我有一些下拉菜单。如果我选择其中一个,我想用与我所选择的内容可靠的东西来改变另一个的内容。我该怎么做呢?使用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了年份,那么在另一个下拉菜单中将只显示该年份的内容。
这是我的内容与下拉菜单
<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>
你在找这样的东西吗?
<html>
<title>dropdownlist</title>
<head>
<script language="Javascript" type="text/javascript" >
function choix(formulaire)
{
var j;
var i = form1.boite1.selectedIndex;
if (i == 0)
for(j = 1; j <3; j++)
form1.boite2.options[j].text="";
else{
switch (i){
case 1 : var text = new Array( "London","Manchester","Birmingham");
break;
case 2 : var text = new Array("Paris","Marseille","Lyon");
break;
case 3 : var text = new Array("Berlin","Munich","Francfort");
break;
}
for(j = 0; j<3; j++)
form1.boite2.options[j+1].text=text[j];
}
form1.boite2.selectedIndex=0;
}
</script>
</head>
<body>
<form name="form1">
<select name="boite1" onChange="choix(this.form)">
<option selected>country</option>
<option>England</option>
<option>France</option>
<option>Germany</option>
</select>
<select name="boite2">
<option selected>cities</option>
<option></option>
<option></option>
<option></option>
</form>
</select>
</body>
</html>
如果你想绝对只使用HTML和PHP,你将需要使用一些ajax,因为PHP是服务器端和HTML是静默端。所以为了方便起见,我推荐上面的代码