我遇到的问题是它在mysqli中给了我一个警告，说明这一点：
Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in ...on line 89
如何摆脱警告？
$questionquery = "
SELECT q.QuestionId, q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_CONCAT(an.Answer ORDER BY an.Answer SEPARATOR ' ') AS Answer, r.ReplyType,
q.QuestionMarks
FROM Answer an
INNER JOIN Question q ON q.AnswerId = an.AnswerId
JOIN Reply r ON q.ReplyId = r.ReplyId
JOIN Option_Table o ON q.OptionId = o.OptionId
WHERE ";
$i=0;
foreach ($terms as $each) {
$i++;
if ($i == 1){
$questionquery .= "q.QuestionContent LIKE ? ";
} else {
$questionquery .= "OR q.QuestionContent LIKE ? ";
}
}
$questionquery .= "GROUP BY q.QuestionId, q.SessionId ORDER BY "; $i = 0; foreach ($terms as $each) {
$i++;
if ($i != 1)
$questionquery .= "+";
$questionquery .= "IF(q.QuestionContent LIKE ? ,1,0)";
}
$questionquery .= " DESC ";
$stmt=$mysqli->prepare($questionquery);
$stmt->bind_param('s', $each = '%' . $each . '%');
$stmt->execute();
$stmt->bind_result($dbQuestionId,$dbQuestionContent,$dbOptionType,$dbNoofAnswers,$dbAnswer,$dbReplyType,$dbQuestionMarks);
$questionnum = $stmt->num_rows();
SQL 的输出：
SELECT q.QuestionId, q.QuestionContent, o.OptionType, q.NoofAnswers, GROUP_CONCAT(an.Answer ORDER BY an.Answer SEPARATOR ' ') AS Answer, r.ReplyType, q.QuestionMarks
FROM Answer an INNER JOIN Question q ON q.AnswerId = an.AnswerId JOIN Reply r ON q.ReplyId = r.ReplyId JOIN Option_Table o ON q.OptionId = o.OptionId
WHERE q.QuestionContent LIKE ?
GROUP BY q.QuestionId, q.SessionId
ORDER BY IF(q.QuestionContent LIKE ? ,1,0) DESC
您似乎将MySQLi的参数绑定方法与PDO的参数绑定方法（PDO接受数组以execute()）混淆了。 在$each上调用bind_param()，并且不向execute()传递任何参数。
$stmt->bind_param('s', $each);
$stmt->execute();
请注意，如果您打算将LIKE查询与通配符一起使用%则这些通配符必须是绑定变量的一部分，以便：
$each = '%' . $each . '%';