小编Blo*_*kas的帖子

在messages.en.yml,我有

confirmed: Congrats %username%, your account is now activated.

但我想用'粗体'用户名示例......我怎么能做到这一点？

confirmed: Congrats <span class='bold'>%username%</span>, your account is now activated.   

当然我可以在这个例子中使用两个句子

first: Congrats
second: , your account ... 

和twig内部使用html标签,但这看起来很脏.

'use strict';

var trkiApp = angular.module('trkiApp', [ 
  'trkiApp.tStatus', 
  'trkiApp.feed'
]);



var tStatus = angular.module('trkiApp.tStatus', [])
    .factory('Status', ['$q']);

var feed = angular.module('trkiApp.feed', []);

现在我不明白我怎么可能访问在另一个模块上定义的服务状态？

'use strict';

feed
    .controller('FeedController', ['$scope','$http','Status']);

我不应该对吗？但不知怎的,我......或者这是一个正确的行为？

我失去了trilion时间谷歌,但没有一个解决方案是好的.

我有这个查询构建器:

        $qb2=$this->createQueryBuilder('s')
        ->addSelect('u')
        ->innerJoin('s.user','u')
        ->where("u.id IN(:followeeIds)")
        ->andWhere('s.admin_status = false')
        ->setParameter('user', $user)
        ->setParameter('followeeIds', $arrayFolloweeIds)
        ->orderBy('s.id','DESC')
        ->setMaxResults(15)
    ;

我可以做第二个查询然后做,$qb->getDQL()但我会缓存查询？

错误:

Invalid parameter number: number of bound variables does not match number of tokens

好的,我有这个代码:

SELECT 
IFNULL(s2.id,s1.id) AS effectiveID, 
IFNULL(s2.status, s1.status) AS effectiveStatus, 
IFNULL(s2.user_id, s1.user_id) as effectiveUser,
IFNULL(s2.likes_count, s1.likes_count) as effectiveLikesCount

FROM statuses AS s1
LEFT JOIN statuses AS s2 ON s2.id = s1.shared_from_id
WHERE s1.user_id = 4310
ORDER BY effectiveID DESC
LIMIT 15

我需要将它重写为querybuilder.那样的东西？

        $fields = array('IFNULL(s2.id,s1.id) AS effectiveID','IFNULL(s2.status, s1.status) AS effectiveStatus', 'IFNULL(s2.user_id, s1.user_id) as effectiveUser','IFNULL(s2.likes_count, s1.likes_count) as effectiveLikesCount');

    $qb=$this->_em->createQueryBuilder()
             ->select($fields)
             ->from('WallBundle:Status','s1')
             ->addSelect('u')
             ->where('s1.user = :user')
             ->andWhere('s1.admin_status = false')
             ->andWhere('s1.typ_statusu != :group')
             ->setParameter('user', $user)
             ->setParameter('group', 'group')
             ->leftJoin('WallBundle:Status','s2', 'WITH', 's2.id=s1.shared_from_id')  
             ->innerJoin('s1.user', 'u')               
             ->orderBy('s1.time', 'DESC') …

好吧,我很绝望.我正在使用FOSRestBundle和NelmioCorsBundle

当我尝试通过不同服务器的角度发布数据时,我仍然收到此错误:

XMLHttpRequest cannot load 
http://IP/app_dev.php/api/v1/pages.json. 
No 'Access-    Control-Allow-Origin' header is present on the requested resource.
Origin      'http://127.0.0.1:9000' is therefore not allowed access.



Remote Address:IP:80
Request URL:http://IP/app_dev.php/api/v1/pages.json
Request Method:OPTIONS
Status Code:200 OK
Request Headers
Accept:*/*
Accept-Encoding:gzip,deflate,sdch
Accept-Language:en-US,en;q=0.8,sk;q=0.6,cs;q=0.4
Access-Control-Request-Headers:accept, authorization, content-type
Access-Control-Request-Method:POST
Cache-Control:no-cache
Connection:keep-alive
Host:192.168.43.195
Origin:http://127.0.0.1:9000
Pragma:no-cache
Referer:http://127.0.0.1:9000/
User-Agent:Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/34.0.1847.116 Safari/537.36

Response Headers
Access-Control-Allow-Headers:X-Requested-With, content-type
Access-Control-Allow-Methods:POST, GET, PUT, DELETE, OPTIONS
Connection:Keep-Alive
Content-Length:573
Content-Type:text/html; charset=iso-8859-1
Date:Sun, 29 Jun 2014 18:07:54 GMT
Keep-Alive:timeout=5, max=100
Server:Apache/2.4.7 (Ubuntu) …

我试图通过我的第一个脚本执行一些命令,但它不起作用.

#!/bin/bash

#connect to server
echo "Connecting to the server..."

ssh -t root@IP '

    #switch user to deploy
    su - deploy

    #switch path
    echo "Switching the path"
    cd /var/www/deploys/bin/app/config

    #run deploy script
    echo "Running deploy script"

    /usr/local/bin/cap -S env=prod deploy

    #restart apache
    sudo /bin/systemctl restart  httpd.service

    bash -l
'

怎么了？我成功连接到服务器,用户被更改,然后我没有看到任何事情发生.当我按下ctrl + c就像在终端中那样,应该执行的命令的一些输出出现,但是有一些错误.

为什么我在启动脚本后看不到终端中发生的一切？我这样做是错误的吗？

顺便说一句:当我尝试手动连接并自己运行命令时,一切都运行良好.

使用CentOS 7.

我将连接的手机用作ADB设备.

→phonegap -v 3.1.0-0.15.0

我的index.html几乎没有任何内容

    <script type="text/javascript" src="phonegap.js"></script>
    <script type="text/javascript" src="cordova.js"></script>
    <script type="text/javascript" src="js/jquery-1.10.2.min.js"></script>
    <script type="text/javascript" src="js/bootstrap.min.js"></script>
    <script type="text/javascript" src="js/angular.min.js"></script>
    <script type="text/javascript" src="js/main.js"></script>
</body>

当我跑

? phonegap local run android  
[phonegap] compiling Android...
[phonegap] successfully compiled Android app
[phonegap] trying to install app onto device
[phonegap] successfully installed onto device

我可以看到我的应用程序运行,之前没有注意到控制台中的错误.然后我尝试了更复杂的东西,如ngRoute,但它没有用.我决定将所有数据删除到几乎为空的项目,如您所见(确保其他内容不会导致此错误).

然后我检查了控制台:

adb logcat | grep -i console

我明白了:

I/Web Console( 3946): Could not find cordova.js script tag. Plugin loading may 

fail.:1511
E/Web Console( 3946): Uncaught module cordova/plugin_list already defined:76 …

我想做的事情并不难.这是我关于symfony的第一个项目,这真的令人困惑.

我正在使用FOSUSerbundle.我不希望登录和注册下面/login和/registration

所以我做了一个捆绑,这是FOSUSerbundle的孩子......它覆盖了它的树枝.

我有:: base.html.twig,其中包括header.html.twig,我在那里:{% render 'FOSUserBundle:Security:login' %}渲染我的teplate(覆盖了FOS)工作gr8.即使提交后的错误也会在:: base模板下呈现"/"路径.

#Security.yml
    form_login:
        check_path: /login_check
        login_path: /
        provider: fos_userbundle

效果很好.

我需要为我的注册做同样的事情.

所以在::base我包含welcome_page.html.twig我编码的地方{% render 'FOSUserBundle:Registration:register' %},我在我的重写模板下面:WelcomePageBundle:Registration:register.html.twig这个:

{% block fos_user_content %}
{% include "FOSUserBundle:Registration:register_content.html.twig" %}
{% endblock fos_user_content %}[/code]

其中还包括我重写的包:WelcomePageBundle:Registration:register_content.html.twig这个:

{% for key, message in app.session.getFlashes() %}
<div class="{{ key }}">
    {{ message|trans({}, 'FOSUserBundle') }}
</div>
{% endfor %}

<form action="{{ path('fos_user_registration_register') }}" {{ form_enctype(form) }} method="POST" id="register_form">
    {{ …

我想知道为什么表单类型中的max_length已弃用？

如何以最干净的方式达到预期的效果呢？

尝试按用户获取喜欢的状态。

public function getLikedStatuses(User $user)
{
    $qb = $this->_em->createQueryBuilder();
                $qb
                ->select('s.id')
                ->from('WallBundle:Likes','l')
                ->innerJoin('l.status', 's')
                ->where('l.user = :user')
                ->setParameter('user', $user)
                ->orderBy('s.id','DESC')
            ;

    $qb2=  $this->_em->createQueryBuilder()
        ->select('st')
        ->from('WallBundle:Status','st');

       $qb2 ->andWhere($qb2->expr()->in('st.id',$qb->getDQL()));


    return $qb2->getQuery()->getResult();
}

错误：无效的参数编号：绑定变量的数量与令牌的数量不匹配

顺便说一句：当我转储$ qb-> getDQL（）时：

string 'SELECT s.id FROM TB\WBundle\Entity\Likes l LEFT JOIN l.status s WHERE l.user = :user' (length=87)

BTW2：当我用（12073）（状态ID）替换'$ qb-> getDQL（）'时，它起作用了...