我在下面有一个“答案”数据库表：
答案表
AnswerId SessionId QuestionId Answer
13 AAA 1 A
14 AAC 1 True
现在您可以看到考试（会话）AAA中的问题1有1个答案，这些是考试（会话）AAC中问题1的1个答案。
以下是“问题”表：
问题表：
SessionId QuestionId QuestionContent NoofAnswers ReplyId QuestionMarks OptionId
AAA 1 What is 2+2? 1 1 5 2
AAC 1 Is 3+3 = 6? 1 1 5 25
现在我有一个搜索功能，用户输入问题中的术语并编译搜索。因此，例如，如果用户输入“2 + 2”，则下面是结果应显示在php / html表中的内容：
QuestionContent Option Type Number of Answers Answer Number of Replies Number of Marks
What is 2+2? A-D 1 A Single 5
但问题是它显示了一个额外的行，它显示如下：
QuestionContent Option Type Number of Answers Answer Number of Replies Number of Marks
What is 2+2? A-D 1 A True Single 5
What is 2+2? A-D 1 A True Single 5
现在我的问题是为什么它显示两行，为什么它在“答案”栏下包含答案“真”，那个问题的答案应该只是“A”？我相信这是因为这两个问题都有相同的QuestionId（问题编号），但它们都属于不同的SessionId，所以这应该不是问题。
下面是代码（我已经减少了以便您更容易阅读并希望看到问题），它执行查询并输出结果：
<?php
//connect to db
// Build the query
$questionquery = "
SELECT DISTINCT q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_CONCAT(DISTINCT Answer SEPARATOR '') AS Answer, r.ReplyType,
q.QuestionMarks
FROM Answer an
INNER JOIN Question q ON q.QuestionId = an.QuestionId
JOIN Reply r ON q.ReplyId = r.ReplyId
JOIN Option_Table o ON q.OptionId = o.OptionId
WHERE ".implode(" AND ", array_fill(0, $numTerms, "q.QuestionContent LIKE ?"))."
GROUP BY an.SessionId, an.QuestionId
ORDER BY ".implode(", ", array_fill(0, $numTerms, "IF(q.QuestionContent LIKE ?, 1, 0) DESC"))."
";
// Make the referenced array
$referencedArray = make_values_referenced(array_merge(
array(str_repeat("ss", $numTerms)), // types
$termArray, // where
$termArray // order by
));
// Bind parameters
if (!call_user_func_array(array($stmt, 'bind_param'), make_values_referenced($referencedArray))) {
die("Error binding parameters: $stmt->error");
}
// This will hold the search results
$searchResults = array();
$searchOption = array();
$searchNoofAnswers = array();
$searchAnswer = array();
$searchReply = array();
$searchMarks = array();
// Fetch the results into an array
if (!$stmt->num_rows()) {
$stmt->bind_result($dbQuestionContent,$dbOptionType,$dbNoofAnswers,$dbAnswer,$dbReplyType,$dbQuestionMarks);
while ($stmt->fetch()) {
$searchResults[] = $dbQuestionContent;
$searchOption[] = $dbOptionType;
$searchNoofAnswers[] = $dbNoofAnswers;
$searchAnswer[] = $dbAnswer;
$searchReply[] = $dbReplyType;
$searchMarks[] = $dbQuestionMarks;
}
}
}
$questionnum = sizeof($searchResults);
// If $searchResults is not empty we got results
if (!empty($searchResults)) {
echo "<p>Your Search: '$inputValue'</p>";
echo"<p>Number of Questions Shown from the Search: <strong>$questionnum</strong></p>";
echo "<table border='1' id='resulttbl'>
<tr>
<th class='questionth'>Question</th>
<th class='optiontypeth'>Option Type</th>
<th class='noofanswersth'>Number of <br/> Answers</th>
<th class='answerth'>Answer</th>
