Welcome to WebmasterWorld Guest from 188.8.131.52
Question 2 Is what more do I need to add to my php script to display them if anything. BELOW IS MY PHP SCRIPT TO RETIREVE DATA.
$conn = mysql_connect("localhost","","");
$person = $_POST['sex'];
$myage = $_POST['dob'];
$myheight = $_POST['size'];
$resultID = mysql_query("SELECT * FROM $person WHERE age in ($myage) and height in ($myheight) ",$conn);
print "<table border = 1><tr><th>id</th>";
while ($row = mysql_fetch_row($resultID))
foreach ($row as $field)
Thanks for help everyone has allways been helpful.
Phil from Orlando Florida
To display the link as an active link you need a line like:
echo "<a href='http://".$yourresult."'>Some explanatory text or repeat the link text</a>";
[edited by: jatar_k at 5:03 pm (utc) on May 22, 2004]
[edit reason] no sigs thanks [/edit]