小编Osc*_*rez的帖子

我正在尝试使用SELECT子句中的子查询执行简单的选择查询,并且根本没有找到方法来执行此操作.我已尝试使用DQL和QueryBuilder,但都不起作用.代码如下,请不要说我只能使用连接,这是一个简单的例子,只是为了说明问题,我有合理的子查询用例.

// With QueryBuilder
$query = $qb->select(array('a', 
                           '(SELECT at.addresstypeName 
                               FROM e:Addresstype at 
                              WHERE at.addresstypeId = a.addresstypeId
                            ) AS addresstypeName'))
            ->from('e:Address', 'a')
            ->where('a.addressId = :addressId')
            ->setParameter('addressId', 1);

// With DQL
$dql = "SELECT a, 
               (SELECT at.addresstypeName 
                  FROM e:Addresstype at 
                 WHERE at.addresstypeId = a.addresstypeId
               ) AS addresstypeName 
          FROM e:Address a 
         WHERE a.addressId = :addressId";
$query = $em->createQuery($dql)->setParameter(':addressId', 1);

地址表中定义了以下关系:

/**
 * @ORM\ManyToOne(targetEntity="Addresstype")
 * @ORM\JoinColumns({
 *   @ORM\JoinColumn(name="addresstype_id", referencedColumnName="addresstype_id")
 * })
 */
protected $addresstype;

在本机SQL中,查询将如下所示:

SELECT
    a.*,
    (
        SELECT at.addresstype_name
        FROM addresstype at
        WHERE …

我正在使用WordPress网站,我创建了一个页面模板,按类别slug显示帖子.为此,我为页面创建一个字段WP_Catid,并将其设置为我想要显示帖子的类别slug.但是,我只希望每页显示五个帖子,并在这些帖子的底部显示分页链接.如何正确显示分页链接？

我的代码如下:

<div id="container">
  <div id="content" role="main">
    <?php
      $btpgid=get_queried_object_id();
      $btmetanm=get_post_meta( $btpgid, 'WP_Catid','true' );
      $paged = (get_query_var('paged')) ? get_query_var('paged') : 1;

      $args = array( 'posts_per_page' => 5,
                     'category_name' => $btmetanm,
                     'paged' => $paged,
                     'post_type' => 'post' );

      $myposts = get_posts( $args );
      foreach ( $myposts as $post ) : setup_postdata( $post ); 
        echo "<div style='border:2px groove black; margin-bottom:5px;'><h3 class='btposth'>";
        the_title(); 
        echo "</h3><div class='btpostdiv'>";
        the_content();
        echo "</div></div>";
      endforeach; 
      next_posts_link( 'Older Entries'); //not displaying
      previous_posts_link('Newer Entries &raquo;'); //not displaying
      wp_reset_postdata();
    ?>
  </div><!-- #content …

我正在研究Jasper Reports,我的查询使用SQL'IN'谓词.

SELECT customer_name AS NAME, 
       id_customer   AS ID 
  FROM customer 
 WHERE customer_role IN ($P{roles})

这里role参数可以有1个或多个整数值,并在生成jasper报告时动态决定.

任何人都可以帮助我如何通过Java程序动态设置'roles'参数的值.

我有一张表格如下:

    6xx     8xx      9xx     11xx      12xx
  1  0.01    0.002    0.004   0.001     0.025
  2  0.025   0.125    0.002   0.01      0.011

我想从列中找到最小值,使该列为绿色.

例如,在第一个中,最小值是0.001,所以我希望它是绿色,第二个0.002是最小值,我希望它是绿色.

任何人都可以指导我如何做到这一点,谢谢

下面是我如何从数据库中选择它并在表中显示int的代码

<?php

$dbHost = 'localhost'; // usually localhost
$dbUsername = 'xxxx';
$dbPassword = 'xxxx';
$dbDatabase = 'xxxx';
$db = mysql_connect($dbHost, $dbUsername, $dbPassword) or die ("Unable to connect to Database Server.");
mysql_select_db ($dbDatabase, $db) or die ("Could not select database.");



$ColumnNames = mysql_query("SELECT column_name 
                              FROM information_schema.COLUMNS 
                             WHERE table_name = 'supplierprice' 
                               AND column_name NOT IN ('supp_price_id',
                                                       'region',
                                                       'country',
                                                       'net_id',
                                                       'networkname', 
                                                       'mcc', …

当我使用PHPTable创建一个简单的表时,行似乎有点太高了.我希望它们与字体具有相同的高度,并且在单元格中文本的下方或上方没有填充/间距..但是没有任何"填充"就无法使它工作...

码:

$styleTable = array('borderSize'=>0, 
                    'borderColor'=>'eeeeee',
                    'cellMargin'=>0, 
                    'spaceBefore' => 0, 
                    'spaceAfter' => 0,
                    'spacing' => 0);

$PHPWord->addTableStyle('myOwnTableStyle', $styleTable);

$table = $section->addTable('myOwnTableStyle');

foreach($aryData['json_data'] as $data) {
  $table->addRow();
  $table->addCell(4000)->addText($data['label'] . ':', array('bold'=>true));
  $table->addCell(8000)->addText($data['value']);
}

我有这个功能测试(我只是显示一个片段):

namespace Just\An\ExampleBundle\Tests\Controller;
use Symfony\Bundle\FrameworkBundle\Test\WebTestCase;

class ExampleControllerTest extends WebTestCase
{
  public function testCrearTipus()
  {
    $client = static::createClient(array(), array(
            'PHP_AUTH_USER' => 'login',
            'PHP_AUTH_PW' => 'pa$$sword'
    ));
    $client->followRedirects();
    $url = '/admin/type/new';
    $crawler = $client->request('GET', $url);
    $token = $crawler->filter('input[name="name_of_form[_token]"]')->first()->getValue();
    .....
  }
}

我只是无法获取我的视图中定义的输入值:

任何帮助,将不胜感激