Html/PHP添加一個下拉與MySQL查詢

Question

我想爲我的項目設計一個網站，我在以下困惑。如果我想添加一個下拉菜單，我用下面的代碼。

<select> 
    <option value="January"<?php if ($row[month] == 'January') echo ' selected="selected"'; ?>>January</option> 
    <option value="February"<?php if ($row[year] == 'February') echo ' selected="selected"'; ?>>February</option> 
</select> 

但是我想做的是following.I有一個數據庫和一堆數據，我將有選擇statement.And查詢我想給所有的人作爲一個選項來做到這一點下拉menu.Iow我可以做到嗎？爲了簡化問題假設我有一個查詢

$sql= "select Name FROM Course WHERE courseName='".$something."'"; 
$resulting=mysqli_query($connection,$sql); 
$resultarray=mysqli_fetch_array($resulting); 

我該如何做到這一點？

編輯：沒有下拉菜單，我通常會以這種方式創建一個表單。

<form action="restaurantAction.php" method="post"> 
Restaurant Name:<input type="text" name="deleteRestaurant"> 
<input type="submit"> 
</form> 

如何申請這一個下拉菜單基於胡志明

的答案

Answer 1

你可以這樣

<select> 
<?php foreach ($resultarray as $key => $value) {?> 
    <option value="<?php echo $value?>"><?php echo $value?></option> 
<?php } ?> 
</select> 

寫你的更新

<form action="restaurantAction.php" method="post"> 
Restaurant Name:<input type="text" name="deleteRestaurant"> 
<select name="restaurantName"> 
<?php foreach ($resultarray as $key => $value) {?> 
    <option value="<?php echo $value?>"><?php echo $value?></option> 
<?php } ?> 
</select> 
<input type="submit"> 
</form> 

Answer 2

<?php 
// List filtered items 
// Fetch data as associated array 
<select> 
while ($row = mysqli_fetch_assoc($resulting)) : ?> 
<option value="<?php echo $resulting['game_name']?>"><?php echo $resulting['game_name']?></option>// $resulting['game_name'], or however called that field in the DB 
</select> 

Answer 3

 <?php      
     $result=mysqli_query($con,"select Name FROM Resturant WHERE ResturantName='".$something."'"); 
     echo '<form action="restaurantAction.php" method="post">' 
     echo 'Resutarant Name:<select name="ResturantName">'; 
      while ($row = mysqli_fetch_array($result)) { 
       echo '<option value="'.$row['Name'].'">'.$row['Name'].'</option>'; 
       } 
     echo '</select>'; 
     echo '<input type="submit">'; 
     echo '</form>'; 
     ?>  

請檢查這個