[英]display a cascading dropdown list
index.php
<div id="container">
<div id="body">
<div id="dropdowns">
<div id="center" class="cascade">
<?php
$sql = "SELECT searchname FROM search_parent";
$query = mysqli_query($con, $sql);
?>
<label>First Option:
<select name="searchname" id = "drop1">
<option value="">Please Select</option>
<?php while ($rs = mysqli_fetch_array($query, MYSQLI_ASSOC )) { ?>
<option value="<?php echo $rs["id"]; ?>"><?php echo $rs["searchname"]; ?></option>
<?php } ?>
</select>
</label>
</div>
<div class="cascade" id="state"></div>
<div id="city" class="cascade"></div>
</div>
</div>
</div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("select#drop1").change(function(){
var country_id = $("select#drop1 option:selected").attr('value');
// alert(country_id);
$("#state").html( "" );
$("#city").html( "" );
if (country_id.length > 0 ) {
$.ajax({
type: "POST",
url: "fetch_state.php",
data: "country_id="+country_id,
cache: false,
beforeSend: function () {
$('#state').html('<img src="loader.gif" alt="" width="24" height="24">');
},
success: function(html) {
$("#state").html( html );
}
});
}
});
});
</script>
fetch_state.php
<?php
include("connection.php");
$country_id = trim(mysqli_escape_string($con,$_POST["country_id"]));
$sql = "SELECT * FROM t1 WHERE parent_id = '".$country_id ."' ";
$count = mysqli_num_rows( mysqli_query($con, $sql) );
if ($count > 0 ) {
$query = mysqli_query($con, $sql);
?>
<label>Select:
<select name="state" id="drop2">
<option value="">Please Select</option>
<?php while ($rs = mysqli_fetch_array($query, MYSQLI_ASSOC)) { ?>
<option value="<?php echo $rs["id"]; ?>"><?php echo $rs["title"]; ?></option>
<?php } ?>
</select>
</label>
<?php
}
?>
<script src="jquery-1.9.0.min.js"></script>
search_parent表
id searchname
1 t1
2 t2
3 t3
當用戶選擇t1時,第二個下拉列表中的值應從t1表中填充,如果選擇t2,則應從t2表中填充,如果選擇t3,則第二個下拉列表應從t3表中填充
t1表
id title parent_id
1 tt1 1
但是,當我運行此代碼時,它不起作用,它只是顯示第一個下拉菜單,當我選擇一個值時,則什么都沒有顯示
我認為您的問題是,您缺少id
字段。
因此,在您的第一個查詢中,您說:
SELECT searchname FROM search_parent
如您所述,可能是t1
, t2
, t3
。 但是您沒有在查詢中選擇id
字段,但是您嘗試將其添加為選項值:
<option value="<?php echo $rs["id"]; ?>"><?php echo $rs["searchname"]; ?></option>
here^^^^^^^^^
因此,沒有任何價值。
您需要將查詢更改為
SELECT id,searchname FROM search_parent
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