只需单击一个按钮，就可以将PHP Drop Box值传递给查询

I have a drop box for Date populated from Oracle table. I need the date from drop box to be passed in a query where clause and then on the click of button records should be displayed on screen and date should be remain selected in a drop box.I have used if(isset($_POST['Date'])). This pass the date to query and data is displayed but it triggers the query when i select the date from drop box without clicking on the button which is not really required. How can i achieve this below is the code snippet:

<form action="" method="post" style="display: inline;"> 
        <input type="submit" name="sub2" value="Rtlogs Status" /> 
    </form> 
 <form action="" method="post" style="display: inline;"> 
  <select name="Date" onchange="this.form.submit()"><option value="">Select   Date</option> 
            <?php 
            $rs = ("select distinct dc_dy_bsn from RTLOGS_BO_CO_FAILED"); 
            $sql=$db->Execute($rs);

            while ($at = $sql->FetchRow()) { 

                echo "<option value='".$at[0]."'>".$at[1]." (".$at[0].")</option>";                 
            }
            ?> 
if(isset($_POST['Date'])){ 
        $rs = $db->Execute("select * from RTLOGS_BO_CO_FAILED where dc_dy_bsn = '".$_POST['Date']."' order by 3 desc"); 
        echo "<table class='mine'>"; 
        $cn = 0; 
        while ($at = $rs->FetchRow()) { 
            if($cn == 0) { 

                echo "<tr>"; 
                foreach ($at as $k => $v) { 
                    if (!is_numeric($k)) { 
                        echo "<th>" . $k . "</th>"; 
                    }   
                } 
                echo "</tr>"; 
            } 
            echo "<tr>"; 
            for ($i = 0; $i < (count($at) / 2); $i++) { 
                echo "<td>" . $at[$i] . "</td>"; 
            } 
            echo "</tr>"; 
            $cn++; 
        } 
        echo "</table>"; 
        $db->Close(); 
    }