从属下拉列表 [英] Dependent drop down list

问题描述

我试图制作一个简单的依赖下拉列表，这是表格

|

id

|

pid

|

类别

|

|

1

|

0

|

动物

|

|

2

|

0

|

水果

|

|

3

|

1

|

兔子

|

|

4

|

2

|

柠檬

|

下拉列表级别1的php代码:

I tried to make a simple dependent drop down list and here is the table

|

id

|

pid

|

category

|

|

1

|

0

|

animal

|

|

2

|

0

|

fruit

|

|

3

|

1

|

rabbit

|

|

4

|

2

|

lemon

|

php code for drop down list level 1:

<select name="level1" class="level1" onChange="get_level2(this.value)">
        <option selected="selected">--level 1--</option>
        <?php

        $sql_level1 = "SELECT * FROM table WHERE pid=0";

        $result_level1 = mysql_query($sql_level1);



        while($row_level1 = mysql_fetch_array($result_level1))

        {

            echo "<option value='".$row_level1['id']."'>".$row_level1['kategori']."</option>";
        }
        ?>
        </select>

1级下拉菜单可以正常工作，但是我不知道如何解析pid级别中的选定ID，以根据pid值使2级下拉菜单..

如何在javascript中将1级onChange设为1级选择的id成为2级pid?

仍在搜寻，但仍无法解决我的问题...

如果我尝试将ID解析为这样的另一个文件，这是正确的吗?
函数get_level2(id)
{
$ .ajax({
类型:"POST"，
网址:"../include/level2.php"，/*水果或动物的ID将发送到此文件*/
beforeSend:function(){
$(#level2").html(< option> Loading ...</option>");
}，
数据:"level2_id =" + id，
成功:功能(msg){
$(#level2").html(msg);
}
});
}

这是level2.php文件:

The level 1 drop down work fine, but I down know how to parsing the chosen id in level 1 to make drop down level 2 based on the pid value..

How to make the level 1 selected id become level 2 pid when level 1 onChange in javascript??

Still googling but still don''t solved my problem...

if i try to parsing the id to another file like this, is it correct??
function get_level2(id)
{
$.ajax({
type: "POST",
url: "../include/level2.php", /* The fruit or animal id will be sent to this file */
beforeSend: function () {
$("#level2").html("<option>Loading ...</option>");
},
data: "level2_id="+id,
success: function(msg){
$("#level2").html(msg);
}
});
}

and this is the level2.php file:

<?php
include "../include/sqlConnect.php";

$level2_id = $_REQUEST['level2_id'];

$sql_level2 = "SELECT * FROM table WHERE pid = '".$level2_id."'";
$result_level2 = mysql_query($sql_level2);
echo "<select name='level2'>";

while($row_level2 = mysql_fetch_array($result_level2))
{
echo "<option value='".$row_level2['id']."'>".$row_level2['kategori']."</option>";
}

echo "</select>";
?>

谁能帮助我..

Can anyone help me please..

推荐答案

sql_level1 SELECT * FROM table WHERE pid = 0" span>

sql_level1 = "SELECT * FROM table WHERE pid=0";

result_level1 > mysql_query(

result_level1 = mysql_query(

sql_level1); span> while(

sql_level1); while(

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