Question

我在这里按照上一个问题获得了一个链接到MySQL的下拉菜单。它的工作方式没有表格，只有形式，所有room_id只是散布在一条线上而不是进入下拉框。关于如何修复它的任何想法？谢谢

//Creates a form for room_id
echo "<form action=''>";
echo "<select name='room_id'>";
//Creates drop down box to show the current rooms vacant
$sql = "SELECT * FROM room";
$sql.= " WHERE room_vacant = 1";
$stmt = $dbh->query($sql);
echo "<select name='room_id'>";
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
    echo "<option value='" . $row['room_id'] . "'>" . $row['room_id'] . "</option>";
} //Closes drop down box
echo "</select>";

//Submit button
echo "<input type='submit' value='Submit'>";

//Closes form
echo "</form>";

Answer 1

原因是你正在回复你的选择两次。删除一个。

//Creates a form for room_id
echo "<form action=''>";
echo "<select name='room_id'>";
//Creates drop down box to show the current rooms vacant
$sql = "SELECT * FROM room";
$sql.= " WHERE room_vacant = 1";
$stmt = $dbh->query($sql);
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
    echo "<option value='" . $row['room_id'] . "'>" . $row['room_id'] . "</option>";
} //Closes drop down box
echo "</select>";

//Submit button
echo "<input type='submit' value='Submit'>";

//Closes form
echo "</form>";

HTML / PHP表单选择带有MySQL链接的选项

1 个答案: