Monk3h Posted April 8, 2008 Share Posted April 8, 2008 How would i get something in my database to be listed on a page? For example, i want every development name in the development table with an ID Less than the current development ID to be listed, seperated by a ,. So if the current development being developed was number 5 i woulod want this to be listed. DEV 1 Name, DEV 2 Name, DEV 3 Name, DEV 4 Name. How would i do this? :/ Quote Link to comment Share on other sites More sharing options...
p2grace Posted April 8, 2008 Share Posted April 8, 2008 Something like this: <?php // assuming there's a variable called $id that has the current id assigned to it $query = "SELECT `id` FROM `table` WHERE `id` < '$id'"; $run = mysql_query($query); if($run){ while($arr = mysql_fetch_assoc($run)){ extract($arr); echo "id = $id<br />"; } }else{ die("Unable to connect to database."); } ?> You should just be able to manipulate the code to your scenario. Is this what you are attempting to do? Quote Link to comment Share on other sites More sharing options...
quickstopman Posted April 8, 2008 Share Posted April 8, 2008 write a simple script like this: <?php include("YOUR CONFIG FILE HERE"); $d_id = $_GET['d_id']; $sql = mysql_query("SELECT * FROM `INSERT YOUR TABLE NAME HERE` WHERE `did` = '. $d_id .'") or die(mysql_error()); $product = mysql_fetch_array($sql); echo $product['name']; ?> please put your own information in in, and also to write any detail from the row, you can use echo $product[' and then any field from that section, then close it off with']; Your Welcome! -Zack PS: I was once like you so don't worry about it. PSS: The guy above me beat me to it =/ Quote Link to comment Share on other sites More sharing options...
Monk3h Posted April 8, 2008 Author Share Posted April 8, 2008 Thank you both VERY much but i cant seem to get it to work.. <?php $title = "Developments"; include("header.php"); ?> <?php $mytribe = mysql_fetch_array(mysql_query("select * from tribes where id=$stat[tribe]")); $devcount = mysql_fetch_array(mysql_query("select * from dev where id=$mytribe[devcount]+1")); Print "Here is where you keep your Tribe up to date with the latest Defensive and Offensive Technologies! Remember.. Nothing is Impossible when you have a Tank!<br><Br><br><br>"; Print "Your Tribe Currently have <b>$mytribe[devcount]</b> Developments Completed.<br><br>"; if ($mytribe[devtime] > 0){ Print "Your Tribe is currently Developing <b>$devcount[name],</b> this development will be completed in Approximately <b>$mytribe[devtime]</b> Hours.<br><br>"; } Print "<b><u>Avalible Developments</u></b><br><br>"; Print" <table border='0' bordercolor='' width='95%' bgcolor=''> <tr> <td><b>Development Name</b></td> <td><b>Time Required</b></td> <td><b>Cost</b></td> <td><b>Action</b></td> </tr> <tr> <td>$devcount[name]</td> <td>$devcount[time]</td> <td>$devcount[cost]</td> <td><a href=development.php?action=start>Start</a> </tr> </table>"; if ($action == start) { if ($stat[id] == !$mytribe[owner]) { Print "You are not the Tribe Owner, you dont have the authority to start a Development"; } elseif ($mytribe[devtime] > 0){ Print "<br><br><br>You are Currently Developing <b>$devcount[name]</b>, you cannot do 2 Developments at once!"; }else{ if ($mytribe[credits] < $devcount[cost]){ Print "<br><br><br>Your Tribe does not have enough Credits for that Development!"; }else{ mysql_query("update tribes set credits=credits-$devcount[cost] where id=$mytribe[id]"); mysql_query("update tribes set devtime=devtime+$devcount[time] where id=$mytribe[id]"); Print "<br><br><br> Your Tribe has started <b>$devcount[name]</b><br>This Development will take Approximately <b>$devcount[time]</b> Hours<br>This development cost your Tribe <b>$devcount[cost]</b> Credits"; }}} ?> <?php include("footer.php"); ?> Thats my Script.. How would i put that code into this Script? one Give me the uneeded $end error the other Says Wrong Syntax. :/ Quote Link to comment Share on other sites More sharing options...
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