch 
}           = require('gulp'),
vinylPaths  = require('vinyl-paths'), // may not be required, see note in clean func.
del         = require('del'),
ts          = require('gulp-typescript');

/*
 * Added a basic TypeScript Project so the example is complete and could run.
 */
var typeScriptProject = ts.createProject({
    declaration: true
});

/*
 * Your tasks converted to plain/standard js functions.
 */
function clean () {
    return src('dbSchema/*')
        .pipe(vinylPaths(del));

    // Looking at your example code the vinylPaths is redundant,
    // as long as you’re using del ^2.0 it will return its promise,
    // so you could replace the above with:
    return del([ 'dbSchema' ]);
}

function copyDb () {
    return src('db/*')
        .pipe(dest('dbSchema'));
}

function scripts () {
    // example src path
    return src('lib/*.ts')
        .pipe(typeScriptProject())
        .pipe(dest('dist'));
}

/*
 * By defining all the tasks separately it makes it really clear how the tasks will run.
 */
var build = gulp.series (
    clean, 
    gulp.parallel (
        copyDb,
        scripts
    )
);

/*
 * Example of using `exports` module notation to declare tasks.
 */
exports.clean = clean;
exports.copyDb = copyDb;
exports.scripts = scripts;
exports.build = build;

/*
 * Good practise to define the default task as a reference to another task.
 */
exports.default = build;