second select of dynamic box

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  • venkatarao
    New Member
    • Apr 2012
    • 1

    #1

    second select of dynamic box

    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>
    Code:
    <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>
    Last edited by Dormilich; Apr 20 '12, 03:54 AM. Reason: Please use [CODE] [/CODE] tags when posting code.
  • Stewart Ross
    Recognized Expert Moderator Specialist
    • Feb 2008
    • 2545

    #2
    What's your question? You have not told us so far.

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