如何删除警告下面的mysqli

Question

我的问题是，它给我的mysqli说明这是一个警告：

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 

我怎样才能摆脱警告？

$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(); 

输出的SQL：

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 

Answer 1

你似乎混淆的MySQLi的参数绑定与PDO的（它接受一个数组​​呼吁$eachbind_param()没有参数传递到​​

方法。

$stmt->bind_param('s', $each); 
$stmt->execute(); 

请注意，如果您打算使用LIKE查询%个通配符，这些通配符必须是绑定变量的一部分，使得：

$each = '%' . $each . '%';