我是gulp-4的新手。我尝试用 php 运行 gulp js,浏览器同步它不起作用?所有其他任务都可以工作,但 php 和浏览器同步根本不起作用。它没有打开浏览器有什么问题吗?是否可以与带有浏览器同步或任何限制的PHP一起使用?我想将浏览器同步与 php 一起使用,但似乎无法让它工作。
这是我的代码...
const browsersync = require("browser-sync").create();
const gulp = require("gulp");
const imagemin = require("gulp-imagemin");
const sass = require("gulp-sass");
const plumber = require("gulp-plumber");
const postcss = require("gulp-postcss");
const del = require("del");
const rename = require("gulp-rename");
const autoprefixer = require("autoprefixer");
const cssnano = require("cssnano");
const newer = require("gulp-newer");
const uglify = require('gulp-uglify');
const concat = require('gulp-concat');
const php = require('gulp-connect-php');
//Php connect
function connectsync() {
php.server({}, function (){
browserSync({
proxy: 'maniadev'
});
});
}
// BrowserSync Reload
function browserSyncReload(done) {
browserSync.reload();
done();
}
// Clean assets
function clean() {
return del(["./dist/assets/"]);
}
// Optimize Images
function images() {
return gulp
.src("./app/assets/img/**/*")
.pipe(newer("./app/assets/img"))
.pipe(
imagemin([
imagemin.gifsicle({ interlaced: true }),
imagemin.jpegtran({ progressive: true }),
imagemin.optipng({ optimizationLevel: 5 }),
imagemin.svgo({
plugins: [
{
removeViewBox: false,
collapseGroups: true
}
]
})
])
)
.pipe(gulp.dest("./dist/assets/img"));
}
// CSS task
function css() {
return gulp
.src("./app/assets/sass/**/*.scss")
.pipe(plumber())
.pipe(sass({ outputStyle: "expanded" }))
.pipe(gulp.dest("./dist/assets/css/"))
.pipe(rename({ suffix: ".min" }))
.pipe(postcss([autoprefixer(), cssnano()]))
.pipe(gulp.dest("./dist/assets/css/"))
.pipe(browsersync.stream());
}
// Transpile, concatenate and minify scripts
function scripts() {
return (
gulp
.src(["./app/assets/js/**/*"])
.pipe(plumber())
.pipe(uglify())
.pipe(concat('main.min.js'))
// folder only, filename is specified in webpack config
.pipe(gulp.dest("./dist/assets/js/"))
.pipe(browsersync.stream())
);
}
// Watch files
function watchFiles() {
gulp.watch("./app/assets/scss/**/*", css);
gulp.watch("./app/assets/js/**/*", gulp.series( scripts));
gulp.watch(
gulp.series(browserSyncReload)
);
gulp.watch("./app/assets/img/**/*", images);
gulp.watch("./app/**/*.php", gulp.series( browserSyncReload ));
}
// define complex tasks
const js = gulp.series(scripts);
const build = gulp.series(clean, gulp.parallel(css, images, js));
const watch = gulp.parallel(watchFiles, connectsync);
// export tasks
exports.images = images;
exports.css = css;
exports.js = js;
exports.clean = clean;
exports.build = build;
exports.watch = watch;
exports.default = build;
我在 oder 中用于创建 PHP 服务器的 gulpfile.js(Gulp 版本 4.0.2)如下:
假设您有一个名为"dist"的文件夹,其中有一个"index.php"文件:
const gulp = require('gulp');
var php = require("gulp-connect-php");
var browsersync = require('browser-sync');
gulp.task('default', function() {
php.server({
// a standalone PHP server that browsersync connects to via proxy
port: 8000,
keepalive: true,
base: "dist"
}, function (){
browsersync({
proxy: '127.0.0.1:8000'
});
});
});
我希望这对您有所帮助作为参考。我只写文件的相关部分,因为您似乎已经让其余部分开始工作了。
为了使用函数而不是任务(如在代码中)并监视 php 文件的更改,您可以执行以下操作:
假设您想观看位于/src 文件夹中的 php 文件,并在有更改时修改/dist 文件夹。
const browsersync = require("browser-sync");
const gulp = require("gulp");
const phpConnect = require('gulp-connect-php');
//Php connect
function connectsync() {
phpConnect.server({
// a standalone PHP server that browsersync connects to via proxy
port: 8000,
keepalive: true,
base: "dist"
}, function (){
browsersync({
proxy: '127.0.0.1:8000'
});
});
}
// BrowserSync Reload
function browserSyncReload(done) {
browsersync.reload();
done();
}
function php(){
return gulp.src("./src/**/*.php").pipe(gulp.dest("./dist"));
}
// Watch files
function watchFiles() {
gulp.watch("src/**/*.php", gulp.series(php, browserSyncReload));
}
const watch = gulp.parallel([watchFiles, connectsync]);
exports.default = watch;
注意:"浏览器同步"仅在有标签"body"时才有效。