grlayouts
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Posts posted by grlayouts
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this site is for php coding help not writing of scripts, best advice would be to hire a web designer no one here will code it for you.
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Code would help, how are you storing the counter entries?
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Hey,
i have a script in my game that picks up the output rate of a unit which if there is more than one unit it devides, i have 3 teams and the script picks up team 1 and 2, but any units in team 3 the rate shows 0?
IE
HEAVY SOLDIER 100%
but with the 3rd team its always showing 0%
<font face="Verdana" color="#000000" size="2"><b>Output Rate</b></font></td></tr>'; $sql = "SELECT id,name FROM units WHERE class={$row[0]} AND visible=1"; $r2 = mysql_query($sql); while ($row2 = mysql_fetch_row($r2)) { echo '<tr><td align="center">'.$row2[1].'</td><td align="center"><input text="text" size="5" style="'.$boxStyle.'" name="a'.$row2[0].'" value='.getRate($userid,$row2[0]).'>%</td></tr>'; } echo '<tr><td colspan="2" align="center"><input type="submit" value="Set Rates for '.$row[1].'"></td></tr>'; echo '</table></form>'; } echo '</center>';DB's
FOR UNITS. id name toground toair hp class evolvesto style cost visible 6 Mulanix 10 10 200 3 0 1 20 1 5 Numani 5 5 100 3 6 1 10 1
any help would be great it's annoying me and i can't solve it.
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im not sure i could group them together, as the outcomes are different ie won, lost, attacker defender. or would i just do it 4 times?
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hey Guys I have a table in a db that stores attack details for my game below
id attacker win defender time deleted 1 17 1 84 1288576720 0 2 22 1 84 1288576989 0
i want to give a stat like,
total attacks made : 20 (times there id is attacker)
percent successful = 20% (if win =1 its a win =0 its a loss)
and the same with times attacked.
I was thinking along the lines of getting the info using
$stat = mysql_fetch_array(mysql_query("select * from attacklog where attacker='$userid' AND win='1'"));would i just do that for every option in the database?
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The code below works and it serves its prupose but i was wondering if there is an easier way to get the info as making it look nice in html seems very challenging??
echo '<table border="0" width="400" bordercolorlight="#808080" bordercolordark="#808080" style="border-collapse: collapse; font-family: Verdana; font-size: 12px" cellspacing="2" cellpadding="2">'; echo '<tr><td bgcolor="#C0C0C0" align="center"><font face="Verdana" color="#000000" size="2"><b>Unit</b></font></td><td bgcolor="#C0C0C0" align="center"><font face="Verdana" color="#000000" size="2"><b>You Have</b></font></td></tr>'; $sql = "SELECT unit,amount FROM army WHERE boss={$userid} AND amount > 0 ORDER BY unit"; $r = mysql_query($sql); $has = false; while ($row = mysql_fetch_row($r)) { $has = true; echo '<tr><td align="center">'.getUnitName($row[0]).'</td><td align="center">'.number_format($row[1]).'</td></tr>'; } if ($has == false) { echo '<tr><td colspan="2" align="center"><B>None</B></td></tr>'; } echo '</table><BR>'; -
Hi,
i want to have a map of the us in which in my game people can control and it turns there colour using co ordinates, i know how to get the coordinates from below but i dont know how to set the borders for each country or save the info into a table ie
where controlledby
texas player1
washington none
and if it's controlled by the player then changed the colour of the square to there colour ie red?
i hope this makes sense and someone can point me in the right direction
<? $foo_x=$_POST['foo_x']; $foo_y=$_POST['foo_y']; echo "X=$foo_x, Y=$foo_y "; ?> <form action='' method=post> <input type="image" alt=' Finding coordinates of an image' src="map.jpg" name="foo" style=cursor:crosshair;/> </form>
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I want to make a map for my online game, i dont know if anyone has seen the map on the game sepica.com i really want to make something like that but dont know where to start?
does anyone play on that site that can help?
basicly i have a world map image that i want each country to be clickable and place the players team colour in it. they have it as loc?282,272 and i dont know how to achive it?
any suggestions?
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I have an in game chatr script which works great apart from one little thing. when a user uses like ' or ; or = or ) the script doesnt post the message in the database can anyone help?
<?php if ($sc == "") { diver ('Battle Of Species Global Game Talk'); $sql3 = "UPDATE users SET gamechat=0 where id=$userid"; mysql_query($sql3); echo '<center><BR>[<a href="index.php?cmd=chat&sc=chatadd">Add Post</a>]</center><BR><BR>'; $sql = "SELECT id FROM comm2 WHERE parent=0 ORDER BY `time` DESC"; $r = mysql_query($sql); while ($row = mysql_fetch_row($r)) { echo '<table border="0"><tr><td>'; showMessage($row[0], 0,$b); echo '</td></tr></table><BR><BR>'; } }else if ($sc =="chatadd") { diver ('Enter Your Post Below'); echo '<center>'; echo '<form method="post" action="index.php?cmd=chat&sc=chatadd2">'; echo '<input type="hidden" name="inPar" value="'.$inPar.'">'; echo '<textarea rows="10" cols="45" name="inText" style="'.$boxStyle.'"></textarea><BR><BR><input type="submit" value="Post">'; echo '</form></center>'; }else if ($sc == "chatadd2") { diver ("Message Posted"); $inText = clean($inText); $sql = "INSERT INTO comm2 (author,parent,time,post) VALUES ('{$userid}','{$inPar}','".time()."','{$inText}')"; mysql_query($sql); $sql2 = "UPDATE users SET gamechat=1"; mysql_query($sql2); echo '<BR><BR><center>You message has been posted!<BR>[<a href="index.php?cmd=chat">Back to Game Chat</a>]'; } function showMessage($m, $l, $b) { global $prefDate; echo '<table border="0" style="font-famile: Verdana; font-size: 12px" cellpadding="4"><tr>'; for ($i = 0; $i < $l; $i++) { echo '<td> </td>'; } echo '<td>'; $sql = "SELECT `post`,`author`,`id`,`time` FROM `comm2` WHERE `id`='{$m}'"; $r = mysql_query($sql); $row = mysql_fetch_row($r); $row[0] = '<B>'.getvalue($row[1],'handle').'</B><BR>'. $row[0]; $row[0] .= '<BR>[<a href="index.php?cmd=chat&sc=chatadd&inPar='.$row[2].'">Reply</a>] [Posted: '. date($prefDate,$row[3]) .']'; echo str_replace(' ','<BR>',$row[0]); echo '</td></tr></table>'; $sql = "SELECT id FROM comm2 WHERE parent={$m} ORDER BY time DESC"; $r = mysql_query($sql); while ($row = mysql_fetch_row($r)) { showMessage($row[0], $l + 1, $b); } } -
yeah i have have but i added a column for team1 and team2.. the glossery should find the word realting to the team and display it but i dont think the databse layout is right.
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function ret11($sql) { $r = mysql_query($sql) or die (mysql_error()); if (mysql_num_rows($r) == 0) { return 0; } $row = mysql_fetch_row($r); return $row[0]; } -
have you tried making a copy to the folder requested ie www.mysite.com/example it may not pick up the htaccess from the main directory
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what parts of the code would i need to provide for it to make sense?
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do you have a code we could look at?
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have you moved the htaccess script file into the folder /example or only in the main directory?
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can anyone help me i have a mysql database for the code
<?php function gloss($word, $team) { $sql = "SELECT `team{$team}` FROM glossary WHERE word='{$word}'"; return ret11($sql); } echo gloss("title", $teamid) .', Welcome to your <B>' .gloss("barracks",$teamid) . '</B>!'; ?>but they way i've made the table in the database it's not working. I think this is more sql help but i can provide the table i have made if it helps? can someone help me out with the layout or point me in the right direction?
[/code]
EDIT
DATaBASE CODE
-- -- Table structure for table `glossary` -- CREATE TABLE IF NOT EXISTS `glossary` ( `title` text NOT NULL, `barracks` text NOT NULL, `word` text NOT NULL, `team` int(11) NOT NULL, ) ENGINE=MyISAM DEFAULT CHARSET=utf8;
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yeah
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yes im sure it's on. i changed the php.ini file
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but even with the globals bring on i should get some sort of reponse from the form?
