使用 Mysqli 时搜索栏不起作用

我创建的搜索栏有问题，我没有错误，我知道我的查询是正确的，因为我的搜索栏曾经在旧的 mysql 代码中正常工作。但是由于我尝试将我的代码更改为 mysqli，它并没有完全起作用。无论用户在搜索栏中输入什么，它始终会指出"请在搜索栏中输入短语"，无论搜索栏是否为空。它也不会显示搜索结果。所以我的问题是，如果找到关键字，为什么它不显示搜索结果？

下面是当前代码：

        <?php
  $questioncontent = (isset($_POST['questioncontent'])) ? $_POST['questioncontent'] : '';
        $searchquestion = $questioncontent;
        $terms = explode(" ", $searchquestion);

        //Query for search bar      
        $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 one term entered in search bar then perform a LIKE statement to look up the term entered       
            if ($i == 1) {
                $questionquery .= "q.QuestionContent LIKE ? ";
            } else {
        //If multiple terms then add an OR statement to check for multiple keywords        
                $questionquery .= "OR q.QuestionContent LIKE ? ";
            }
        }
        //Order by terms entered in ascending order they have been entered               
        $questionquery .= "GROUP BY q.QuestionId, q.SessionId ORDER BY ";
        $i = 0;
        foreach ($terms as $each) {
            $i++;
        //if there are multiple terms, then for example there are 2 terms then display content which contains both terms first, then display content which contains only one of those terms                  
            if ($i != 1)
                $questionquery .= "+";
            $questionquery .= "IF(q.QuestionContent LIKE ? ,1,0)";
        }
        $questionquery .= " DESC ";
        //prepare query, bind the terms and execute query    
        $stmt = $mysqli->prepare($questionquery);
        $stmt->bind_param('ss', $each = '%' . $each . '%', $each = '%' . $each . '%');
        $stmt->execute();
        $stmt->bind_result($dbQuestionId, $dbQuestionContent, $dbOptionType, $dbNoofAnswers, $dbAnswer, $dbReplyType, $dbQuestionMarks);
        $questionnum = $stmt->num_rows();
        //If search bar is empty and user submits the search bar, then below is the phrase it should display:

        if (empty($questioncontent)) {
            echo "Please enter in a phrase in the text box in able to search for a question";
        }
        //Below is the code if no results are found from the search:
        else if ($questionnum == 0) {
            echo "<p>Your Search: '$searchquestion'</p>";
            echo "<p>Number of Questions Shown from the Search: <strong>$questionnum</strong></p>";
            echo "<p>Sorry, No Questions were found from this Search</p>";
        }
        //Finally below is the code that displays the results of the search if search is successful:
        else {
            echo "<p>Your Search: '$searchquestion'</p>";
            echo"<p>Number of Questions Shown from the Search: <strong>$questionnum</strong></p>";
            $output = "";
            $output .= "
                <table border='1' id='resulttbl'>
                  <tr>
                  <th class='questionth'>Question</th>
                  </tr>
            ";
            while ($stmt->fetch()) {
                $output .= "
                  <tr>
                  <td class='questiontd'>{$dbQuestionContent['QuestionContent']}</td>
                  <td class='addtd'><button type='button' class='add' onclick='"parent.addwindow('{$dbQuestionContent['QuestionContent']}');'">Add</button></td>
                  </tr>";
            }
            $output .= "        </table>";
            echo $output;
        }
    ?>

这是它自己测试的应用程序的链接。如果您在搜索栏中输入"AAA"，它应该显示结果，但它一直指出需要输入短语。如果您输入随机内容以使搜索找不到结果，它仍然会指出请输入一个短语。应用