Question

我目前正在努力对网站进行一些修改。网站目前的工作方式是页面为config.inc.php页面调用一次require。配置页面还具有SQL调用，用于显示从数据库中显示的内容。虽然这很好用，但我需要创建一个显示同一个表的新页面，但SQL已更改为显示不同的数据。

我遇到的麻烦是config.inc.php页面上有SQL数据，可以提供所需内容。我想在不同的页面上显示不同的结果，但网站要求config.inc.php显示所有内容。我更改了SQL查询以测试我想要显示的内容是否有效并且它有效，但是我如何在不必使用原始SQL命令的情况下显示它。

这是原始代码;

    $config = require_once 'link to admin section';
$db_username = $config['components']['db']['username'];
$db_password = $config['components']['db']['password'];

$db_link = mysql_connect('nosey', $db_username, $db_password) or die(mysql_error());
mysql_select_db('nosey', $db_link);

if ( $_SERVER['REQUEST_METHOD'] == 'GET' ) {
  if ( isset($_GET['id']) && $_GET['id'] ) {
    $selectSqlJSONString = 'SELECT * FROM vacancies WHERE id = '.$_GET['id'].' ORDER BY created ASC LIMIT 1';
    $jsonResult = mysql_query($selectSqlJSONString, $db_link);
    $jsonRow = mysql_fetch_object($jsonResult);

    if ( !isset($_SERVER['HTTP_X_FANCYBOX']) )
        include 'details_ajax_header.php';

    include 'filename.php';

    if ( !isset($_SERVER['HTTP_X_FANCYBOX']) )
        include 'details_ajax_footer.php';

    exit;
  }
}
$region = '1';
if ( isset($_GET['region']) && $_GET['region'] ) {
  $region .= ' AND ' . sprintf('county LIKE "%%%s%%"', mysql_real_escape_string(str_replace('-', ' ',$_GET['region']), $db_link ));
}

$selectSqlString = 'SELECT * FROM vacancies WHERE '.$region.' ORDER BY created DESC'; 
$result = mysql_query($selectSqlString, $db_link) or die(mysql_error());

$vacancies = array();
while ($row = mysql_fetch_object($result)) {
  $vacancies[] = $row;
}

这就是我要在另一个页面上显示的内容，以显示不同的结果集。

    $config = require_once 'link to admin section';
    $db_username = $config['components']['db']['username'];
    $db_password = $config['components']['db']['password'];

    $db_link = mysql_connect('nosey', $db_username, $db_password) or die(mysql_error());
    mysql_select_db('nosey', $db_link);

    if ( $_SERVER['REQUEST_METHOD'] == 'GET' ) {
      if ( isset($_GET['id']) && $_GET['id'] ) {
        $selectSqlJSONString = 'SELECT * FROM vacancies WHERE id = '.$_GET['id'].' ORDER BY created ASC LIMIT 1';
        $jsonResult = mysql_query($selectSqlJSONString, $db_link);
        $jsonRow = mysql_fetch_object($jsonResult);

        if ( !isset($_SERVER['HTTP_X_FANCYBOX']) )
            include 'details_ajax_header.php';

        include 'filename.php';

        if ( !isset($_SERVER['HTTP_X_FANCYBOX']) )
            include 'details_ajax_footer.php';

        exit;
      }
    }
    $region = '1';
    if ( isset($_GET['region']) && $_GET['region'] ) {
      $region .= ' AND ' . sprintf('county LIKE "%%%s%%"', mysql_real_escape_string(str_replace('-', ' ',$_GET['region']), $db_link ));
    }

   $selectSqlString = 'SELECT * FROM vacancies WHERE featured = "1" ORDER BY created DESC';
    $result = mysql_query($selectSqlString, $db_link) or die(mysql_error());

    $vacancies = array();
    while ($row = mysql_fetch_object($result)) {
      $vacancies[] = $row;
    }

请注意我想要改变的是SELECT * FROM vacancies WHERE featured = "1"。

这就是我所需要的！但我想包括这个，而不必改变config.inc.php页面！（此信息来自哪里）我该怎么做？

如何包含文件，但不执行所有内容？

0 个答案: