Re: [PHP-DB] Resource id #2
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 $row['fieldname'} rows do? Call me ignorant
> (you wouldn't be the first), but a statement that simply has a
> variable name doesn't DO anything. Should these have echos in front
> of them?
>
> --- Adam Voigt <[EMAIL PROTECTED]> wrote:
> > Umm, he is putting them into an array, I quote:
> >
> > while ($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
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > _
> > > > The new MSN 8: advanced junk
Re: [PHP-DB] Resource id #2
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt <[EMAIL PROTECTED]> wrote:
> Umm, he is putting them into an array, I quote:
>
> while ($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 $result to find the
> "resouce id
> > > #2".
> > >
> > > Thanks for any help you can provide.
> > >
> > > --joel
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> _
> > > The new MSN 8: advanced junk mail protection and 2 months FREE*
>
> > > http://join.msn.com/?page=features/junkmail
> > >
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> --
> Adam Voigt ([EMAIL PROTECTED])
> The Cryptocomm Group
> My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc
>
> ATTACHMENT part 2 application/pgp-signature name=signature.asc
=
Mark Weinstock
[EMAIL PROTECTED]
***
Re: [PHP-DB] Resource id #2
Umm, he is putting them into an array, I quote:
while ($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
> >
> >
> >
> >
> >
> >
> >
> > _
> > The new MSN 8: advanced junk mail protection and 2 months FREE*
> > http://join.msn.com/?page=features/junkmail
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
--
Adam Voigt ([EMAIL PROTECTED])
The Cryptocomm Group
My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc
signature.asc
Description: This is a digitally signed message part
Re: [PHP-DB] Resource id #2
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
>
>
>
>
>
>
>
> _
> The new MSN 8: advanced junk mail protection and 2 months FREE*
> http://join.msn.com/?page=features/junkmail
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
signature.asc
Description: This is a digitally signed message part
Re: [PHP-DB] Resource id #2
On Wednesday 27 November 2002 02:09, 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'%\"";
It's always a good idea to print out your query so you can visually check
whether it looks OK. So:
print $search;
> $result = mysql_query($search, $connection) or die("Couldn't search the
> library.");
It's also a good idea to see what really went wrong by using mysql_error().
IE:
if (($result = mysql_query($search, $connection) === FALSE)) {
print mysql_error();
die("Couldn't search the library.");
}
--
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Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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RE: [PHP-DB] Resource id #2
Try this...
either...
while ($row = mysql_fetch_array($result)) {
$title = $row['Books.Title'];
$author = $row['Books.Author'];
...
print $title;
}
or...
while ($row = mysql_fetch_array($result)) {
print $row['Title'];
...
}
-Original Message-
From: The Cossins Fam [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 26, 2002 1:10 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource id #2
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:
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 $result to find the "resouce id #2".
Thanks for any help you can provide.
--joel
_
The new MSN 8: advanced junk mail protection and 2 months FREE*
http://join.msn.com/?page=features/junkmail
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Re: [PHP-DB] Resource id #2
Add: print_r($row)
In your while loop, that will show you everything that is being returned
with both it's numeric and text based position.
On Tue, 2002-11-26 at 13:09, 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
>
>
>
>
>
>
>
> _
> The new MSN 8: advanced junk mail protection and 2 months FREE*
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The Cryptocomm Group
My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc
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Description: This is a digitally signed message part
[PHP-DB] Resource id #2
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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Re: [PHP-DB] resource id#2 - ????
Thanks a ton, I was really looking right past that variable.
Jas
"Paul Dubois" <[EMAIL PROTECTED]> wrote in message
news:p05111763b92d3addd3b4@[192.168.0.33]...
> At 11:42 -0600 6/12/02, Jas wrote:
> >Not sure how to over come this, the result of a database query keeps
giving
> >me this:
> > >/* Get Ip address, where they came from, and stamp the time */
> >if (getenv(HTTP_X_FORWARDED_FOR)){
> > $ipaddy = getenv(HTTP_X_FORWARDED_FOR);
> >} else {
> > $ipaddy = getenv(REMOTE_ADDR); }
> >
> >$referrer = $HTTP_REFERER;
> >$date_stamp = date("Y-m-d H:i:s");
> >
> >/* Start session, and check registered variables */
> >session_start();
> >if (isset($HTTP_SESSION_VARS['ipaddy']) ||
> >isset($HTTP_SESSION_VARS['referrer']) ||
> >isset($HTTP_SESSION_VARS['date_stamp'])) {
> > $main = "Video clips stored in Database";
> >
> >/* Insert client info to database table */
> > require('/path/to/connection/class/con.inc');
> > $table_sessions = "dev_sessions";
> > $sql_insert = "INSERT INTO $table_sessions
> >(ipaddy,referrer,date_stamp) VALUES
('$ipaddy','$referrer','$date_stamp')";
> > $result = @mysql_query($sql_insert,$dbh) or die("Couldn't execute
> >insert to database!");
> >
> >/* Pull video info from database table into an array */
> > $table_content = "dev_videos";
> > $sql_content = @mysql_query("SELECT * FROM $table_content",$dbh);
> > while ($row = @mysql_fetch_array($record)) {
>
> $record isn't defined anywhere. Do you mean $sql_content?
