Question

我正在为学校建立一个关于MLB和数据库的网站。但现在我希望数据库中的信息作为下拉列表中的一个选项。这是我的HTML代码：

<!DOCTYPE html>
<html>
<head>
<link rel="icon" type="image/png" href="./images/favicon-32x32.png" 
sizes="32x32" />
<link rel="icon" type="image/png" href="./images/favicon-16x16.png" 
sizes="16x16" />
<title>MLB: Major League Baseball</title>
<link href="css folder/MLBstylesheet.css" rel="stylesheet" 
type="text/css"/>
</head> 
<body>
<div id="container">
<div id="titel">    

    <img class="MLBTitel"  src="./images/MLBtitel.jpg" alt="MLBTitel" >
<div id="titeltekst">
MAJOR LEAGUE BASEBALL
    <br>    
</div>

<nav>
    <ul>
        <li><a class= "menu" href="index.html">Home</a></li>
        <li><a class= "menu" href="spelers.php">Spelers</a></li>
        <li><a id = "active" class= "menu" href="teams.php">Teams</a></li>
        <li><a class= "menu" href="wedstrijden.html">Wedstrijden</a></li>
        <li><a class= "menu" href="contact.html">Contact</a></li>
    </ul>
</nav>

<br><br>        
</div>
<div id="teamtabel">
<?php
$servername = "localhost";
$username = "id1419279_root";
$password = "*******";
$dbname = "id1419279_mlb";

// Create connection
$conn = new mysqli($localhost, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM teams";
$result = $conn->query($sql);

echo "<select name='naamteam'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['teamnaam'] ."'>" . $row['teamnaam'] ."
</option>";
}
echo "</select>";
?>

</div>
</div>
</body>
</html>

选项可能是团队名称，但似乎我的下拉列表仍为空：picture of the dropdownlist on the website 这是我的数据库的样子： picture of my database

我希望有人可以帮助我，我真的很感激。

Answer 1

Dependencies

使用数据库信息作为下拉列表选项

1 个答案: