php 表警告：为 foreach 提供的参数无效

Question

我想根据下面的查询输出一个 5 列表，phpmyadmin 上的输出是正确的，但我收到错误：

为 php 页面上的 foreach() 提供的参数无效。任何帮助将不胜感激。感谢

码：

<?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 jos_jquarks_users_profiles AS users_profiles    
  LEFT JOIN jos_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 jos_jquarks_quizzes_answersessions     

  WHERE quizsession_id = quizSession.id              AND status <> -1 ) AS score, 

( SELECT count(distinct question_id)              FROM jos_jquarks_quizzes_answersessions            
  WHERE quizsession_id = quizSession.id ) AS maxScore,            
DATE_FORMAT(quizSession.finished_on,'%M %d, %Y') AS FinishedOn
 FROM jos_jquarks_quizsession AS quizSession  

LEFT JOIN jos_jquarks_users_quizzes AS users_quizzes ON users_quizzes.id = quizSession.affected_id    
LEFT JOIN jos_jquarks_quizzes AS qui ON users_quizzes.quiz_id = qui.id     
LEFT JOIN jos_jquarks_quizzes_answersessions AS quizSessAns ON quizSessAns.quizsession_id = quizSession.id 
LEFT JOIN jos_jquarks_sessionwho AS sessionWho ON sessionWho.session_id = quizSession.id        
LEFT JOIN jos_jquarks_users_profiles AS users_profiles ON users_profiles.user_id = sessionWho.user_id 
LEFT JOIN jos_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 "; 
$thStyle = "padding:7px ";

echo '<table style="' . $tableStyle . '" cellpadding="7" cellspacing="7">'; 
echo "<tr> <th style=align:center>Quiz Title </th><th style=align:center> 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>";

?>

Answer 1

这通常意味着您的查询中存在错误，并且它没有返回任何结果，因此没有对象可以通过 foreach 运行。调试此问题的最简单方法是从 joomla 管理员面板（位于全局设置 -> 系统中）打开调试模式，然后转到引发此错误的页面，它应该显示 SQL 错误。

无论如何，查看查询我能找到的唯一错误（假设所有字段/表都正确）是最后你有：

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

这应该是：

WHERE sessionWho.user_id = $id " ) ;

或

WHERE sessionWho.user_id = ". intval($id)  ) ;