Question

我试图在MySQL数据库的网页上显示一个表，但它现在正在工作！这是我的代码：

<?php 
    function list_schools() {
       $conn = mysqli_connect("localhost", "root", "", "Database_Project");
       if ($conn === false) {
          die("Could not connect:"  . mysqli_connect_error()); 
       } 

       $output = "";
       $result = mysqli_query($conn, "SELECT school_id, name, address, phone_number, email, type FROM Schools;") or die('cannot show tables');       
       while ($row = mysqli_fetch_array($result)){
          $output .= '
          <tr>
          <td>' . $row['school_id'] . '</td>
          <td>' . $row['name'] . '</td>                                                   
          <td>' . $row['address'] . '</td>
          <td>' . $row['phone_number'] . '</td>
          <td>' . $row['email'] . '</td> 
          <td>' . $row['type'] . '</td>                                    
          </tr>';
       }
       return $output;

    }

    $exec_func = list_schools();
?>

<table cellpadding="0" cellspacing="0" width="100%" class="sortable">
    <thead>
       <tr>
          <th>ID</th>
          <th>Name</th>
          <th>Address</th>
          <th>Phone Number</th>
          <th>Email</th>
          <th>Type</th>
          <td>&nbsp;</td>
       </tr>
    </thead>
    <tbody>
       <?php echo exec_func; ?>
    </tbody>
</table>

每当我尝试在浏览器上显示它时，我都会得到以下信息：

1

无法显示1

中的列

我做错了什么？

Answer 1

从您的查询中，很明显您不会获得ID列，而是school_id。

因此，在构建$output时只需更正您的陈述：

     <td>' . $row['school_id'] . '</td>

Answer 2

为什么不尝试：

<?php 
    function list_schools() {
       $conn = mysqli_connect("localhost", "root", "", "Database_Project");
       if ($conn === false) {
          die("Could not connect:"  . mysqli_connect_error()); 
       } 

       $output = "";
       $result = mysqli_query($conn, "SELECT school_id, name, address, phone_number, email, type FROM Schools;") or die('cannot show tables');       



    }

    $exec_func = list_schools();
?>

<table cellpadding="0" cellspacing="0" width="100%" class="sortable">
    <thead>
       <tr>
          <th>ID</th>
          <th>Name</th>
          <th>Address</th>
          <th>Phone Number</th>
          <th>Email</th>
          <th>Type</th>
          <td>&nbsp;</td>
       </tr>
    </thead>
    <tbody>
       <?php while ($row = mysqli_fetch_array($result)){
          echo '
          <tr>
          <td>' . $row['school_id'] . '</td>
          <td>' . $row['name'] . '</td>                                                   
          <td>' . $row['address'] . '</td>
          <td>' . $row['phone_number'] . '</td>
          <td>' . $row['email'] . '</td> 
          <td>' . $row['type'] . '</td>                                    
          </tr>'; } ?>
    </tbody>
</table>

Answer 3

您可以使用以下代码

   <table cellpadding="0" cellspacing="0" width="100%" class="sortable">
    <thead>
       <tr>
          <th>ID</th>
          <th>Name</th>
          <th>Address</th>
          <th>Phone Number</th>
          <th>Email</th>
          <th>Type</th>
          <td>&nbsp;</td>
       </tr>
    </thead>
    <tbody>
   <?php 
       $conn = mysqli_connect("localhost", "root", "", "Database_Project");
       if ($conn === false) {
          die("Could not connect:"  . mysqli_connect_error()); 
       } 

        $results = $mysqli->query("SELECT school_id, name, address, phone_number, email, type FROM Schools");
        while($obj = $results->fetch_object()) { 
         ?>
          <tr>
          <td> <?php echo $row['school_id']; ?> </td>
          <td> <?php echo $row['name']; ?></td>                                                   
          <td> <?php echo $row['address']; ?></td>
          <td> <?php echo $row['phone_number']; ?></td>
          <td> <?php echo $row['email']; ?></td> 
          <td><?php echo $row['type']; ?></td>                                    
          </tr>
      <?php
       }
?>


    </tbody>
</table>

Answer 4

你的问题在于：

$result = mysqli_query($conn, "SELECT school_id, name, address, phone_number, email, type FROM Schools;") or die('cannot show tables');

您无法在此处使用or运算符。请查看此Logical Operators以获取解释。

示例：

echo 't' or 'f'; //will output 1

使用此代码：

$result = mysqli_query($conn, "SELECT school_id, name, address, phone_number, email, type FROM Schools;")
$num_rows = mysqli_num_rows($result);
if (!$num_rows) {
die('cannot show tables');   
}

使用MySQL数据库中的数据无法使用HTML创建表

4 个答案: