dthomas31uk Posted October 26, 2008 Share Posted October 26, 2008 I have a web page that I have created that allows a user to select info from database http://www.gostoke.eu/eu_home.php When the submit button is pressed the results are shown on the same page, but what I want to do is to display the results on a new page. Can anyone help??? Here is the code that handles the form information <?php include 'eu_home.php'; ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> <title>Demo Multiple drop down list box from plus2net</title> </head> <body> <?php $pickup=$_POST['pickup']; $cat=$_POST['cat']; $subcat=$_POST['subcat']; $load=$_POST['load']; $appointment_date = sprintf('%02d-%02d-%04d',$_POST['day'], $_POST['month'], $_POST['year']); //mm-dd-yyyy echo "Date required $appointment_date"."<BR>"; echo "Picking up from $pickup" ."<BR>"; echo "Going to $subcat"."<BR>"; $result = mysql_query("SELECT full_price, half_price, luton FROM eu_place WHERE city = '" . mysql_real_escape_string($subcat) . "'") or die(mysql_error()); $row = mysql_fetch_array($result); switch($load) { case 'full_load' : $total = $row['full_price']; break; case 'half_load' : $total = $row['half_price']; break; case 'luton': $total = $row['luton']; break; } $rResult = mysql_query("SELECT price FROM uk_place WHERE city = '" . mysql_real_escape_string($pickup) . "'") or die(mysql_error()); while ($aRow = mysql_fetch_array($rResult, MYSQL_ASSOC)) { $iPrice = $aRow['price']; echo "£"; echo ($iPrice) + ($total)."<BR>"; } echo "<form method=post name=f2 action='confirm.php'>"; echo 'To confirm the above details, please enter your details below and click submit' . "<br />"; echo 'Full Name' . "<input type='text' name='name' id='name'/>" . "<br />"; echo 'Telephone Number' . "<input type='text' name='telephone' id='telephone'/>" . "<br />"; echo 'Email' . "<input type='text' name='email' id='email'/>" . "<br />"; echo 'Comments:' ."<textarea id='textareainput' name='textareainput' rows='10' cols='22'></textarea>"; echo "<input type='hidden' name='date' id='date' value='$appointment_date'/>"; echo "<input type='hidden' name='pickUpPoint' id='pickUpPoint' value='$pickup'/>"; echo "<input type='hidden' name='destination' id='destination' value='$subcat'/>"; echo "<input type='hidden' name='loadType' id='loadType' value='$load'/>"; echo "<input type='hidden' name='price' id='price' value='($iPrice) + ($total)'/>" . "<br />"; echo "<input type=submit value=Confirm>"; ?> </div> </body> </html> Link to comment https://forums.phpfreaks.com/topic/130158-help-showing-form-results-on-a-new-page-postposted-sun-oct-26-2008-159-pm/ Share on other sites More sharing options...
GKWelding Posted October 28, 2008 Share Posted October 28, 2008 You should use IF statements to set which code is displayed like so... if(isset($_POST['submit'])){ $pickup=$_POST['pickup']; $cat=$_POST['cat']; $subcat=$_POST['subcat']; $load=$_POST['load']; $appointment_date = sprintf('%02d-%02d-%04d',$_POST['day'], $_POST['month'], $_POST['year']); //mm-dd-yyyy echo "Date required $appointment_date"."<BR>"; echo "Picking up from $pickup" ."<BR>"; echo "Going to $subcat"."<BR>"; $result = mysql_query("SELECT full_price, half_price, luton FROM eu_place WHERE city = '" . mysql_real_escape_string($subcat) . "'") or die(mysql_error()); $row = mysql_fetch_array($result); switch($load) { case 'full_load' : $total = $row['full_price']; break; case 'half_load' : $total = $row['half_price']; break; case 'luton': $total = $row['luton']; break; } $rResult = mysql_query("SELECT price FROM uk_place WHERE city = '" . mysql_real_escape_string($pickup) . "'") or die(mysql_error()); while ($aRow = mysql_fetch_array($rResult, MYSQL_ASSOC)) { $iPrice = $aRow['price']; echo "£"; echo ($iPrice) + ($total)."<BR>"; } echo "<form method=post name=f2 action='confirm.php'>"; echo 'To confirm the above details, please enter your details below and click submit' . "<br />"; echo 'Full Name' . "<input type='text' name='name' id='name'/>" . "<br />"; echo 'Telephone Number' . "<input type='text' name='telephone' id='telephone'/>" . "<br />"; echo 'Email' . "<input type='text' name='email' id='email'/>" . "<br />"; echo 'Comments:' ."<textarea id='textareainput' name='textareainput' rows='10' cols='22'></textarea>"; echo "<input type='hidden' name='date' id='date' value='$appointment_date'/>"; echo "<input type='hidden' name='pickUpPoint' id='pickUpPoint' value='$pickup'/>"; echo "<input type='hidden' name='destination' id='destination' value='$subcat'/>"; echo "<input type='hidden' name='loadType' id='loadType' value='$load'/>"; echo "<input type='hidden' name='price' id='price' value='($iPrice) + ($total)'/>" . "<br />"; echo "<input type=submit value=Confirm>"; }ELSE{ SHOW FORM FOR SELECTING OPTIONS; } Link to comment https://forums.phpfreaks.com/topic/130158-help-showing-form-results-on-a-new-page-postposted-sun-oct-26-2008-159-pm/#findComment-676550 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.