<th class='noofrepliesth'>Number of <br/> Replies</th>
<th class='noofmarksth'>Number of <br/> Marks</th>
</tr>\n";
$script = '';
foreach ($searchResults as $key=>$question) {
$script .= 'var key_' . $key . '="' . str_replace('"','\"', $question) . '";' . PHP_EOL;
echo '<tr class="questiontd">'.PHP_EOL;
echo '<td>'.htmlspecialchars($question).'</td>' . PHP_EOL;
echo '<td class="optiontypetd">'.htmlspecialchars($searchOption[$key]).'</td>' . PHP_EOL;
echo '<td class="noofanswerstd">'.htmlspecialchars($searchNoofAnswers[$key]).'</td>' . PHP_EOL;
echo '<td class="answertd">'.htmlspecialchars(implode(' ', $searchAnswer)).'</td>' . PHP_EOL;
echo '<td class="noofrepliestd">'.htmlspecialchars($searchReply[$key]).'</td>' . PHP_EOL;
echo '<td class="noofmarkstd">'.htmlspecialchars($searchMarks[$key]).'</td>' . PHP_EOL;
echo "<td class='addtd'><button type='button' class='add' onclick=\"parent.addwindow(key_$key,'$searchMarks[$key]','$searchNoofAnswers[$key]','$searchOption[$key]','$searchReply[$key]','$searchAnswer[$key]');\">Add</button></td></tr>";
}
echo "</table>" . PHP_EOL;
echo '<script type="text/javascript">' . PHP_EOL;
echo $script;
echo '</script>' . PHP_EOL;
}
?>
答案 0 :(得分：1)
这里的问题似乎与您定义的db模式不是您使用它的方式，我的意思是答案表上的唯一值似乎应该是answerID或SessionId + QuestionId，所以如果您在查询中使用该表还应该在你的连接条件中包含SessionId和Question表，我的意思是至少在不同的会话中看起来两个不同的问题可以有相同的questionID是关键问题。
所以你的代码应该如下：
SELECT DISTINCT q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_CONCAT(DISTINCT Answer SEPARATOR '') AS Answer, r.ReplyType,
q.QuestionMarks, q.SessionId
FROM Answer an
INNER JOIN Question q ON q.QuestionId = an.QuestionId and an.SessionId = q.SessionId
JOIN Reply r ON q.ReplyId = r.ReplyId
JOIN Option_Table o ON q.OptionId = o.OptionId
WHERE ".implode(" AND ", array_fill(0, $numTerms, "q.QuestionContent LIKE ?"))."
GROUP BY an.SessionId, an.QuestionId
ORDER BY ".implode(", ", array_fill(0, $numTerms, "IF(q.QuestionContent LIKE ?, 1, 0) DESC"))."
";
UPDATED 上方的SQL查询以反映更新的问题
新问题仍然与查询未指定会话相同，因此跨会话的不同问题可能会混合在一起，我建议您在选择列表中查询sessionID以获取正确的数据，并将其隐藏在页面，如果你不想显示它，但请记住，你需要记录，如果你有一些其他详细信息页面来显示问题，你仍然需要将sessionID和QuestionId完全传递到该页面以识别唯一数据
答案 1 :(得分：0)
在选择查询中加入an.QuestionId。
$questionquery = "
SELECT DISTINCT an.QuestionId,q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_CONCAT(DISTINCT Answer SEPARATOR '') AS Answer, r.ReplyType,
q.QuestionMarks
FROM Answer an
INNER JOIN Question q ON q.QuestionId = an.QuestionId
JOIN Reply r ON q.ReplyId = r.ReplyId
JOIN Option_Table o ON q.OptionId = o.OptionId
WHERE ".implode(" AND ", array_fill(0, $numTerms, "q.QuestionContent LIKE ?"))."
GROUP BY an.SessionId, an.QuestionId
ORDER BY ".implode(", ", array_fill(0, $numTerms, "IF(q.QuestionContent LIKE ?, 1, 0) DESC"))."
";
您已将{by}用于an.QuestionId，但未将其包括在选择列中，因为此数据即将到来。