Question

 <?php
 require ('mysql_connect.php');
 $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
 ?>
 <html>
 <head> <title> Training booking </title> </head >
 <body>
 <p > <b> Student Training Booking Form </b> </p >

 <form action = "training.php " method = "get " >
     <label> Name :
         <input name=" name " placeholder=" Name " type="  "><br />
             Email :
         <input name=" email " placeholder=" example@example.com " 
 type=" "><br />
     </label>
  </form >

  <p > Please Select a Training session </p >
  <select name="top">
  <option value="0">--TOPICS--</option>
  <?php
  $query = ("SELECT topic_name FROM topic");
  while ($row = mysqli_fetch_array($query))
   ?>
  {
   <option> <?php echo $query ['topic_name']   ?> </option>
   }
   </select>

嗨，我试图用MySQL填充一个简单的下拉列表，并且下拉列表没有填充它只是保持空白。有人可以帮帮我吗。（我知道代码到处都是。）

更新

代码现在看起来像这样，它仍然无效。

<p > Please Select a Training session </p >
<?php
$sql = "SELECT topic_name FROM topic";
$query = mysqli_query($sql);
?>

<select name="topic">
<option value="0">--TOPICS--</option>
<?php
while ($row = mysqli_fetch_array($query)) {
?><option><?php echo $row ['topic_name']   ?></option><?php
}
?>
</select>

任何人都知道哪里出错了？

Answer 1

你正在看＆＃34; S＆＃34;在您的下拉列表中，因为您将查询字符串视为一个数组并显示第一个字母：

echo $query['topic_name']

必须先执行查询才能获取结果行比较mysqli_query和mysqli_fetch_array。

$query = "SELECT topic_name FROM topic";
$result = mysqli_query($mysqli,$query);

while ($row = mysqli_fetch_array($result)) {
   ?><option><?=$row['topic_name']?></option><?php
}

顺便说一句，我还建议使用prepared statements。

PHP MySQL下拉列表没有填充

1 个答案: