2011-08-11 101 views
0

我已经写了下面的代码,但不知何故头不匹配呈现表中生成的列,,有人可以给我一个提示如何改善它?从php代码格式化输出表

<?php 

$database =& JFactory::getDBO(); 

//Declare Variables 
$user = JFactory::getUser(); 
$id = $user->get('id'); 
$name = $user->get('name'); 

// Display quizzes 
echo "</br>"; 
echo "Quizzes History for : " ; 
echo "<b>"; 
echo $name; 
echo "</b>"; 


echo "</br>"; 
echo "</br>"; 

$database->setQuery('SELECT distinct qui.title AS name,' . 
     ' (SELECT GROUP_CONCAT(profiles.title) 
       FROM #__jquarks_users_profiles AS users_profiles 
       LEFT JOIN #__jquarks_profiles AS profiles ON users_profiles.profile_id = profiles.id 
       WHERE users_profiles.user_id = sessionWho.user_id) AS profile, ' . 
     ' (SELECT sum(score) 
       FROM #__jquarks_quizzes_answersessions 
       WHERE quizsession_id = quizSession.id 
       AND status <> -1) AS score,' . 
     ' (SELECT count(distinct(question_id)) 
       FROM #__jquarks_quizzes_answersessions 
       WHERE quizsession_id = quizSession.id) AS maxScore,' . 

      ' (SELECT count(id) 
       FROM #__jquarks_quizzes_answersessions 
       WHERE status=-1 
       AND quizsession_id = quizSession.id) AS evaluate,' . 
' quizSession.finished_on,sessionWho.email' .  
     ' FROM #__jquarks_quizsession AS quizSession' . 
     ' LEFT JOIN #__jquarks_users_quizzes AS users_quizzes ON users_quizzes.id = quizSession.affected_id' . 
     ' LEFT JOIN #__jquarks_quizzes AS qui ON users_quizzes.quiz_id = qui.id' . 
     ' LEFT JOIN #__jquarks_quizzes_answersessions AS quizSessAns ON quizSessAns.quizsession_id = quizSession.id' . 
     ' LEFT JOIN #__jquarks_sessionwho AS sessionWho ON sessionWho.session_id = quizSession.id' . 
     ' LEFT JOIN #__jquarks_users_profiles AS users_profiles ON users_profiles.user_id = sessionWho.user_id' . 
' LEFT JOIN #__jquarks_profiles AS profiles ON profiles.id = users_profiles.profile_id '. 

' WHERE sessionWho.user_id =' .$id) ; 

if (!$database->query()) { //write data and if error occurs alert 
    echo "<script> alert('".$database->getErrorMsg()."'); </script>"; 
} 

//var_dump($database); 
$tableStyle = "padding: 5px;border:1px"; 
$tdStyle = "padding:5px "; 

echo '<table style="' . $tableStyle . '" cellpadding="7" cellspacing="7">'; 
echo "<tr> <th> Quiz Title </th><th> Score </th><th>Maximum Score </th><th> Unanswered </th> <th>Finished On </th></tr>"; 

$row = $database->loadRowList(); 
foreach($row as $valuearray) 
{ 
echo '<tr style=" align="center">'; 
foreach($valuearray as $field) 
{ 

echo "<td>$field</td>"; 
} 
echo "</tr>"; 
} 
echo "</table>"; 
?> 
+0

研究使用[HEREDOCs](http://php.net/heredoc)来构建这些SQL查询。它将为您节省重复字符串连接的麻烦,并允许您使用缩进“很好”地写出它。 –

+1

不知道它是否对某些东西有影响,但是在这个字符串中“echo”';”你有style =“(双引号)而不是style =''(两个单引号)。即这是生成无效的HTML(我确信浏览器将修复它) – llamerr

+0

哇,什么是查询你有4个子查询和6个连接,你能告诉我这是什么性能?只是想知道 – WooDzu

回答

2

您有5 <th>标题,但从表中选择6列。显式打印您想要的表格中的字段(您应该),或更改查询以仅选择您想要的5。

+0

我需要的6个领域,你可以告诉我如何可以增加我的表到6,并可能分开标题来匹配列,,,? – Tony77

+0

在你想要的表的第一行写'第六字段'。标题将按您在HTML中编写的顺序显示。 –

+0

标题现在没有与输出内联,如何可以给生成的HTML给一个静态测量 – Tony77