我正在尝试将3个调试任务汇集到1个中来完成。由于gulp的本质是异步的,所以我遇到了一些问题。所以我搜索并找到了一个解决方案,使用run-sequence模块来解决这个问题。我尝试了下面的代码,但它似乎没有按预期工作。它无法同步。
以下是我的尝试,请问大家有什么想法吗?我不想运行所有这三个命令才能完成所有任务。我该怎么办?
var gulp = require('gulp'),
useref = require('gulp-useref'),
gulpif = require('gulp-if'),
debug = require('gulp-debug'),
rename = require("gulp-rename"),
replace = require('gulp-replace'),
runSequence = require('run-sequence'),
path = '../dotNet/VolleyManagement.UI';
gulp.task('debug', function () {
gulp.src('client/*.html')
.pipe(debug())
.pipe(gulp.dest(path + '/Areas/WebAPI/Views/Shared'));
});
gulp.task('rename', function () {
gulp.src(path + '/Areas/WebAPI/Views/Shared/index.html')
.pipe(rename('/Areas/WebAPI/Views/Shared/_Layout.cshtml'))
.pipe(gulp.dest(path));
gulp.src(path + '/Areas/WebAPI/Views/Shared/index.html', {read: false})
.pipe(clean({force: true}));
});
gulp.task('final', function(){
gulp.src([path + '/Areas/WebAPI/Views/Shared/_Layout.cshtml'])
.pipe(replace('href="', 'href="~/Content'))
.pipe(replace('src="', 'src="~/Scripts'))
.pipe(gulp.dest(path + '/Areas/WebAPI/Views/Shared/'));
});
gulp.task('debugAll', runSequence('debug', 'rename', 'final'));