这适用于 phpmyadmin ,但当我在代码上使用时,会出现错误,错误显示,解析错误:语法错误,意外的“%”....
我的语法有什么问题>?
<?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>";
$database->setQuery(" SELECT distinct qui.title AS name,
( SELECT GROUP_CONCAT(profiles.title)
FROM jos_jquarks_users_profiles AS users_profiles
LEFT JOIN jos_jquarks_profiles AS profiles ON users_profiles.profile_id = profiles.id
WHERE users_profiles.user_id = sessionWho.user_id ) AS profile,
( SELECT sum(score)
FROM jos_jquarks_quizzes_answersessions
WHERE quizsession_id = quizSession.id AND status <> -1 ) AS score,
( SELECT count(distinct question_id) FROM jos_jquarks_quizzes_answersessions
WHERE quizsession_id = quizSession.id ) AS maxScore,
DATE_FORMAT(quizSession.finished_on,'%M %d, %Y')
FROM jos_jquarks_quizsession AS quizSession
LEFT JOIN jos_jquarks_users_quizzes AS users_quizzes ON users_quizzes.id = quizSession.affected_id
LEFT JOIN jos_jquarks_quizzes AS qui ON users_quizzes.quiz_id = qui.id
LEFT JOIN jos_jquarks_quizzes_answersessions AS quizSessAns ON quizSessAns.quizsession_id = quizSession.id
LEFT JOIN jos_jquarks_sessionwho AS sessionWho ON sessionWho.session_id = quizSession.id
LEFT JOIN jos_jquarks_users_profiles AS users_profiles ON users_profiles.user_id = sessionWho.user_id
LEFT JOIN jos_jquarks_profiles AS profiles ON profiles.id = users_profiles.profile_id
WHERE sessionWho.user_id = ' .$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>";
?>
最佳答案
你必须避开 ' 标记。 所以而不是
DATE_FORMAT(quizSession.finished_on,'%M %d, %Y')
尝试
DATE_FORMAT(quizSession.finished_on,\'%M %d, %Y\')
或者您可以只使用双引号 "来开始和结束字符串。
关于php - 如何在我的查询中使用日期格式?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7072002/