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bolagani venkatarao
second selectbox of dynamic
0 Answer(s)      a year and a month ago
Posted in : JavaScript Questions

hello friends i hava a code like this and could any body help me with editing code and send me code. i am getting city values but i can't location values of city from database.thanks in advance.

<html>
<head>
<title>demo</title>
<body>
<form method="POST" action="http://localhost/vtigercrm/modules/Webforms/
post.php">
<input type="hidden" value="Leads" name="moduleName" />
<table>
<tbody>
<tr>
  <td><label>Last Name</label></td>
  <td><input type="text" name="lastname" value="" /></td>
</tr>
<tr>
  <td><label>First Name</label></td>
  <td><input type="text" name="firstname" value="" /></td>
</tr>
<tr>
  <td><label>Company</label></td>
  <td><input type="text" name="company" value="" /></td>
</tr>
<tr>
  <td><label>email</label></td>
  <td><input type="text" name="email" value="" /></td>
</tr>
<tr><td><label>city</label></td><td colspan=1>
   <select name="city">
    <option>city</option>
<?php
$connect=mysql_connect("localhost","root","venkataz");
if(!$connect) die("access failed");
$select=mysql_select_db("vtigercrm530");
if(!$select) die("select failed");
echo "selected vtigercrm530";
$query="SELECT DISTINCT address_city FROM vtiger_users";
$result=mysql_query($query);
if(!$result) die("database access failed:" . mysql_error());
while($row = mysql_fetch_array($result))
{
echo "<option value>".$row['address_city']."</option>";
}
?>
</select>
</tr>
<tr><td><label>lane</label></td><td colspan=1>
<select id="lane" onselect="show();">
<option>location</option>
<script type="text/javascript">
function show()
{
var lane=document.getElementById["lane"].selectedIndex;
alert(lane);
}
</script>
<?php
$connect1=mysql_connect("localhost","root","venkataz");
if(!$connect1) die("access failed");
$select1=mysql_select_db("vtigercrm530");
if(!$select1) die("select failed");
echo "";
echo "<br/>";
$query1="SELECT address_city FROM vtiger_users";
$result1=mysql_query($query1);
if(!$result1) die("database access failed:" . mysql_error());
$lcity=$_REQUEST["city"];
$rows=mysql_num_rows($result1);
for($j=0;$j<$rows;$j++)
if(mysql_result($result1,$j,"address_city")==$lcity)
{
$query2="SELECT address_street FROM vtiger_users WHERE address_city='".mysql_result($result1,$j,'address_city')."'";
$result2=mysql_query($query2);
while($row= mysql_fetch_array($result2))
{
echo "<option value>".$row['address_street']."</option>";
}
}
?>
</select>
</td></tr>
</tbody>
<table>
<input type="submit" value="Submit" />
</form>
</body>
</head>
</html>
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