如何从数据库获取值以显示在下拉列表中？

Question

This..

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

yeilds this..

<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>

How can I make the first part of this, so that the values are in a drop down list?

Answer 1

This will make what you want:

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>

On the next page, fetch your data with

$_POST['departmentName'];