[已解决]php+mysql+js实现二级联动下拉列表，二级列表的内容不显示？？

发布于 2022-10-15 04:55:19 字数 3832 浏览 13 评论 0

本帖最后由 kettysong 于 2011-06-07 22:00 编辑

菜鸟求救，该怎么改啊？
下面是源码
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gbk">
<title>全动态二级联动下拉列表</title>
</head>
<body>
<?
/***********************************************
** 功能：PHP+mysql实现二级级联下拉框
** 数据库：数据库名（db_city）、数据表（t_provice、t_city）
** 表t_provice中字段：id（id编号）、proviceName（省份名）
** 表t_city中的字段：id（id编号）、proviceId（省份ID）、cityName（城市名）
***********************************************/

//****************** 连接选择数据库 ***************
$link = mysql_connect("localhost", "root", "")
or die("Could not connect : " . mysql_error());mysql_query("set names 'gbk'");//
mysql_select_db("db_city") or die("Could not select database");

//******************提取省份信息******************
$sqlSel = "select * from t_provice order by id ";
$result = mysql_query($sqlSel) or die("Query failed : " . mysql_error());

$forum_data = array();
while( $row = mysql_fetch_array($result) )
{
$forum_data[] = $row;
}
//print_r ($forum_data);
mysql_free_result($result);

//**************获取城市信息**************
$sqlSel2 = "select * from t_city order by proviceId desc";

if( !($result2 = mysql_query($sqlSel2)) )
{
die('Could not query t_city list');
}

$forum_data2 = array();
while( $row2 = mysql_fetch_array($result2) )
{
$forum_data2[] = $row2;
}

mysql_free_result($result2);
?>

<script language = "JavaScript">
var onecount2;
subcat2 = new Array();
<?php
$num2 = count($forum_data2);
?>
onecount2=<?echo $num2;?>;
<?
for($j=0;$j<$num2;$j++)
{
?>
subcat2[<?echo $j;?>] = new Array("<?echo $forum_data2[$j]['id'];?>","<?echo $forum_data2[$j]['proviceId'];?>","<?echo $forum_data2[$j]['cityName'];?>");
<?}?>
function changelocation(id)
{
document.myform.city.length = 0;
var id=id;
var j;
document.myform.city.options[0] = new Option('==选择城市==','');
for (j=0;j < onecount2; j++)
{
if (subcat2[j][1] == id)
{
document.myform.city.options[document.myform.city.length] = new Option(subcat2[j][2], subcat2[j][0]);
}
}
}
</script>

<form name="myform" method="post">
<select name="bigClass" onChange="changelocation(document.myform.bigClass.options[document.myform.bigClass.selectedIndex].value)" size="1">
<option selected>请选择省份</option>

<?php
$num = count($forum_data);

for($i=0;$i<$num;$i++)
{
?>
<option value="<?echo $forum_data[$i]['id'];?>"><?echo $forum_data[$i]['proviceName'];?></option>
<?
}
?>
</select>
<select name="city" >
<SELECT name=city size=1 id="city">
<option selected value="">==选择城市==</option>
</select>
</form>
</body>
</html>