Question

我想用PHP填充MySQL数据库中的数据表，但是代码执行时表格单元格保持为空，我没有收到任何错误

以下是代码：

<?php
$host = "localhost"; // Host name 
$username = ""; // Mysql username 
$password = ""; // Mysql password 
$db_name = "test"; // Database name 
$tbl_name = "test_mysql"; // Table name 
$server_name = "localhost";

// Create connection
$con = new mysqli($server_name, $username, $password, $db_name, 3306);
if($con->connect_error){
    die("Connection failed: ".$con->connect_error);
}

// Check connection
if($con->connect_error){
    die("Connection failed: ".$conn->connect_error);
}

$sql = "SELECT * FROM $tbl_name";
$result = $con->query($sql);
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
    <tr>
        <td>
            <table width="400" border="1" cellspacing="0" cellpadding="3">
                <tr>
                    <td colspan="4"><strong>List data from mysql</strong></td>
                </tr>
                <tr>
                    <td align="center"><strong>Name</strong></td>
                    <td align="center"><strong>Lastname</strong></td>
                    <td align="center"><strong>Email</strong></td>
                    <td align="center"><strong>Update</strong></td>
                </tr>
                <?php
                if($result->num_rows > 0){
                    // output data of each row
                    while($row = $result->fetch_assoc()){ ?>
                        <tr>
                            <td><? echo $rows['name']; ?></td>
                            <td><? echo $rows['lastname']; ?></td>
                            <td><? echo $rows['email']; ?></td>
                            <td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td>
                        </tr>
                        <?php
                    }
                }
                ?>
            </table>
        </td>
    </tr>
</table>
<?php
$con->close();
?>

我认为它可能缺少代码，我感谢你能给我的任何帮助！

Answer 1

您在while循环声明中使用了$row而不是$rows。

while($rows = $result->fetch_assoc()){
    echo"<tr>
           <td>{$rows['name']}</td>
           <td>{$rows['lastname']}</td>
           <td>{$rows['email']}</td>
           <td align='center'><a href='update.php?id={$rows['id']}'>update</a></td>
         </tr>"
}

Answer 2

试试这个......

<?php
$host = "localhost"; // Host name 
$username = ""; // Mysql username 
$password = ""; // Mysql password 
$db_name = "test"; // Database name 
$tbl_name = "test_mysql"; // Table name 
$server_name = "localhost";

// Create connection
$con = new mysqli($server_name, $username, $password, $db_name, 3306);
if($con->connect_error){
   die("Connection failed: ".$con->connect_error);
}

// Check connection
if($con->connect_error){
 die("Connection failed: ".$conn->connect_error);
}

$sql = "SELECT * FROM $tbl_name";
$result = $con->query($sql);
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
    <td>
        <table width="400" border="1" cellspacing="0" cellpadding="3">
            <tr>
                <td colspan="4"><strong>List data from mysql</strong></td>
            </tr>
            <tr>
                <td align="center"><strong>Name</strong></td>
                <td align="center"><strong>Lastname</strong></td>
                <td align="center"><strong>Email</strong></td>
                <td align="center"><strong>Update</strong></td>
            </tr>
            <?php
            if($result->num_rows > 0){
                // output data of each row
                while($rows = $result->fetch_assoc()){ ?>
                    <tr>
                        <td><?php echo $rows['name']; ?></td>
                        <td><?php echo $rows['lastname']; ?></td>
                        <td><?php echo $rows['email']; ?></td>
                        <td align="center"><a href="update.php?id=<?php echo $rows['id']; ?>">update</a></td>
                    </tr>
                    <?php
                }
            }
            ?>
        </table>
    </td>
</tr>
</table>
<?php
$con->close();
?>

Answer 3

我的解决方案是您在下面添加

$sql = "SELECT * FROM $tbl_name";
$result = $con->query($sql);

var_dump（$ result）; die; //包含这行代码，看看它是否真正获得了您从数据库中选择的内容。如果您可以在var_dump（）中获取这些记录，那么您可以告诉我们接下来要做什么

Answer 4

您遇到的一个问题是不同的变量名称

这是变量$row：

while($row = $result->fetch_assoc()){

这是变量$rows：

<td><? echo $rows['name']; ?></td>

我希望至少在某处发出警告

Answer 5

您使用$rows代替$row这是您正确的代码：

<tr>
    <td><? echo $row['name']; ?></td>
    <td><? echo $row['lastname']; ?></td>
    <td><? echo $row['email']; ?></td>
    <td align="center"><a href="update.php?id=<? echo $row['id']; ?>">update</a></td>
</tr>

使用mysql中的数据填充HTML表格会提供空表格单元格

5 个答案: