twilitegxa Posted September 15, 2009 Share Posted September 15, 2009 Can anyone help me? I'm wanting to have a page where people could train their characters (this is the beginning of an RPG game). What I want to do is if they press the train button, a query is run for whatever option they have selected. Say they chose test1 as the option and pressed train. I want to update the table: energy - 2 experience + 50 The first thing I need to know how to do is have this script run on the same page only when they press the train button (it's a submit button actually; does that matter? or does it need to be a regular button?). The second thing is how to write the script to where it runs the script I want when the button is pressed. Here is the page I have so far: <?php include("connect_db.php"); $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; } $display_block = " <h3>Train Your Character To Level Up</h3> <table> <tr> <td valign=top> <form> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> </form> </td> <td valign=top>Player: $identity<br /> Level: $level<br /> Energy: $energy<br> Current Experience: $experience<br /> Experience To Next Level:<br /> </td> </tr> </table>"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Sailor Moon RPG - Training Board</title> <style type="text/css" media="screen"> /*<![CDATA[*/ @import url(global.css); /*]]>*/ </style> </head> <body> <!-- HEADER --> <h1 class="logo">Sailor Moon RPG</h1> <!-- /HEADER --> <?php include("topnav.php"); ?> <div id="main"> <?php include("includes/log.php"); ?> <?php include("mainnav.php"); ?> <h1>Sailor Moon RPG - Training Board</h1> <?php print $display_block; ?> </div> <?php include("bottomnav.php"); ?><!-- FOOTER --> <!-- FOOTER --> <div id="footer_wrapper"> <div id="footer"> <p>Sailor Moon and all characters are<br /> trademarks of Naoko Takeuchi.</p> <p>Copyright © 2009 Liz Kula. All rights reserved.<br /> A product of <a href="#" target="_blank">Web Designs By Liz</a> systems.</p> <div id="foot-nav"> <ul> <li><a href="http://validator.w3.org/check?uri=http://webdesignsbyliz.com/digital/index.php" target="_blank"><img src="http://www.w3.org/Icons/valid-xhtml10-blue" alt="Valid XHTML 1.0 Transitional" height="31" width="88" /></a></li> <li><a href="http://jigsaw.w3.org/css-validator/validator?uri=http://webdesignsbyliz.com/digital/global.css" target="_blank"><img class="c2" src="http://jigsaw.w3.org/css-validator/images/vcss-blue" alt="Valid CSS!" /></a></li> </ul> </div> </div> </div> <!-- /FOOTER --> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/ Share on other sites More sharing options...
seventheyejosh Posted September 15, 2009 Share Posted September 15, 2009 use AJAX Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918687 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Can this not be done with PHP? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918689 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 For what you wish to do is change text on a page that has already been sent from the server. PHP is a server side scripting language so can't do stuff once the process has left the server. So what you need to do is to use Javascript (AJAX) to make a request and then display the result. Using JQuery's ajax functionality will make the task extremely easy. Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918692 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Okay, well if I run it on a second page instead, how do I write the if statement? I tried this: training_board.php: <?php include("connect_db.php"); $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; } $display_block = " <h3>Train Your Character To Level Up</h3> <table> <tr> <td valign=top> <form action=train.php> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> </form> </td> <td valign=top>Player: $identity<br /> Level: $level<br /> Energy: $energy<br> Current Experience: $experience<br /> Experience To Next Level:<br /> </td> </tr> </table>"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Sailor Moon RPG - Training Board</title> <style type="text/css" media="screen"> /*<![CDATA[*/ @import url(global.css); /*]]>*/ </style> </head> <body> <!-- HEADER --> <h1 class="logo">Sailor Moon RPG</h1> <!-- /HEADER --> <?php include("topnav.php"); ?> <div id="main"> <?php include("includes/log.php"); ?> <?php include("mainnav.php"); ?> <h1>Sailor Moon RPG - Training Board</h1> <?php print $display_block; ?> </div> <?php include("bottomnav.php"); ?><!-- FOOTER --> <!-- FOOTER --> <div id="footer_wrapper"> <div id="footer"> <p>Sailor Moon and all characters are<br /> trademarks of Naoko Takeuchi.</p> <p>Copyright © 2009 Liz Kula. All rights reserved.<br /> A product of <a href="#" target="_blank">Web Designs By Liz</a> systems.</p> <div id="foot-nav"> <ul> <li><a href="http://validator.w3.org/check?uri=http://webdesignsbyliz.com/digital/index.php" target="_blank"><img src="http://www.w3.org/Icons/valid-xhtml10-blue" alt="Valid XHTML 1.0 Transitional" height="31" width="88" /></a></li> <li><a href="http://jigsaw.w3.org/css-validator/validator?uri=http://webdesignsbyliz.com/digital/global.css" target="_blank"><img class="c2" src="http://jigsaw.w3.org/css-validator/images/vcss-blue" alt="Valid CSS!" /></a></li> </ul> </div> </div> </div> <!-- /FOOTER --> </body> </html> train.php: <?php include("connect_db.php"); $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; } if ($_POST['train'] == 'test1') { $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; $update_energy = ($player_info['energy'] - 2); } $accept_scout_username = mysql_query("UPDATE training SET energy ='$update_energy' WHERE identity = '$identity'"); } ?> But I am getting this error: Notice: Undefined index: train in C:\wamp\www\train.php on line 18 What am I doing wrong? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918707 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 It's not an error it's a notice, most of us suppress our errors so we can be sloppy A notice is just to warn you that you have done some sloppy code. Such as call on non existancing variables etc if ((isset($_POST['train'])) && ($_POST['train'] == 'test1')) will get rid of the notice. Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918710 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Thank you, that eliminated the error, but my update is not working. Can you tell me why? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918712 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 Prolly because you are running the update outside the loop. If you are wanting to just decrease the int value of 'energy' you can have mysql do it with one query and save some processing time. UPDATE training SET energy =energy+2 WHERE 1; Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918713 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 I don't understand what you mean. i tried putting the update statement inside of the whole loop, but that didn't work either. How would the statement: UPDATE training SET energy =energy+2 WHERE 1; take two away from the energy field? And how would I change my code? I don't quite understand what you mean. Sorry. Can you help explain it better? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918718 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 <?php include("connect_db.php"); /****** * Code doesn't do anything. ****** $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; } */ if ($_POST['train'] == 'test1') { /******* * This code only updates the last identity in the table training. ******* $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; $update_energy = ($player_info['energy'] - 2); } $accept_scout_username = mysql_query("UPDATE training SET energy ='$update_energy' WHERE identity = '$identity'"); ********** * This query updates all the identities in the table training. **********/ mysql_query("UPDATE training SET energy=energy-2 WHERE 1"); } ?> Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918722 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 If I take out the if statement, the script is working, so the if statement seems to be the problem. What am I doing wrong? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918723 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 The script you sent didn't work either :-( Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918724 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 Are you sure $_POST['train'] == 'test1' ? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-918728 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Well, I choose test1 from the select list named train. Is this the right coding? Here's the code from the form: <form action=train.php> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> </form> Am I writing something wrong? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919036 Share on other sites More sharing options...
backie Posted September 15, 2009 Share Posted September 15, 2009 I am pretty sure the default method for forms is GET. So it would be in the $_GET['train'] and $_REQUEST['train'] but not $_POST['train']. Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919039 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Okay, that did the trick! I didn't realize I was using the wrong method. Thank you for letting me know! Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919047 Share on other sites More sharing options...
mikesta707 Posted September 15, 2009 Share Posted September 15, 2009 Your PHP makes absolutely no sense. You have a worthless loop in there that just gets information from the table for no reason. Then you do the same thing, and for some reason you just update the last entry in the table. This won't work how you expect it to at all. If you want to grab a users info, then you have to set a where clause or something. <?php include("connect_db.php"); //why do you do this... $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; } //the above code makes no sense at all. you are overwriting the variable everytime the loop runs //you will only have the data from the last loop run... if ($_POST['train'] == 'test1') { //again, this makes no sense at all $get_player_info = "select * from training"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; $update_energy = ($player_info['energy'] - 2); }//makes no sense at all //this is going to update the very last entry... not whoever is trying to train. $accept_scout_username = mysql_query("UPDATE training SET energy ='$update_energy' WHERE identity = '$identity'"); } ?> I very strongly suggest that you try a simpler problem before you tackle an RPG. you seem to not know the basics of pulling the right information from a mysql table. To be completely honest, this code will probably run, but won't do anything useful, and to make it do something useful will probably take 3 pages of explanation. here is a php/mysql tutorial Another php/mysql tutorial I suggest you read up on those tutorials, try a simpler project, and then tackle the RPG. It will be much less of a headache Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919056 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 I am working on the where clause now. But for some reason, I am doing the last page wrong. I have a page where they choose which identity to train, then the page that displays the identity with appropriate info, then the third page should update the values, but I have a flaw in my where clause. Could I not use $_GET? Here are the pages I have: choose_train.php: <?php session_start(); if(!isset($_SESSION['loggedIn'])) { header("Location: login.php"); } include ("connect_db.php"); //show scouts characters $get_scouts = "select * from scouts where username = '".$_SESSION['userName']."'"; $get_scouts_res = mysql_query($get_scouts, $conn) or die(mysql_error()); while ($list_scouts = mysql_fetch_array($get_scouts_res)) { $identity = ucwords($list_scouts['identity']); $scout_id = $list_scouts['id']; echo "<ul class=\"character_list\"><li><a href=\"training_board.php?identity=$identity\">$identity</li></ul> "; } //show knights characters $get_knights = "select * from knights where username = '".$_SESSION['userName']."'"; $get_knights_res = mysql_query($get_knights, $conn) or die(mysql_error()); while ($list_knights = mysql_fetch_array($get_knights_res)) { $identity = ucwords($list_knights['identity']); $knight_id = $list_knights['id']; echo "<ul class=\"character_list\"><li><a href=\"showprofile_knights.php?id=$knight_id\">$identity</li></ul> "; } //show fdark_warriors characters $get_fdark_warriors = "select * from fdark_warrior where username = '".$_SESSION['userName']."'"; $get_fdark_warriors_res = mysql_query($get_fdark_warriors, $conn) or die(mysql_error()); while ($list_fdark_warriors = mysql_fetch_array($get_fdark_warriors_res)) { $identity = ucwords($list_fdark_warriors['identity']); $topic_id = $list_fdark_warriors['id']; echo "<ul class=\"character_list\"><li><a href=\"showprofile_fdark_warrior.php?id=$topic_id\">$identity</li></ul> "; } //show mdark_warriors characters $get_mdark_warriors = "select * from mdark_warrior where username = '".$_SESSION['userName']."'"; $get_mdark_warriors_res = mysql_query($get_mdark_warriors, $conn) or die(mysql_error()); while ($list_mdark_warriors = mysql_fetch_array($get_mdark_warriors_res)) { $identity = ucwords($list_mdark_warriors['identity']); $topic_id = $list_mdark_warriors['id']; echo "<ul class=\"character_list\"><li><a href=\"showprofile_mdark_warrior.php?id=$topic_id\">$identity</li></ul> "; } $display_block = ""; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> </body> </html> <?php include("connect_db.php"); $display_block = "<h3>Train Your Character To Level Up</h3> <table> <tr>"; $get_player_info = "select * from training where identity = '$_GET[identity]'"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; $display_block .= " <td valign=top> <form action=train.php> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> </form> </td> <td valign=top>Player: $identity<br /> Level: $level<br /> Energy: $energy<br> Current Experience: $experience<br /> Experience To Next Level:<br /> </td> </tr>"; } $display_block .