Question

我已經有一個while循環顯示我的表單中的選擇輸入，具體取決於用戶預訂的乘客數量。每個下拉菜單都應該顯示來自數據庫中列的所有數據。問題是，只有第一個下拉顯示我拿來的數據是座位1至30

查詢來獲得當前飛行的航空公司ID

$result = mysqli_query($conn,"select * from flight_seat_status where flight_number = '$_SESSION[departure]'"); 

變量我曾經開始計數循環

$print = 1; $name = 0; 

我while循環的問題

echo "<h1 class='adminContent_h1'>Please choose a seat</h1>"; 
    while($print <= $_SESSION['passengers']){ 
    echo "<label style = 'width:170px;'>". $_SESSION['firstname'][$name]."'s Seat</label>"; 
    echo "<select class = 'content_dropdown' style = 'width:15%; margin-bottom:20px;' name = 'passengers[]'>"; 
    while($row = mysqli_fetch_assoc($result)){ 
    echo "<option value='".$row['seat_id']."'>".$row['seat_number']."</option>"; 
     } 
    echo "</select><br>"; 
     $print++; 
     $name++; 
    } 

它能夠根據用戶打印出多個下拉菜單，但只有第一個下拉菜單包含數據。

我的代碼有什麼問題？請詳細解釋

Answer 1

必須設置指針回到0再次

// set the pointer back to the beginning 
    mysqli_data_seek($result, 0); 
    while($row = mysqli_fetch_assoc($result)){ 
    // do whatever here... 
    } 

Answer 2

請使用下面的代碼。所有座位記錄您必須在乘客循環外取一次並存儲在另一個陣列中，然後每次都必須使用乘客循環內的座位排列。

echo "<h1 class='adminContent_h1'>Please choose a seat</h1>"; 
$seats = []; 
while($row = mysqli_fetch_assoc($result)){ 
    $seats[] = $row; 
} 
while($print <= $_SESSION['passengers']){ 
    echo "<label style = 'width:170px;'>". $_SESSION['firstname'][$name]."'s Seat</label>"; 
    echo "<select class = 'content_dropdown' style = 'width:15%; margin-bottom:20px;' name = 'passengers[]'>"; 
    foreach($seats as $res_seat){ 
     echo "<option value='".$res_seat['seat_id']."'>".$res_seat['seat_number']."</option>"; 
    } 
    echo "</select><br>"; 
    $print++; 
    $name++; 
}