Wilsee

Thank you very much for your kind help, I have found it to be of much benefit, Just one more thing, could I have a brief breakdown of how that code actually works, I think I know, I just would like some clarification. Many thanks!

Good day, I am currently trying to build a mock PHP Website upon where a user can query for an insurance quote ([premium]) by using a form that interacts with a MysQL database. The fields for the database are [age] [grp] [region] and [premium] (premium being what I want to return to the user). Ok so I need a form, the form I have is as followed; <form action="capture_car_input.php" method = "post"> <P> Please fill in the form below with your details to generate your car insurance query; </P> <P class="P2"> Your insurance group - </P> <input type=Radio name=insuregroup> 1 <input type=Radio name=insuregroup> 2 <input type=Radio name=insuregroup> 3 <input type=Radio name=insuregroup> 4 <input type=Radio name=insuregroup> 5 <input type=Radio name=insuregroup> 6 <P class="P2"> Your age (17-70 Only) - </P> <input type=text name=custage size ="2" maxlength = "2"> <P class="P2"> Your region of residence - </P> <input type=Radio name=custregion> South East <input type=Radio name=custregion> South West <input type=Radio name=custregion> Midlands <input type=Radio name=custregion> North <input type=Radio name=custregion> Wales <input type=Radio name=custregion> Scotland <P> When you are ready please proceed and click the submit button below. </P> <input type="submit" VALUE = "Submit your query"> <input type= "reset" VALUE = "Resets your details"> </form> Nothing wrong with that I may add? So then I tried to create another PHP file that captures the form's data <?php //Connect to MySql This is the code I have so far: $server=mysql_connect("(server address)","demo",""); if (!$server): print ("error connecting to server"); exit; endif; //Open cp1082 database mysql_select_db("cp1082",$server); $sql = "SELECT grp, age, region, premium FROM insurance"; $result = mysql_query($sql); if (mysql_error()): print "There has been an error<BR>".mysql_error(); exit; endif; $row=mysql_fetch_array($result); print $row ["grp"];<"br"> print $row ["age"];<"br"> print $row ["region"];<"br"> print "Your premium is" . $row["premium"]; mysql_close($server); ?> So I gather inside this code I need While and If statements to return the correct insurance quote to the user, for example if a user inputted group 2, age 18, and region Wales or enter data for just one field as the form parameters it would return so many £'s or $'s for the insurance quote / premium. But how do I go about implementing this code, I've tried using various if's and when statements but with no success, what if a user only entered his/her age how do I go about implementing code for this so only the premiums for age 18 for example would be retured? Help greatly appreciated, cheers, W.