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php table warning : Invalid argument supplied for foreach

开发者 https://www.devze.com 2023-03-28 03:07 出处:网络
I want to output a 5 column table based on the query below , the output on phpmyadmin is correct butI am gettingerror :

I want to output a 5 column table based on the query below , the output on phpmyadmin is correct but I am getting error :

Invalid argument supplied for foreach() on the php page. Any help would be highly appreciated. Thanks

code :

<?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 = quizSes开发者_Python百科sion.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>";

?>


This typically means that there is an error in your query, and it's not returning any results so there is no object to run through the foreach. The easiest way to debug this is to turn on debug mode from the joomla administrator panel (it's in global settings -> system), and then going to the page this error is being thrown, and it should show the SQL error.

Anyway, looking at the query the only error I can find (assuming all fields/tables are correct) is that at the end you have:

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

This should be:

WHERE sessionWho.user_id = $id " ) ;

or

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