我有一些下拉菜单。如果我选择其中一个,我想用我选择的内容改变其他内容。我怎样才能做到这一点 ?用html和php。
例如,我有一个表
Year with id_year and year
Stuff with id_stuf and stuff
如果我在第一个下拉菜单中选择一年,则在另一个drodown菜单中将仅显示该年的内容。
这是我的下拉菜单
的内容<div class="view">
<form name="tabel" method="post" action="insertexamen.php">
<table>
<tr>
<td>Data</td>
<td><input type="date" name="data" value="data" required="required"/><br></td>
</tr>
<tr>
<td>An</td>
<td>
<?php
$sql_year="SELECT * FROM an";
$rez_year = mysqli_query($link,$sql_year);
echo "<select name=\"year\" >";
while($year=mysqli_fetch_array($rez_year))
{
echo "<option value=\"".$year['id_an']."\">".$year['grupa']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Materie</td>
<td>
<?php
$sql_mat="SELECT * FROM materii";
$rez_mat = mysqli_query($link,$sql_mat);
echo "<select name=\"mat\" >";
while($mat=mysqli_fetch_array($rez_mat))
{
echo "<option value=\"".$mat['id_mat']."\">".$mat['numemat']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Profesor</td>
<td>
<?php
$sql_proff="SELECT * FROM profesor";
$rez_proff = mysqli_query($link,$sql_proff);
echo "<select name=\"proff\" >";
while($proff=mysqli_fetch_array($rez_proff))
{
echo "<option value=\"".$proff['id_prof']."\">".$proff['numep']." ".$proff['prenumep']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Asistent</td>
<td>
<?php
$sql_profff="SELECT * FROM profesor";
$rez_profff = mysqli_query($link,$sql_profff);
echo "<select name=\"profff\" >";
while($profff=mysqli_fetch_array($rez_profff))
{
echo "<option value=\"".$profff['id_prof']."\">".$profff['numep']." ".$profff['prenumep']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Sala</td>
<td>
<?php
$sql_room="SELECT * FROM sala";
$rez_room= mysqli_query($link,$sql_room);
echo "<select name=\"room\" >";
while($room=mysqli_fetch_array($rez_room))
{
echo "<option value=\"".$room['id_s']."\">".$room['salaa']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Tip</td>
<td>
<?php
$sql_type="SELECT * FROM examen";
$rez_type= mysqli_query($link,$sql_type);
echo "<select name=\"type\" >";
while($type=mysqli_fetch_array($rez_type))
{
echo "<option value=\"".$type['id_tip']."\">".$type['tip']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td><input name="submit" type="submit" value="Trimite"/></td>
<td><input name="reset" type="reset" value="Reset"/></td>
</tr>
</table>
答案 0 :(得分:0)
你在寻找类似的东西:
count(distinct Real (Color))
如果你只想使用html和php完成它,你将需要使用一些ajax,因为php是服务器端,而html是slient side。所以,为了你的东西我会推荐上面的代码