Calling a Range From MySQL Table With Drop Down Menu

**kjm7267** · Jul 4, 2007 22:00

Good evening from Palo Alto, California. I have four items in a MYSQL database table. Each has 2 fields: name and weight, as shown here:

Bob 149
Sue 98
Jack 155
Sally 110

I want users to be able to visit a drop-down menu and ask to see the names associated with one of three weight ranges: up to 100, between 100 and 150, and over 150. If, for example, they selected 100-150, I'd like the code to return:

Sally
Bob

But I'm stuck on how to do this. I imagine that the HTML page with the drop down menu would look something like:

<HTML>
<HEAD></HEAD>
<BODY>
<form action="generatelist.php" method="POST">
Show me the names of people with weights of
<select name="weight" size="1">
<option value> </option>
<option value="upto100">up to 100</option>
<option value="100-150">100-150</option>
<option value="150+">over 150</option>
</select>
<INPUT TYPE="submit" value="Search"></FORM>
</BODY>
</HTML>

I also imagine that "generatelist.php" should look something like:

<?
$weight = $_POST['weight'];
mysql_connect("ip.address.goes.here", "sql@nameofmysite.com", "password") or die(mysql_error());
@mysql_select_db("nameofdatabase") or die( "Unable to select database");
$query="SELECT * FROM landforsale WHERE weight='WeightRange'";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
$i=0;
while ($i < $num) {
$weight=mysql_result($result,$i,"weight");
$name=mysql_result($result,$i,"name");
?>
<? echo "$name";?><br><br>
<?
++$i;
}
?>

Please, do you have some guidance for me? Thank you!

**Raju Gautam** · Jul 4, 2007 23:50

Try out this code which working fine with me although this might not be the good way if someone expert finds easy and short way.

PHP Code:


if($_POST['Submit']){

    $weight = $_POST['weight'];

    

    if($weight == "100"){

        $param = " AND weight<100";

    }

    else if($weight == "100-150"){

        $param = " AND weight>100 AND weight<150";

    }else if($weight == "150"){

        $param = " AND weight>150";

    }else{

        $param = "";

    }

    $sql = "SELECT * FROM tbltest WHERE 1 $param";

    $result = mysql_query($sql);

    while($rows = mysql_fetch_array($result)){

        echo $rows['name1'] . "<br>";

    }

}

HTML Code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Test</title>
</head>
<body>
<form name="frm1" method="post">
<table width="350" border="0" cellspacing="1" cellpadding="3">
  <tr>
    <td colspan="2" align="left" valign="middle">Select Weight </td>
    </tr>
  <tr>
    <td align="right" valign="middle">&nbsp;</td>
    <td><select name="weight" size="1">
        <option value="0">Select</option>
        <option value="100">Up To 100</option>
        <option value="100-150">100-150</option>
        <option value="150">Over 150</option>
      </select>    </td>
  </tr>
  <tr>
    <td align="right" valign="middle">&nbsp;</td>
    <td><input type="submit" name="Submit" value="Send Mail"></td>
  </tr>
  <tr>
    <td colspan="2" align="left" valign="middle">&nbsp;</td>
  </tr>
</table>
</form>
</body>
</html>

**kjm7267** · Jul 5, 2007 07:56

Thank you, rajug, very much for your reply. The code seems not to be working for me. I wonder -- does someone see a simpler solution?
Thank you!

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Other suggestions? And thanks, rajug!

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