我的select
获取所有数据,然后在单独的下拉菜单中显示所有数据
问题:如何将我所有的侍从都放在同一个下拉列表中?
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "categorys";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM all_category";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = mysqli_fetch_array( $result )) {
echo "<select>";
echo '<option>' . $row['category'] . '</option>';
echo "</select>";
}
}
?>
从循环中取出选择标签,只使用选项标签:
if ($result->num_rows > 0) {
echo "<select>";
while($row = mysqli_fetch_array( $result )) {
echo '<option>' . $row['category'] . '</option>';
}
echo "</select>";
}
此外,缩进代码会使其看起来更好。
这是因为在while循环中打开和关闭新的<select>
标记。这需要在循环本身之外完成:
echo "<select>";
while($row = mysqli_fetch_array( $result )) {
echo '<option>' . $row['category'] . '</option>';
}
echo "</select>";