我正在使用 gulp。任务正在运行创建所需的文件夹。但是在浏览器中运行时出现Cannot GET/错误。我附上了我的项目结构的图像以及输出
我的 index.html 包含以下内容：
<!DOCTYPE html>
<html class="no-js" lang="en">
<head>
<title>Bootstrap 4 Layout</title>
<meta http-equiv="x-ua-compatible" content="ie=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Raleway:400,800">
<link rel='stylesheet' href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css">
<link rel="stylesheet" href="/css/bootstrap.css">
<link rel="stylesheet" href="/css/styles.css">
</head>
<body>
<script src="/js/jquery.min.js"></script>
<script src="/js/popper.min.js"></script>
<script src="/js/bootstrap.min.js"></script>
</body>
</html>
以下是我的 gulpfile.js
var gulp = require('gulp');
var browserSync = require('browser-sync').create();
var sass = require('gulp-sass');
// Compile sass into CSS & auto-inject into browsers
gulp.task('sass', function() {
return gulp.src(['node_modules/bootstrap/scss/bootstrap.scss', 'src/scss/*.scss'])
.pipe(sass())
.pipe(gulp.dest("src/css"))
.pipe(browserSync.stream());
});
// Move the javascript files into our /src/js folder
gulp.task('js', function() {
return gulp.src(['node_modules/bootstrap/dist/js/bootstrap.min.js', 'node_modules/jquery/dist/jquery.min.js', 'node_modules/popper.js/dist/umd/popper.min.js'])
.pipe(gulp.dest("src/js"))
.pipe(browserSync.stream());
});
// Static Server + watching scss/html files
gulp.task('serve', ['sass'], function() {
browserSync.init({
server: "./src"
});
gulp.watch(['node_modules/bootstrap/scss/bootstrap.scss', 'src/scss/*.scss'], ['sass']);
gulp.watch("src/*.html").on('change', browserSync.reload);
});
gulp.task('default', ['js','serve']);