Re: Resource id #2

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They don't do anything, but my point was, he said that what he pulled
from the DB needed to be put into an array, and I was pointing out, it
already was.

On Wed, 2002-11-27 at 12:23, Mark wrote:
> But what do all those $row['fieldname'} rows do? Call me ignorant
> (you wouldn't be the first), but a statement that simply has a
> variable name doesn't DO anything. Should these have echos in front
> of them?
> 
> --- Adam Voigt <adam@cryptocomm.com> wrote:
> > Umm, he is putting them into an array, I quote:
> > 
> > while ($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'];
> > > > 
> > > > }
> > 
> > See the while condition?
> > 
> > On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
> > > You need to put your $result into an array. you can use:
> > > 
> > > $result_array = mysql_fetch_array($result);
> > > 
> > > then, if you know the field names in the array, print them like
> > so:
> > > 
> > > echo $result_array["field1"];
> > > echo $result_array["field2"];
> > > 
> > > or if you dont know their names you can refer to their position
> > numbers
> > > starting from 0 e.g.
> > > 
> > > echo $result_array[0];
> > > echo $result_array[1];
> > > 
> > > using the position numbers you could put together a quick script
> > to
> > > crawl through the array and print all the fields with a few lines
> > of
> > > code.
> > > 
> > > On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
> > > > 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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> > > > 
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> > > 
> > -- 
> > Adam Voigt (adam@cryptocomm.com)
> > The Cryptocomm Group
> > My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc
> > 
> 
> > ATTACHMENT part 2 application/pgp-signature name=signature.asc
> 
> 
> 
> =====
> Mark Weinstock
> mark_weinstock@yahoo.com
> ***************************************
> You can't demand something as a "right" unless you are willing to fight to death to defend everyone else's right to the same thing.
> -Stolen from the now-defunct Randy's Random mailing list.
> ***************************************
> 
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-- 
Adam Voigt (adam@cryptocomm.com)
The Cryptocomm Group
My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc

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