如何从数据库中获取值以显示在下拉列表中?
问题描述:
while ($line = sqlsrv_fetch_array($query_result, SQLSRV_FETCH_ASSOC)) {
$departmentName = trim($line['Department']);
echo "<input type=\"submit\" value=\"$departmentName\" name=\"departmentName\"/>";
echo "<br>";
yeilds这个..
<form action="process_case2.php" method="post">
<input type="submit" value="Accounting" name="departmentName"/><br>
<input type="submit" value="Administration" name="departmentName"/><br>
<input type="submit" value="Finance" name="departmentName"/><br>
<input type="submit" value="Human Resources" name="departmentName"/><br>
<input type="submit" value="InfoSystems" name="departmentName"/><br>
<input type="submit" value="Legal" name="departmentName"/><br>
<input type="submit" value="Marketing" name="departmentName"/><br>
<input type="submit" value="Production" name="departmentName"/><br>
</form>
我怎样才能让这第一部分,使值在一个下拉列表?
答
这会让你想要什么:
while ($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>
在接下来的页面,与
$_POST['departmentName'];
获取你的数据我已经全部其他位,我只需要知道我需要做的到这一行echo“”;“,为了使它成为一个下拉列表。 – user3342038 2014-11-05 22:53:58
哪一个? ---- – baao 2014-11-05 22:54:34
“; – user3342038 2014-11-05 22:55:37