如何從數據庫中獲取值並在下拉列表中顯示

Question

我已經嘗試了很多方法，但值並沒有顯示在下拉列表中
這裏，這是我的代碼。你能告訴我任何我錯的東西嗎

<?php 
$result = mysqli_query($con,"SELECT * FROM project"); 

    if(mysqli_num_rows($result)==0){ 
    echo "<tr><td>No Rows Returned</td></tr>"; 
    }else{ 
    $row = mysqli_fetch_assoc($result); 
     $pos = 0; 
     echo "<select name=Pname >"; 
     while($pos <= count ($row)){ 
     echo "<option value="$row["project_no"]">"$row["project_name"]"</option>"; 
      $pos++; 
      } 
echo "</select>";?> 

而我寫爲.php文件。謝謝你的幫助。

Answer 1

嘗試了這一點：

$output = ''; 
if(mysqli_num_rows($result) == 0){ 
    // echo error; 
} else { 
    while($row = mysqli_fetch_assoc($result)){ 
     $project_no = $row['project_no']; 
     $project_name = $row['project_name']; 

     $output .= '<option value="' . $project_no . '">' . $project_name . '</option>"; 
    } 
} 

那麼你的HTML裏面，打印<select>元素中的$output變量：

<select> 
<?php 
    print("$output"); 
?> 
</select> 

應該打印所有選項您所請求的每一行來自數據庫。

希望這有助於:)

Answer 2

試試這個：

$result = mysqli_query($con,"SELECT * FROM project"); 

if(mysqli_num_rows($result)==0){ 
    echo "<tr><td>No Rows Returned</td></tr>"; 
}else{ 
    echo "<select name=Pname >"; 
    while ($row = mysqli_fetch_assoc($result)) { 
     echo "<option value="$row["project_no"]">"$row["project_name"]"</option>"; 
    } 
    echo "</select>"; 
} 

Answer 3

這是結果的代碼，我可以運行它。我把這個代碼的形式代碼的HTML

的

$result = mysqli_query($con,"SELECT * FROM project"); ?> <?php $output = ''; if(mysqli_num_rows($result) == 0){ // echo error; } else { echo " <select name = Pname>"; while($row = mysqli_fetch_assoc($result)){ $project_no = $row['project_no']; $project_name = $row['project_name']; $output = "<option value=" . $project_no . "> ". $project_name ." </option>"; print("$output"); } echo " </select>"; } ?>

謝謝每一個幫助我^^