php - SQL 中的 LIKE 不起作用

Question

我正在使用 PDO 来运行此查询，有没有办法查看查询是否返回 false？为什么当我期望结果时我的查询什么也没有返回？

                $selectsql = "SELECT `firstname`,`surname`,`phone`,`address`,`username`,`StudentID` FROM `student_info` WHERE ? LIKE ?";
                $selectstmt = $conn->prepare($selectsql);
                $selectstmt->execute(array($searchby,"%{$searchval}%"));
                while($data = $selectstmt->fetch(PDO::FETCH_ASSOC)){
                  echo "
                  <tr>
                    <td>{$data['firstname']}</td>
                    <td>{$data['surname']}</td>
                    <td>{$data['phone']}</td>
                    <td>{$data['address']}</td>
                    <td>{$data['username']}</td>
                    <td>Delete Account</td>                    
                  </tr>
                  ";
                  var_dump($data);
                }   

我把 $searchby 和 $searchval 转储了，它们对我来说似乎很好，没有打字错误或任何问题。

Answer 1

您的查询存在问题:

$selectsql = "SELECT firstname`,`surname`,`phone`,`address`,`username`,`StudentID` FROM `student_info` WHERE ? LIKE ?";

where 之后的占位符弄乱了您的查询，在 where 子句之后应该有一个列名称，顺便说一句，我看到 $searchby 根据项目进行更改由用户选择。您应该注意的一件事是:

Table and Column names cannot be replaced by parameters in PDO.

您的查询应该如下所示。

<?php

$search = "%".$searchval."%";
$selectsql = "SELECT `firstname`,`surname`,`phone`,`address`,`username`,`StudentID` FROM `student_info` WHERE $searchby LIKE ?";
    $selectstmt = $conn->prepare($selectsql);
    $selectstmt->execute([$search]);
    $results = $selectstmt->fetch((PDO::FETCH_ASSOC));

    if(count($results) > 0){
      var_dump($results);
        foreach($results as $data):?>
          <tr>
              <td><?php echo $data['firstname'];?></td>
              <td><?php echo $data['surname'];?></td>
              <td><?php echo $data['phone'];?></td>
              <td><?php echo $data['address'];?></td>
              <td><?php echo $data['username'];?></td>
              <td>Delete Account</td>                    
        </tr>

      <?php
          endforeach;
    }else{

      echo "no results found";
    }
?>