[英]How can I do to reload a symfony2 server with gulp?
I would like to reload automatically my server symfony2 to see in real time the changes on my html pages, when I do some change with gulp, 我想自动重新加载我的服务器symfony2,以便在对gulp进行一些更改时实时查看html页面上的更改,
CASE A: 案例A:
for example: 例如:
php app/console server:run
启动我的symfony2服务器php app/console server:run
gulp
and deploy the task watch that focus on reload every change on the code 启动我的gulp文件gulp
并部署任务监视程序,该监视程序专注于重新加载代码上的每个更改 CASE B: 案例B:
For example 例如
gulp
, in this case setup the symfony2 server from my gulp file and start it when I call the command gulp, and trigger the watch task that focus on relod every change on the code. 星我一饮而尽文件gulp
从我一饮而尽文件,在这种情况下设置了Symfony2的服务器并启动它,当我调用命令一饮而尽,并触发专注于relod上的代码的每一个变化表任务。 Test 1: gulp-connect-php
测试1: gulp-connect-php
var connect = require('gulp-connect-php');
gulp.task('connect', function() {
connect.server();
});
/** here, please don't say me use assets, is a special requirement and is an example*/
gulp.task('productstemplates',function(){
gulp.src( './appjs/products/templates/**/*.html')
.pipe( gulp.dest( './web/products/') )
.pipe( connect.reload() );
});
/** another task to watch the changes over css,less,html pages, php controllers ..*/
gulp.task('default', ['connect','another_task']);
What fails here, when I run this simple configurations not works because this deploy a server over the current folder, then symfony2 server is not deployed, and I get the following 失败的原因是,当我运行此简单配置时,由于在当前文件夹上部署服务器,因此没有部署symfony2服务器,因此无法正常工作,我得到了以下信息
127.0.0.1:55669 [404]: / - No such file or directory
Test 2: gulp-connect
测试2: gulp-connect
'use strict';
var gulp = require('gulp');
var connect = require('gulp-connect');
var concat = require('gulp-concat');
var uglify = require('gulp-uglify');
var livereload = require('gulp-livereload');
var browserSync = require('browser-sync');
var reload = browserSync.reload;
gulp.task('appjs',function(){
gulp.src( './appjs/app.js')
.pipe( gulp.dest('./web/') );
});
gulp.task('app-index',function(){
console.log("chaning something on index.html... ");
gulp.src( './appjs/index.html')
.pipe( gulp.dest( './web/') )
.pipe( connect.reload() );
});
gulp.task('watch',function(){
gulp.watch( ['./appjs/app.js'], ['appjs']);
gulp.watch( ['./appjs/index.html'], ['app-index']);
});
var tasks = [ 'appjs','app-index','watch'];
gulp.task('default',tasks );
Test 3: livereload
测试3: livereload
'use strict';
var gulp = require('gulp');
var connect = require('gulp-connect');
var livereload = require('gulp-livereload');
gulp.task('appjs',function(){
gulp.src( './appjs/app.js')
.pipe( gulp.dest('./web/') );
});
gulp.task('app-index',function(){
console.log('chaning something on index.html...');
gulp.src( './appjs/index.html')
.pipe( gulp.dest( './web/') )
.pipe( livereload({ start: true }) );
});
gulp.task('watch',function(){
livereload.listen();
gulp.watch( ['./appjs/app.js'], ['appjs']);
gulp.watch( ['./appjs/index.html'], ['app-index']);
});
var tasks = [ 'appjs','app-index','watch'];
gulp.task('default',tasks );
then I don't know how to do the rigth setup. 那我就不知道如何进行严格的设置
any body can help me to do the right setup or show me a right way to do it? 任何机构都可以帮助我进行正确的设置或向我展示正确的方法?
Afeter some atempts I got it, so the following is the code to auto refresh my pages when some change is detect 经过一些尝试,我明白了,因此以下是在检测到某些更改时自动刷新页面的代码
1. Settup your gulp file: 1.设置您的gulp文件:
'use strict';
var gulp = require('gulp');
var livereload = require('gulp-livereload');
gulp.task('appjs',function(){
gulp.src( './appjs/app.js')
.pipe( gulp.dest('./web/') );
});
gulp.task('app-index',function(){
console.log('chaning something on index.html...');
gulp.src( './appjs/index.html')
.pipe( gulp.dest( './web/') )
.pipe( livereload({ start: true }) );
});
gulp.task('watch',function(){
livereload.listen();
gulp.watch( ['./appjs/app.js'], ['appjs']);
gulp.watch( ['./appjs/index.html'], ['app-index']);
});
var tasks = [ 'appjs','app-index','watch'];
gulp.task('default',tasks );
2. in the page that you wanna do a change monitoring you must put it: 2.在您要进行更改监视的页面中,必须输入以下内容:
The following code is dev mode. 以下代码是开发人员模式。
<script type="text/javascript">document.write('<script src="//localhost:35729/livereload.js?snipver=1" type="text/javascript"><\/script>')</script>
Source: Integrating Grunt/Gulp and Livereload to existing Apache server serving PHP/Zend 来源: 将Grunt / Gulp和Livereload集成到服务PHP / Zend的现有Apache服务器
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