Question

基本上，我正在创建一个拥有7种电影类型的电影订购系统，我希望每个电影选项卡根据类型填充。在这一点上，我甚至无法让桌子出现。任何人都可以帮助我吗？

    <div id="content">
    <table>
    <tr><th>Name</th> <th>Genre</th> <th>Year</th> <th>Rating</th></tr>
<?php
//connect to database
$dbc = mysql_connect('localhost' , 'username' , 'password');
$test = mysql_select_db('movies', $dbc);

if ($test = mysql_select_db('movies', $dbc))//test
{
    $query = "SELECT * FROM 'movies' WHERE 'genre' = 'Action'";

    //call query
    $result = mysql_query($query, $dbc);

    while ($row = mysql_fetch_array($result))
    {
        ?>

        <tr>
        <td><?php print $row['name']; ?></td>
        <td><?php print $row['genre']; ?></td>
        <td><?php print $row['year']; ?></td>
        <td><?php print $row['rating'];?></td>
        </tr>
<table>

        <?php

    }//close the while loop

}//close the if statement

mysql_close($dbc);

?>

Answer 1

首先，不要因为安全问题而使用mysql，以后不会在PHP中使用它。阅读http://php.net/manual/en/function.mysql-connect.php。

尝试使用PDO或mysqli，例如：

$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM 'movies' WHERE 'genre' = 'Action'");

while($row = mysqli_fetch_array($result))
  {
  echo $row['name'];
  echo "<br>";
  }

mysqli_close($con);

使用MySql使用特定列填充html表

1 个答案: