BillyBoB

Anyone Here know whats wrong?

I have a login form that works in Firefox but it doesnt work in IE. If somebody could help me that would be great! <?php ob_start(); include("config.php"); if(!$logged) { if(!$_POST['login']) { ?> <table cellspacing="0" cellpadding="0" border="0" width="124"> <tr> <td align="left"> <form method="POST"> <table cellspacing="0" cellpadding="0" border="0" width="124"> <tr> <td> Username: </td> </tr> <tr> <td style="background: url('images/textbox.gif') no-repeat;" height="16"> <input type="text" height="16" name="username" class="login"> </td> </tr> <tr> <td> Password: </td> </tr> <tr> <td style="background: url('images/textbox.gif') no-repeat;" height="16"> <input type="password" height="16" name="password" class="login"><br/><br/> </td> </tr> <tr> <td> <div align="center"> <input type="image" src="images/loginbutton.gif" name="login" value="Login"> </div> </td> </tr> </table> </form> </td> </tr> </table> <br/> <?php }else{ $username = $_POST['username']; $password = md5($_POST['password']); $info = mysql_query("SELECT * FROM members WHERE username = '$username'") or die(mysql_error()); $data = mysql_fetch_array($info); if($data[password]!=$password) { echo("Wrong Password! Try again. <meta http-equiv=\"Refresh\" content=\"2;\"/>"); }else{ $query = mysql_query("SELECT * FROM members WHERE username = '$username'") or die(mysql_error()); $user = mysql_fetch_array($query); setcookie("id", $user[id],time()+(60*60*24*1), "/", ""); setcookie("pass", $user[password],time()+(60*60*24*1), "/", ""); $newtime = time(); $time = gmdate("YmdHis", $newtime); $ip = $_SERVER['REMOTE_ADDR']; $lquery = mysql_query("UPDATE members SET ip = '$ip' , lastlog = '$time' WHERE username = '$username' ") or die(mysql_error()); echo ("<meta http-equiv=\"Refresh\" content=\"3; \"/>Login Successful"); } } }else{ echo(" CONSOLE "); } ?> Thanks in advance EDIT: This site is still in beta so, I can only allow you to the login page. http://animeblogsite.com/beta/ htaccess username and password: Username: BetaTester Password: happyhelper234 Login Page Username and Password: Username: Beta Password: happyhelper234

try <?php include("http://yoursite.com/content/rss/weather/shg_weather_tmp.php"); ?> somthing along those lines becuase i think that all the ../ are messing up the server

well i think it might be a very simple error replace: $query = "INSERT INTO order_data (customer_number, comp_brand, comp_model, comp_serial_number, comp_processor, comp_hdd_size, comp_ram, comp_problem, repair_price, date_of_order) values( '$customer_number', '$comp_brand', '$comp_model', '$comp_serial_number', '$comp_processor', '$comp_hdd_size', '$comp_ram', '$comp_problem', '$repair_price', '$date_of_order', '$order_status',)"; With: $query = "INSERT INTO order_data (customer_number, comp_brand, comp_model, comp_serial_number, comp_processor, comp_hdd_size, comp_ram, comp_problem, repair_price, date_of_order) values( '$customer_number', '$comp_brand', '$comp_model', '$comp_serial_number', '$comp_processor', '$comp_hdd_size', '$comp_ram', '$comp_problem', '$repair_price', '$date_of_order', '$order_status')"; there is an extra comma in yours i was reviewing the coding over again and it seems that u have no order status table inputed u have the value but no table to put it in.... your code stops at date_of_order there should be a order_status table

maybe i could help i have had this error b4 try this code and tell my wat u get out of it exactly: <?php $news = mysql_query("SELECT * FROM news WHERE id= 1"); $newsarr = mysql_fetch_array($news); echo $newsarr[date]; ?> replace 1 with a valid id!!! replace date with the correct value like whatever the the date is stored as!!! Then reply if u get the same thing as i did then i will tell u from there

How would i take string out of a database such as 19" Samsung and put a \ infront of the " ??

nvm i got it fixed up now ty for helping me dude

idk it still isnt working :( the link is http://fbi.ds-studios.net

i am build a cs based website for a clan but i was wondering how would i do this [code] while($rinfo=mysql_fetch_array($rsql)&&$i!=0) { if($i==4) { echo(" </tr> <tr> "); } if($i==8) { echo(" <tr> "); } echo(" <td width=\"45px\" height=\"64px\"> <center> <a href=\"members.php?id=$rinfo[membersid]\"><img src=\"roster/$rinfo[pic]\" width=\"35px\" height=\"40px\" title=\"$rinfo[name]\"/></a> </center> </td> "); if($i==1) { echo(" </tr> "); } $i--; } [/code] $i is declared in the earlier script but it shows how many ppl are on the roster and im trying to do somin like if $i != 0 then keep going but if it equals 0 then stop but as of now im getting nothing showing up

it took me forever to make that up too .... i had to do a lot of research because i don't do this kind of stuff

but wont he get the first one not the second ? because he metioned <img src> and </img> and what if he does like setting the text like <font> or somin? i was thinking soming like [code] <?php #start by putting your info into a array so we can search $array = array($article); if(in_array("<img",$array)) {   #cut it down to 100 maybe?   $pos = strpos($article,"</img>");   $article2 = substr($article,0,$pos);   $article = substr($article,$pos,100);   $article = $article . $article2; }else{   $article = substr($article,0,200); } ?> [/code]

ok guys this is good i figured it out our time was saving with "-" " " and ":" so we just replaced them with "" so it fixed

This worked before my host got switched idk if its me or not but i have my file as timestamp(14) http://dreamshowstudios.net the news on there ^^ is where heres coding: [code] <?php $datefromdb = $row['whenposted']; $year = substr($datefromdb,0,4); $mon  = substr($datefromdb,4,2); $day  = substr($datefromdb,6,2); $hour = substr($datefromdb,8,2); $hour = $hour - 6; $min  = substr($datefromdb,10,2); $sec  = substr($datefromdb,12,2); $time = date("l F dS, Y h:i A",mktime($hour,$min,$sec,$mon,$day,$year)); echo '<center>' . $row['title'] . '<br />by ' . $row['poster'] . ' on ' . $time . ' CST<br />'; echo $row['textmessage'] . '<br /><hr /></center>'; ?> [/code]

there is a way but i dont know if its against the rules so if so plz moderator tell me so i can change hxxp://milw0rm.com its a cracker site

im trying for somin like it searches a array then it tells me if there is a site from searching for http:// or .com or .net etc... then adding

http://dreamshowstudios.net thats wat im getting on home page not in shout idk why shout works ??

Page loaded in 0.00005 secs thats wat it said

this is wat happens to me ... :(

no problem dude

I have a news database and i was wondering if i could somehow make it where when if fetches the array if it could search it and put it in a <a href=$link >$link</a> and make it where if it allready has the tag <a href= > and </a> then do nothing this is probably a longer anwser but if som1 could plz help that would be much appreciated

try this maybe ? [code] <?php $array = array(".jpg",".jpeg",".gif",".png",".tiff"); if(in_array($file,$array)) ?> [/code]

nope i tryed the mentioned way and it worked for me here check it out http://dreamshowstudios.net its at bottom

yea i know thats why you shouldnt use them both in a file... i mean why?

hope this helps [code] <?php $query = mysql_query("SELECT * FROM tablename"); $array = mysql_fetch_array($query); if (in_array("mp3", $array) ) {   echo("this is a mp3"); } if (in_array("jpg", $array) ) {   echo("this is a jpg"); } ?> [/code]

try  [code] $results = mysql_query("SELET * FROM table"); $array = mysql_fetch_array($results); [/code]