php - 在 PHP 中使用对 MySQL 的单个 mysqli_fetch_array() 查询填充多个下拉列表

Question

我在填充多个下拉菜单时遇到问题。我需要通过对数据库的单个查询来完成，是否可以在单个 while 循环中完成？这是代码。只有第一个下拉列表(Player)被填充。数据库表(playerDB)有 2 列 - Player 和 Game。并且需要填充相应的下拉列表

 <form name="form1" action="" method="post">
    <fieldset>
      <legend>Selecting report</legend>
       <p>
          <?php
             $connect = mysqli_connect('localhost','root','','mydatabase');

             if(mysqli_connect_errno($connect))
             {
                echo 'Failed to connect';
             }
            else{
              echo '';
            }

            if (!$res=mysqli_query($connect, "SELECT * from playerDB")){
                  echo ("Error description: "  .mysqli_error($connect));
            }
         ?>

         <label>Select Player</label>
         <select>
           <?php
             while($row=mysqli_fetch_array($res))
             {
           ?>
            <option> <?php echo $row["Player"]; ?> </option           
           <?php
             }
          ?>
        </select>

          <label>Select Game</label>
          <select id = "myGameList">          
             <?php
             while($row=mysqli_fetch_array($res))
             {
           ?>
            <option> <?php echo $row["Game"]; ?> </option>                
           <?php
             }
           ?>
          </select>

任何帮助将不胜感激，谢谢!

Answer 1

1. 不要使用 while() 来填充选择框，而是尝试将记录保存在数组中，然后将该数组用作多次。

2. 还尝试仅获取那些您实际进一步需要的列。它将使查询更轻，记录数组也更轻。

像下面这样:-

<?php
     $connect = mysqli_connect('localhost','root','','mydatabase');

     if(mysqli_connect_errno($connect))
     {
        echo 'Failed to connect';
     }
    else{
        if (!$res=mysqli_query($connect, "SELECT Player,Game from playerDB")){ // get that much column only which you want not all
              echo ("Error description: "  .mysqli_error($connect));
        }
    }
    $data = []; //create array 
    while($row=mysqli_fetch_array($res))
    {
        $data['Player'][] = $row["Player"]; //assign values to array
        $data['Game'][] = $row["Game"];//assign values to array
    }
?>


<form name="form1" action="" method="post">
    <fieldset>
        <legend>Selecting report</legend>
        <label>Select Player</label>
        <select>
        <?php
            foreach ($data['Player'] as $player){//use array as many times you want 
        ?>
            <option> <?php echo $player; ?> </option           
        <?php
            }
        ?>
        </select>
        <label>Select Game</label>
        <select id = "myGameList">          
        <?php
            foreach ($data['Game'] as $game){//use array as many times you want 
        ?>
            <option> <?php echo $game; ?> </option>                
        <?php
            }
        ?>
</select>