如何从数据库获取值以显示在下拉列表中？

Question

这个..

while ($line = sqlsrv_fetch_array($query_result, SQLSRV_FETCH_ASSOC)) {
    $departmentName = trim($line['Department']);
    echo "<input type=\"submit\" value=\"$departmentName\" name=\"departmentName\"/>";  
    echo "<br>";

这样。

<form action="process_case2.php" method="post">
    <input type="submit" value="Accounting" name="departmentName"/><br>
    <input type="submit" value="Administration" name="departmentName"/><br>
    <input type="submit" value="Finance" name="departmentName"/><br>
    <input type="submit" value="Human Resources" name="departmentName"/><br>
    <input type="submit" value="InfoSystems" name="departmentName"/><br>    
    <input type="submit" value="Legal" name="departmentName"/><br>
    <input type="submit" value="Marketing" name="departmentName"/><br>      
    <input type="submit" value="Production" name="departmentName"/><br> 
</form>

我如何才能做到这一点的第一部分，以便这些值在下拉列表中？

Answer 1

这将使您想要：

while ($line = sqlsrv_fetch_array($query_result, SQLSRV_FETCH_ASSOC)) {
$departmentName[] = trim($line['Department']);
}
?>
<form action="process_case2.php" method="post">

// paste other fields here

<select id="what" class="ever" name="departmentName">
<?php
foreach ($departmentName as $a)
{
?>
<option value="<?= $a; ?>"><?= $a; ?></option>
<?php
}
?>

</select>

// or here

<input type="submit" name="submit" value="submit"/>
</form>

在下一页上，使用

$_POST['departmentName'];