Question

当它现在显示在我的表格中时，它会使用我的数据将对齐中心回显到表格中。我该如何摆脱这个

<?php
                mysqli_connect("localhost", "root", "") or die(mysql_error()); //connect to server
                mysqli_select_db("first_ db") or die("Can not connect to database");
                $query = mysqli_query("Select * from volunteer");
                while($row = mysqli_fetch_array($query))
                {
                    echo "<tr>";
                        echo '<td> align="center">'. $row['StaffID']."</td>";
                        echo '<td> align="center">'. $row['Name']."</td>";
                        echo '<td> align="center">'. $row['Address']."</td>";
                        echo '<td> align="center">'. $row['PhoneNumber']."</td>";
                        echo '<td> align="center">'. $row['Gender']."</td>";
                        echo '<td> align="center">'. $row['StaffStatus']."</td>";
                        echo '<td> align="center"><a hred="Edit.php">Edit</a> </td>'
                        echo '<td> align="center"><a href="Delete.php">Delete</a> </td>';
                        echo "</tr>";
                }
?>      


            </table>
        </body>     
</html>

Answer 1

首先，总是将你的会话放在任何其他代码之前

第二

  mysql_connect
  mysql_query 
  mysql_select_db
  mysql_fetch_array

已被弃用

 mysqli_connect     
 mysqli_query
 mysqli_select_db
 mysqli_fetch_array

第三，你的桌子没有

tbody or thead

用echo

替换打印

Answer 2

你在同一时间混合双引号和单引号的这个问题就是问题所在。

<?php

$servername = "localhost";
$username   = "username";
$password   = "";
$dbname     = "first_db";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}


$sql   = "Select * from volunteer";
$query = mysqli_query($conn, $sql);
if (mysqli_num_rows($query) > 0) {

    while ($row = mysqli_fetch_assoc($query)) {
        echo "<tr>";
        echo "<td> align=\"center\">" . $row['StaffID'] . "</td>";
        echo "<td> align=\"center\">" . $row['Name'] . "</td>";
        echo "<td> align=\"center\">" . $row['Address'] . "</td>";
        echo "<td> align=\"center\">" . $row['PhoneNumber'] . "</td>";
        echo "<td> align=\"center\">" . $row['Gender'] . "</td>";
        echo "<td> align=\"center\">" . $row['StaffStatus'] . "</td>";
        echo "<td> align=\"center\"><a hred=\"Edit.php?id=" . $row['StaffID'] . "\">Edit</a> </td>";
        echo "<td> align=\"center\"><a href=\"Delete.php?id=" . $row['StaffID'] . "\">Delete</a> </td>";
        echo "</tr>";
    }

} else {

    echo "no results";
}

?>      


            </table>
        </body>     
</html>

Answer 3

将Print替换为echo

                echo "<tr>";
                    echo '<td> align="center">'. $row['StaffID]."</td>";
                    echo '<td> align="center">'. $row['Name']."</td>";
                    echo '<td> align="center">'. $row['Address']."</td>";
                    echo '<td> align="center">'. $row['/PhoneNumber']."</td>";
                    echo '<td> align="center">'. $row['Gender']."</td>";
                    echo '<td> align="center">'. $row['StaffStatus']."</td>";
                    echo '<td> align="center"><a hred="Edit.php">Edit</a> </td>'
                    echo '<td> align="center"><a href="Delete.php">Delete</a> </td>';
                    echo "</tr>";

使用html和php将sql数据添加到表格中

3 个答案: