1

我正在尝试在包含用户 SQL 数据的 html 页面上显示一个表格。

假设我可以连接并且表中有信息。

<?php
echo "<table>
            <tr>
            <th>First Name</th>
            <th>Last Name</th>
            <th>Hair Color</th>
        </tr>";
    $result = mysql_query("SELECT * FROM `table`");
    while($row = mysql_fetch_assoc($result))
    {
        echo "<tr>";
        echo "<td>" .$row['first']."</td>";
        echo "<td>".$row['last']."</td>";
        echo "<td>".$row['color']."</td>";
        echo "</tr>";
    }
    echo "</table>";

    echo "<input type="button" value="updateTable" id="btn">";
  ?>

因为没有数据输出,我哪里出错了?

4

2 回答 2

1

你可以试试这个,

下面的行导致错误,

 echo "<input type="button" value="updateTable" id="btn">";

应该是这样的

 echo '<input type="button" value="updateTable" id="btn">';

PHP代码:

  <?php     

        echo "<table>
                <tr>
                <th>First Name</th>
                <th>Last Name</th>
                <th>Hair Color</th>
            </tr>";
        $result = mysql_query("SELECT * FROM `table`");
        while($row = mysql_fetch_assoc($result))
        {
            echo "<tr>";
            echo "<td>" .$row['first']."</td>";
            echo "<td>".$row['last']."</td>";
            echo "<td>".$row['color']."</td>";
            echo "</tr>";
        }
        echo "</table>";

        echo '<input type="button" value="updateTable" id="btn">';          

?>
于 2013-11-06T19:03:09.297 回答
1

首先,请考虑切换到MySQLi_和/或 PDO 和准备好的语句。MySQL_已弃用。

请阅读这两篇文章(在直播之前):

有一行语法错误,这一行:

echo "<input type="button" value="updateTable" id="btn">";

应该写成:

echo '<input type="button" value="updateTable" id="btn">';

或者:

echo "<input type='button' value='updateTable' id='btn'>";

或转义双引号:

echo "<input type=\"button\" value=\"updateTable\" id=\"btn\">";

在我自己的服务器上测试:

<?php
echo "<table>
    <tr>
    <th>First Name</th>
    <th>Last Name</th>
    <th>Hair Color</th>
</tr>";

// CONNECT TO THE DATABASE

$db = mysql_connect("localhost", "username", "password");
if($db->connect_errno > 0) {
  die('Connection failed [' . $db->connect_error . ']');
}

$mydb = "your_db";
$table = "your_table";
mysql_select_db($mydb) or die ('Unable to select database');

$result = mysql_query("SELECT * FROM `$table`");

while($row = mysql_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td>" .$row['first']."</td>";
        echo "<td>".$row['last']."</td>";
        echo "<td>".$row['color']."</td>";
        echo "</tr>";
    }
    echo "</table>";

    echo "<input type=\"button\" value=\"updateTable\" id=\"btn\">";
?>
于 2013-11-06T19:38:00.613 回答