小编Zoh*_*han的帖子

我有以下ajax调用

var prev_sibling = $(this).prev().attr("value");
var next_sibling = $(this).next().attr("value");
var order = (prev_sibling + next_sibling)/2; 
var data = {PID:element_id, TGID:parent_id, ORD:order};
$.ajax({
        type: "POST",
        data: data,
        url:"{{ path('v2_pm_patents_dragpatents') }}",
        cache: false
});

在我的行动中,我得到了订单并将其设置为这样

$order = $request->get('ORD');
$patent->setOrder($order);

但是ajax调用给了我以下错误

Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order = '750' WHERE id = '0d0c0810-bc75-11e1-96a5-9787dec335c2'' at line 1 (500 …

我安装了wamp服务器和Symfony2框架的副本.我正在尝试使用以下命令创建一个Bundle:

php app/console generate:bundle --nampespace=IDP/IDP_Bundle --format=yml

我的PHP在 C:/wamp/bin/php/php5.3.10

但是当我运行命令时,它说:

could  not open input file app/console

谁能告诉我出了什么问题？

嘿,我的symfony2项目有两个捆绑包.一个是Bundle,另一个是PatentBundle.

我的app/config/route.yml文件是

MunichInnovationGroupPatentBundle:
resource: "@MunichInnovationGroupPatentBundle/Controller/"
type:     annotation
prefix:   /
defaults:  { _controller: "MunichInnovationGroupPatentBundle:Default:index" }

MunichInnovationGroupBundle:
resource: "@MunichInnovationGroupBundle/Controller/"
type:     annotation
prefix:   /v1
defaults:  { _controller: "MunichInnovationGroupBundle:Patent:index" }

login_check:
pattern:   /login_check

logout:
pattern:   /logout

在我的控制器里面我有

<?php
namespace MunichInnovationGroup\PatentBundle\Controller;

use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\Request;
use JMS\SecurityExtraPatentBundle\Annotation\Secure;
use Symfony\Component\Security\Core\Exception\AccessDeniedException;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
use Symfony\Component\Security\Core\SecurityContext;

use MunichInnovationGroup\PatentBundle\Entity\Log;
use MunichInnovationGroup\PatentBundle\Entity\UserPatent;
use MunichInnovationGroup\PatentBundle\Entity\PmPortfolios;
use MunichInnovationGroup\PatentBundle\Entity\UmUsers;
use MunichInnovationGroup\PatentBundle\Entity\PmPatentgroups;
use MunichInnovationGroup\PatentBundle\Form\PortfolioType;
use MunichInnovationGroup\PatentBundle\Util\SecurityHelper;
use Exception;
/**
 * Portfolio controller.
 * @Route("/portfolio")
 */
class PortfolioController extends …

我的代码中有一个ajax调用.我希望通过该调用实现的功能正常.我想删除数据库中的一些记录,这些记录在通过ajax调用方法时实际被删除但是在symfony方法中它必须返回一个响应,这就是为什么当执行该方法时它会给我错误

我的Ajax调用是

  $.ajax({
        type: "POST",
        data: data,
        url:"{{ path('v2_pm_patents_trashpatents') }}",
        cache: false,
        success: function(){
        document.location.reload(true);
        }
    });

并且执行的方法是

 public function trashpatentAction(Request $request){
    if ($request->isXmlHttpRequest()) {
        $id = $request->get("pid");
        $em = $this->getDoctrine()->getEntityManager();
        $patent_group = $em->getRepository('MunichInnovationGroupPatentBundle:PmPatentgroups')->find($id);
        if($patent_group){
            $patentgroup_id = $patent_group->getId();
            $em = $this->getDoctrine()->getEntityManager();
            $patents = $em->getRepository('MunichInnovationGroupPatentBundle:SvPatents')
            ->findBy(array('patentgroup' => $patentgroup_id));
            if($patents){
                foreach($patents as $patent){
                    if($patent->getIs_deleted()==false){
                        $patent->setIs_deleted(true);
                        $em->flush();
                    }
                }
            }
            $patent_group->setIs_deleted(true);
            $em->flush();
        }
        else{
            $em = $this->getDoctrine()->getEntityManager();
            $patent = $em->getRepository('MunichInnovationGroupPatentBundle:SvPatents')->find($id);
            if ($patent) {
                $patent->setIs_deleted(1);
                $em->flush();
            }
        }
        return true;
    }
}

如何从此方法成功返回？有任何想法吗？谢谢

如何在Symfony2控制器中定义构造函数.我想在我的控制器的所有方法中获取登录的用户数据,目前我在我的控制器的每个操作中执行类似的操作以获取登录用户.

$em = $this->getDoctrine()->getEntityManager("pp_userdata");
$user = $this->get("security.context")->getToken()->getUser();

我想在构造函数中执行一次,并使我的所有操作都可以使用此登录用户

如何使用symfony2中的Doctrine检查记录是否已成功插入数据库？

我在控制器中的动作是

public function createAction(){
    $portfolio = new PmPortfolios();
    $portfolio->setPortfolioName('Umair Portfolio');
    $em = $this->getDoctrine()->getEntityManager();
    $em->persist($portfolio);
    $em->flush();
    if(){
         $this->get('session')->setFlash('my_flash_key',"Record Inserted!");
    }else{
         $this->get('session')->setFlash('my_flash_key',"Record notInserted!");
    }
}

我应该在if声明中写些什么？

我试图在symfony2中使用注释来定义我的路线.我的Bundle名称是PatentBundle.但是我得到了一个错误

No route found for "GET /portfolio/

我的app/config/routing.yml

MunichInnovationGroupPatentBundle:
resource: "@MunichInnovationGroupPatentBundle/Controller/"
type:     annotation
prefix:   /
defaults:  { _controller: "MunichInnovationGroupPatentBundle:Default:index" }

我的组合控制器看起来像

<?php
 namespace MunichInnovationGroup\PatentBundle\Controller;

 use MunichInnovationGroup\PatentBundle\Entity\Log;

 use MunichInnovationGroup\PatentBundle\Entity\UserPatent;
 use Symfony\Component\HttpFoundation\Response;
 use Symfony\Component\HttpFoundation\Request;
 use JMS\SecurityExtraBundle\Annotation\Secure;
 use Symfony\Component\Security\Core\Exception\AccessDeniedException;
 use Symfony\Bundle\FrameworkBundle\Controller\Controller;
 use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method;
 use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
 use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
 use MunichInnovationGroup\PatentBundle\Entity\SvPatents;
 use MunichInnovationGroup\PatentBundle\Entity\PmPortfolios;
 use MunichInnovationGroup\PatentBundle\Entity\UmUsers;
 use MunichInnovationGroup\PatentBundle\Form\PatentType;
 use MunichInnovationGroup\PatentBundle\Entity\PmPatentgroups;
 use Symfony\Component\Security\Core\SecurityContext;
 use MunichInnovationGroup\PatentBundle\Util\SecurityHelper;
 use Exception;

/**
 * Portfolio controller.
 * @Route("/portfolio")
*/
class PortfolioController extends Controller {

/**
 * Index action.
 *
 * @Route("/", name="portfolio") …

是否可以使用twitter bootstrap创建嵌套左侧边栏？我检查了他们的教程但找不到它.我想制作带有一些列表的侧栏,每个列表包含嵌套列表,默认情况下这些列表是隐藏的,当用户将鼠标悬停在列表上时,隐藏选项应该变为可见.

<div class="span2">  
    <div class="well sidebar-nav">  
        <ul class="nav nav-list">  
            <li class="nav-header">Categories</li>  
            <li class="divider"></li>
            <li><a href="#">Books</a> 
                              <ul>
                                  <li><a href="#">Science Books</a></li>
                                  <li><a href="#">Computer Books</a></li>
                                  <li><a href="#">History Books</a></li>
                               </ul>
                            </li>  
            <li><a href="#">Electronics</a></li> 
                                <ul>
                                  <li><a href="#">TV</a></li>
                                  <li><a href="#">Freezer</a></li>
                                  <li><a href="#">Radio</a></li>
                                </ul> 
            <li><a href="#">Computer</a></li>
        </ul>  
    </div> 
</div>

可以用twitter bootstarp做到吗？如果没有其他解决方案？

谢谢

我有一个非常简单的问题,如何从get-> Request()获取POST值？

public function emptytrashAction(){
    $request = $this->getRequest();
    $portfolio_id = $_POST["test"];
}

我不想使用$_POST变量,我提交的表单只包含这个隐藏的字段测试.表格是,

 <form name="empt_trash" action="{{ path('MunichInnovationGroupPatentBundle_portfolio_emptytrash') }}" method="post" >
    <input type="hidden" name="test" value={{ selected_portfolio.id }}>
    <input class="button3 tooltip" name = "submit" type="submit" value="Empty"></a>
 </form>

如何在不使用的情况下获取隐藏字段的值$_POST？

编辑

如果一个方法同时使用GET和POST请求,对于Post请求,我会检查我的代码

            if ($request->getMethod() == 'POST')

但它不是symfony2方式,那么检查POST请求的正确方法是什么？

我需要得到以前兄弟姐妹的名字.保持简单我有一些示例代码

<html>

 <head>
   <script type="text/javascript">
     function myFunction()
     {
      var itm=document.getElementById("item2");  
      alert(itm.previousSibling.name);
     }
   </script>
 </head>

 <body>
  <p name='pn'>paragraph</p>
  <button id='item2' onclick="myFunction()">Try it</button>
 </body>
</html>

编辑:

<table id="sort">
    <tr name="nodrag nodrop">
        <td colspan=3><strong><a style="cursor:pointer;" class="toggle">Group 1</a></strong>  </td>
        <td style="text-align: right;"><a class="button3" href="#" ><span> Edit </span></a> <a class="button3" href="#" ><span> Delete </span></a></td>
    </tr>
    <tr id="1" class="tr_group"'>
        <td style="width:10px;" class="dragHandle">&nbsp;</td>
        <td><a href=# style="margin-left: 20px;">Umair Iqbal</a></td>
        <td><span style="font-size: 12px; color: #999; line-height: 100%;">A Student at TUM</span></td>
        <td style="text-align: right;"><a class="button3" href="#" ><span> Edit </span></a> <a …