pranav_kavi

Read this snippet frm http://www.php.net

i suppose u wud b ving a memberID in ur 'Members' table and committeeID in ur 'Committee' table. Have a join table like 'committee_members' that has 2 columns-memberID & committeeID .so this table wud store the mapping information. hope tis helps..

but that depends on the datatype of the column-playerID. If playerID is of numeric type in db,then query wud b SELECT * FROM players WHERE playerID IN (1, 2) If playerID is of varchar type in db,then query needs 2 b SELECT * FROM players WHERE playerID IN ('1', '2')

do u want ur query to be of the form SELECT * FROM players WHERE playerID IN (-1, -2) ? did u mean hyphens or single quotes arnd i.e '1','2' ?

chk tis out... http://www.dbforums.com/t1044828.html

try, $players = $_POST['matchTeam']; $playerGames = "SELECT * FROM players WHERE playerID IN (". implode(", ",$players). ")"; echo "query: ".$playerGames; $playerGames_query = mysql_query($playerGames) or die(mysql_error()); ......(rest of the code)...

Print out your query and check if the query is valid. In the code u ve givn,r u using a do-while loop or just a while loop?

r u using a \n sumwher in ur code or file that comes widin the limit of the php tags?

so twas it..instead of a normal numeric comparision,the string comparision had failed the condition.

wat s the databse type for column 'Category'?

have ur chkd ur column name-'Category' i.e whethr it s 'category' or 'Category'??? i think u can includ the code for db query.

try readin tis article.. http://www.thescripts.com/forum/thread2524.html

once d user has logged in,set his username in session.in the following pages u can access the session variable to find out the current logged in user.

initally try tis, mysql_connect("localhost","root","suigion") or die("Unable to connect to database..."); mysql_select_db("blogzone") or die("Unable to select database...");

what is the error ur getting??

u can do, $que="select col1 from table where col2= ( select MAX(col2) from table ); mysql_query($que);

i think u can do tat....

i suppose u can use str_split() method that can split the string into an array of characters.then echo the array contents till a limit of 150.

These links ll help u, http://www.phpfreaks.com/tutorial_cat/19/Object-Oriented-Programming.php OR http://www.codewalkers.com/c/a/Programming-Basics/Beginning-Object-Oriented-Programming-in-PHP/

try this, while($data = mysql_fetch_assoc($result2)) { $oid = $data['oid']; echo $oid; }

Instead of $name,ve u just tried putting 'Joe Smith' directly (just 2 check..)

Based on, Double quotes are not properly closed here properly,

This may help u, 1)Open a temporary file in write mode. 2)Read each line of the file using fgets. 3)'explode' each line using '|' as delimiter 4)Only if the first token is not equal to 45,write that line to temporary file 5)Once all the lines have been checked,unlink the original file & rename the temporary file to name of the original file

query to create the table is as follows frm backend, create table `test`.`colours` ( `name` text NULL , `favoriteColor` text NULL , `county` text NULL )

i cant figure out y it is not workin 4 u... but ur code s workin for me well..xcept that the value in 'favoritecolor' is not being displayed as the column names do not match. ve u chkd frm the backend...r d records being inserted properly???