[PHP-DB] Resource id #2

2002-11-26 Thread The Cossins Fam 
Hello.

I am using MySQL as a database for a departmental library.  I have written
a quick search script, but keep getting resource id #2 as a result to my
search.  I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything.  However, I've only started programming, much less working with
PHP, so perhaps someone can help me out.  Here's my code:

?

$quickSearch = mcse;

$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;

$connection = mysql_connect(localhost, root) or die(Couldn't connect
to the library database.);

$db_select = mysql_select_db(library, $connection) or die(Couldn't
select the library database.);

$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
   LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
   LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
   LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
   LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
   LEFT JOIN $table7 ON Books.CD = CD.CD_ID
   WHERE Books.Title LIKE \%'$quickSearch'%\
   OR Books.Author LIKE \%'$quickSearch'%\
   OR Books.ISBN LIKE \%'$quickSearch'%\
   OR BookList.dbase LIKE \%'$quickSearch'%\
   OR BookList.dbase_user LIKE \%'$quickSearch'%\
   OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
   OR Category.Category LIKE \%'$quickSearch'%\
   OR Category.Sub_category LIKE \%'$quickSearch'%\
   OR Publisher.Publisher LIKE \%'$quickSearch'%\;

$result = mysql_query($search, $connection) or die(Couldn't search the
library.);

while ($row = mysql_fetch_array($result)) {
   $row['Books.Title'];
   $row['Books.Author'];
   $row['Books.ISBN'];
   $row['BookList.dbase'];
   $row['BookList.dbase_user'];
   $row['BoxSet.BoxSet'];
   $row['Category.Category'];
   $row['Category.Sub_category'];
   $row['Publisher.Publisher'];
   $row['AuthUsers.email'];

}


?

I then have some HTML to display the result of the search.  I don't
receive any error messages - I just see an empty table from the HTML
code I wrote.  I added an echo of the $result to find the resouce id
#2.

Thanks for any help you can provide.

--joel







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[PHP-DB] User Defined Function Problem

2003-03-17 Thread The Cossins Fam 
Does anyone see anything I might be missing as to why this won't work?

From the file calling the function:
?php

$round = Round 1;
$region = East;
include check.php;
isAuthorized($_COOKIE['username'], $_COOKIE['password']);
include checkEntry.php;
check4Entry($username, $round, $region);
...

From checkEntry.php:
?php

function check4Entry($name, $reg, $rnd) {

   $connection = mysql_connect(localhost, root) or 
die(mysql_error());

   $db = mysql_select_db(marchmadness, $connection) or 
die(mysql_error())
;

   $checking = SELECT * FROM pick WHERE userID = \$uname\ AND  round 
= \$rnd\ AND region = \$reg\;

   $result = mysql_query($checking, $connection) or die(mysql_error());

   $numRows = mysql_num_rows($result) or die(mysql_error());

   if ($numRows  0) {
   print(htmlheadlink rel=\stylesheet\ 
type=\text/css\ href=\bbb.css\/headbodycentertable 
class=\outTbl\trth align=\center\ class=\outTh\ Oops! There are 
already selections made for $uname in $reg - $rnd. /th/tr 
trtd$uname, if you'd like to modify your selections, please click a 
href=\modifyEntry.php\here/a./td/tr/table/center/body/html);

   exit;

   }

}

?

==

When I submit the form, I get a blank screen as a result.  There aren't any 
errors ouput to the screen.  I've compared this with another function I've 
created, and there doesn't appear to be anything different.  The latter 
works.

BTW, I bet you'll never guess what this is for...;-)

Thanks for your help!

--Joel

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[PHP-DB] Reading File Contents

2003-08-26 Thread The Cossins Fam 
Hello.

I have what should be a simple fix for the more advanced programmers out 
there.  I'm having problems with the display format of a text file.  I've 
been reading several resources, and I'm just not having much success with 
it.

How would you be able to display the following file named test.txt:

1. Test
2. Test
3. Test
...without it displaying like:

1. Test 2. Test 3. Test?

I've tried the nl2br function, as well as adding \n to the end of each 
line and using fgets to replace \n with br.  However, nothing seems to 
be working.  I'm probably missing something very simple, but if someone 
wouldn't mind helping a rookie out, I'd appreciate it.

--Joel

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