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Hi,
I have created a login that once worked on another server and i have changed the register globals on in the php.ini.. when I hit login the page does nothing..
the page is located at www.ycreate.com/sos to see what i mean the login code is below..
any idea's?
LOGIN FORM
<form method="post" action="index.php?cmd=login"> <font size="3"><B>PLAYERS LOG IN BELOW</B><br /> <br /> <BR> </font><table border="0" style="font-family = Verdana; font-size = 12px"> <tr><td><font size="1">Username:</font></td><td><font size="1"><input type="text" size="15" name="inUser" style="<?php echo $boxStyle; ?>"></font></td></tr> <tr><td><font size="1">Password:</font></td><td><font size="1"><input type="password" size="15" name="inPass" style="<?php echo $boxStyle; ?>"></font></td></tr> <tr><td colspan="2" align="center"><font size="1"><input type="submit" value="Login"></font></td></tr> </table></form>
LOGIN SCRIPT
(DB CONNECT) <? if ($cmd == "login") { setcookie('sosuser' , $inUser); setcookie('sospass' , $inPass); if ($_COOKIE['sossess'] == "") { $sql = "SELECT val FROM tickers WHERE id=1"; $r = mysql_query($sql); $row = mysql_fetch_row($r); $next = $row[0]; setcookie('sossess', $next, time()+60*60*24*365); $sql = "UPDATE tickers SET val=val+1 WHERE id=1"; mysql_query($sql); } header('location: index.php?cmd=home'); }else if ($cmd == "logout") { setcookie('sosuser' , ' '); setcookie('sospass' , ' '); header('location: index.php'); } if ($cmd == '') { $cmd = 'default'; } $userid = -1; if (in_array($cmd, $iPages)) { $sql = "SELECT id,team FROM users WHERE username='{$_COOKIE['sosuser']}' AND password='{$_COOKIE['sospass']}'"; $r = mysql_query($sql); if (mysql_num_rows($r) == 0) { $cmd = "badlogin"; }else { $row = mysql_fetch_row($r); $userid = $row[0]; $teamid = $row[1]; $sql = "UPDATE users SET lastaction=".time()." WHERE id={$userid}"; mysql_query($sql); } } include ("serv.php"); if (!in_array($cmd, $noHeader)) { echo 'LOGGED IN' {; ?> -
thank you so much.. =)
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Okay I have a login and every time someone tries to use it they get the error
"Please fill out all fields."
the code i have is
<?php print"<form method=post action=login.php> <td width=100% align=center> <center><table > <tr align=center><td> Username </tD><td> Password </td><tr align=center> <tD> <input type=text name=user size=17 class=dashb> </tD><tD> <input type=password name=pass size=17 class=dashb> </tD></tr> </table> </td></tR><tR> <td> <center> <input type=submit value=Login>"; ?>
and for the login.php
<?php if (!$user || !$pass) { include("headernologin.php"); print "Please fill out all fields."; exit; } include("headernologin.php"); $logres = mysql_num_rows(mysql_query("select * from players where user='$user' and pass='$pass'")); $stat = mysql_fetch_array(mysql_query("select * from players where user='$user' and pass='$pass'")); if ($logres <= 0) { print "Login failed. If you have not already, please signup. Otherwise, check your spelling and login again."; exit; } else { session_register("user"); session_register("pass"); } $ip = " $_SERVER[REMOTE_ADDR] "; mysql_query("update players set ip='$ip' where id=$stat[id]"); print"<br><center>You are being logged at IP $ip"; ?>the header file only have the database connect.
any idea's why it keeps saying that?
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okay i have updated the code to the new standards and codded out the session checks now im getting the error fill in both fields.
can someone suggest a simpler way to login than what i am doing?
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and sorry header1 is
<? if (!session_is_registered("user") || !session_is_registered("pass")) { print "<center><font color=silver>Your session is over you will need to log back in....<a href=index.php>CLICK HERE</a>."; exit; } $stat = mysql_fetch_array(mysql_query("select * from players where user='$user' and pass='$pass'")); if (empty ($stat[id])) { print "Invalid login."; exit; }
stupid question from a newbie
in PHP Coding Help
Posted
$client = mysql_fetch_array(mysql_query("select * from clients'")); print"$client[name]";That would display if you had a field called name in the clients table. edit as appropriate.