使用 PHP 和 mysqli 从数据库中获取选择框值

Question

我想从 phpmyadmin 数据库中提取人名并将它们放置在 HTML 选择框中，当用户从选择框中选择名称时，它应该在表格中显示该人的数据库中的详细信息。我似乎无法让它工作，我可以从数据库中获取名称并显示在选择框中，但是当您单击名称时，它似乎会显示数据库中的每条记录，而不仅仅是一个为那个人。我使用的是 mysql 而不是 mysql。这是我的代码

这是我的后端资料

<?php 
$conn = mysqli_connect("localhost", "root", "root") or die ("No connection"); 
mysqli_select_db($conn, "flat") or die("db will not open"); 
$query = "select FlatCode, Address from FLAT"; 
$result = mysqli_query($conn, $query) or die("Invalid query"); 
echo "<table border='1'><tr><th>modulecode</th><th>studentnum</th></tr>";  
while($row = mysqli_fetch_array($result)) 
{ 
    echo "<tr><td>" . $row[0] .  "</td><td>" .  $row[1] .  "</td></tr>";
} 
echo "</table>"; 
mysqli_close($conn); 
?> 

这是我的前端工作

<font size="4"> Choose an Owner Name</font><br><br> 
<form action="flat.php" method="post"> 
<select name="name"> 
<?php 
$con = mysqli_connect("localhost", "root", "root") or die ("No connection"); 
mysqli_select_db($con , "flat") or die ("db will not open"); 
$query = "SELECT distinct OwnerName, FlatCode, Address from FLAT"; 
/*$query= $_POST ("name")
function change_guery($query)
mysqli_use_result*/

$result = mysqli_query($con, $query) or die("Invalid query");  
while($rows = mysqli_fetch_array($result))

{ 
      echo "<option value=\"" . $rows[0] . "\">" . $rows[0] . "</option>"; 
} 

echo "</select>"; 
	       

mysqli_close($con); 
?> 
<input type="submit" value="Submit Value"> 
</form></body></html> 

Answer 1

你的 flat.php 代码有问题。您正在通过表单正确发布信息，但您忘记通过 flat.php 中的 $_POST 接收信息。

看下面的代码和里面的cmets，应该可以了-

<?php 
$n = $_POST["name"];//we receive the name passed by the form
$conn = mysqli_connect("localhost", "root", "root") or die ("No connection"); 
mysqli_select_db($conn, "flat") or die("db will not open"); 
$query = "select FlatCode, Address from FLAT WHERE `OwnerName` = '$n' LIMIT 1";//see the changes here 
$result = mysqli_query($conn, $query) or die("Invalid query"); 
echo "<table border='1'><tr><th>modulecode</th><th>studentnum</th></tr>";  
$row = mysqli_fetch_array($result);
//as the result will return 1 row only so we dont need while loop here
echo "<tr><td>" . $row[0] .  "</td><td>" .  $row[1] .  "</td></tr>";
echo "</table>"; 
mysqli_free_result($result);//dont forget to free result
mysqli_close($conn); 
?>