= "</table>"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Sailor Moon RPG - Training Board</title> <style type="text/css" media="screen"> /*<![CDATA[*/ @import url(global.css); /*]]>*/ </style> </head> <body> <!-- HEADER --> <h1 class="logo">Sailor Moon RPG</h1> <!-- /HEADER --> <?php include("topnav.php"); ?> <div id="main"> <?php include("includes/log.php"); ?> <?php include("mainnav.php"); ?> <h1>Sailor Moon RPG - Training Board</h1> <?php print $display_block; ?> </div> <?php include("bottomnav.php"); ?><!-- FOOTER --> <!-- FOOTER --> <div id="footer_wrapper"> <div id="footer"> <p>Sailor Moon and all characters are<br /> trademarks of Naoko Takeuchi.</p> <p>Copyright © 2009 Liz Kula. All rights reserved.<br /> A product of <a href="#" target="_blank">Web Designs By Liz</a> systems.</p> <div id="foot-nav"> <ul> <li><a href="http://validator.w3.org/check?uri=http://webdesignsbyliz.com/digital/index.php" target="_blank"><img src="http://www.w3.org/Icons/valid-xhtml10-blue" alt="Valid XHTML 1.0 Transitional" height="31" width="88" /></a></li> <li><a href="http://jigsaw.w3.org/css-validator/validator?uri=http://webdesignsbyliz.com/digital/global.css" target="_blank"><img class="c2" src="http://jigsaw.w3.org/css-validator/images/vcss-blue" alt="Valid CSS!" /></a></li> </ul> </div> </div> </div> <!-- /FOOTER --> </body> </html> train.php: <?php session_start(); include("connect_db.php"); if ((isset($_GET['train'])) && ($_GET['train'] == 'test1')) { $display_block = "<h3>Train Your Character To Level Up</h3> <table> <tr>"; $get_player_info = "select * from training WHERE identity = '$identity'"; $get_player_info_res = mysql_query($get_player_info, $conn) or die(mysql_error()); while ($player_info = mysql_fetch_array($get_player_info_res)) { $id = $player_info['id']; $identity = $player_info['identity']; $level = $player_info['level']; $energy = $player_info['energy']; $experience = $player_info['experience']; $update_energy = ($energy - 2); $update_experience = ($experience + 50); $lose_energy = mysql_query("UPDATE training SET energy ='$update_energy' WHERE identity = '$identity'"); $gain_experience = mysql_query("UPDATE training SET experience ='$update_experience' WHERE identity = '$identity'"); $display_block .= " <td valign=top> <form action=train.php> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> </form> </td> <td valign=top>Player: $identity<br /> Level: $level<br /> Energy: $energy<br> Current Experience: $experience<br /> Experience To Next Level:<br /> </td> </tr>"; } $display_block .= "</table>"; } else { print "didn't work :-("; } ?> <html> <head> <title></title> </head> <body> <?php print $display_block; ?> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919129 Share on other sites More sharing options...
mikesta707 Posted September 15, 2009 Share Posted September 15, 2009 ahh ok this code looks much better. one thing $get_player_info = "select * from training where identity = '$_GET[identity]'"; should be $get_player_info = "select * from training where identity = '".$_GET['identity']."'"; you have to surround associative array keys with single quotes. also in train.php I dont see anywhere where $identity is set before this line $get_player_info = "select * from training WHERE identity = '$identity'"; which needs identity to be set for the query to run correctly Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919133 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 That's the problem I'm having. I can't figure out how I need to set $identity on the train.php page to grab the identity from the previous pages. Can you help? Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919143 Share on other sites More sharing options...
mikesta707 Posted September 15, 2009 Share Posted September 15, 2009 create a hidden field with the value of the identity, than using the get value, assign identity. like $display_block .= " <td valign=top> <form action=train.php> <select name=train> <option>test1</option> <option>test2</option> </select> <input type=submit name=submit value=Train> <input type='hidden' name='id' value='$indentity' /> </form> </td> <td valign=top>Player: $identity<br /> Level: $level<br /> Energy: $energy<br> Current Experience: $experience<br /> Experience To Next Level:<br /> </td> </tr>"; Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919144 Share on other sites More sharing options...
twilitegxa Posted September 15, 2009 Author Share Posted September 15, 2009 Thank you so much! That did it! Thanks for pointing out the hidden idea! I wasn't thinking! :-) Quote Link to comment https://forums.phpfreaks.com/topic/174278-solved-update-run-on-same-page-when-form-submited/#findComment-919151 Share on other sites More sharing options...
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