>
> > $id = $row['id'];
> > $file_name = $row['file_name'];
> > $file_size = $row['file_size'];
> > $file_properties = $row['file_properties'];
> > $file_codec = $row['file_codec'];
> > $file_author = $row['file_author'];
> > $result = @mysql_query($sql_content,$dbh) or die("Couldn't execute
> >query on database!");
> >
> >/* loop through records and print results of array into table */
> > $current .=
>
>"$id$file_name$file_size$file_properties
<
> >/td>$file_codec$file_author"; }
> > } else {
> > /* Start a new session and register variables */
> > session_start();
> > session_register('ipaddy');
> > session_register('referrer');
> > session_register('date_stamp'); }
> >?>
> >Just to make sure everything is working correctly I have echoed every
> >statement to the screen so I may see what's going on like so:
> >
> >
> > // my results should be here but instead I get
> >Resource ID #2 printed wtf?
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >Any help would be great!
> >Jas
> >
> >
> >
> >--
> >PHP Database Mailing List (http://www.php.net/)
> >To unsubscribe, visit: http://www.php.net/unsub.php
>
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RE: [PHP-DB] resource id#2 - ????
I totally didn't even see that section of code. I am sorry I wrote back
like you were an idiot! I think we know who the real idiot is today... :-)
Sorry again!
-Natalie
-Original Message-
From: Paul DuBois [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 1:53 PM
To: Jas; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] resource id#2 -
At 11:42 -0600 6/12/02, Jas wrote:
>Not sure how to over come this, the result of a database query keeps
>giving me this: /* Get Ip address, where they came from, and stamp the time */
>if (getenv(HTTP_X_FORWARDED_FOR)){
> $ipaddy = getenv(HTTP_X_FORWARDED_FOR);
>} else {
> $ipaddy = getenv(REMOTE_ADDR); }
>
>$referrer = $HTTP_REFERER;
>$date_stamp = date("Y-m-d H:i:s");
>
>/* Start session, and check registered variables */ session_start();
>if (isset($HTTP_SESSION_VARS['ipaddy']) ||
>isset($HTTP_SESSION_VARS['referrer']) ||
>isset($HTTP_SESSION_VARS['date_stamp'])) {
> $main = "Video clips stored in Database";
>
>/* Insert client info to database table */
> require('/path/to/connection/class/con.inc');
> $table_sessions = "dev_sessions";
> $sql_insert = "INSERT INTO $table_sessions
>(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp')";
> $result = @mysql_query($sql_insert,$dbh) or die("Couldn't
>execute insert to database!");
>
>/* Pull video info from database table into an array */
> $table_content = "dev_videos";
> $sql_content = @mysql_query("SELECT * FROM $table_content",$dbh);
> while ($row = @mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
> $id = $row['id'];
> $file_name = $row['file_name'];
> $file_size = $row['file_size'];
> $file_properties = $row['file_properties'];
> $file_codec = $row['file_codec'];
> $file_author = $row['file_author'];
> $result = @mysql_query($sql_content,$dbh) or die("Couldn't
>execute query on database!");
>
>/* loop through records and print results of array into table */
> $current .=
>"$id$file_name$file_size$file_proper
>ties<
>/td>$file_codec$file_author"; }
> } else {
> /* Start a new session and register variables */
> session_start();
> session_register('ipaddy');
> session_register('referrer');
> session_register('date_stamp'); }
>?>
>Just to make sure everything is working correctly I have echoed every
>statement to the screen so I may see what's going on like so:
>
>
> // my results should be here but instead I get
>Resource ID #2 printed wtf?
>
>
>
>
>
>
>
>
>
>
>Any help would be great!
>Jas
>
>
>
>--
>PHP Database Mailing List (http://www.php.net/)
>To unsubscribe, visit: http://www.php.net/unsub.php
--
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Re: [PHP-DB] resource id#2 - ????
At 11:42 -0600 6/12/02, Jas wrote:
>Not sure how to over come this, the result of a database query keeps giving
>me this:
>/* Get Ip address, where they came from, and stamp the time */
>if (getenv(HTTP_X_FORWARDED_FOR)){
> $ipaddy = getenv(HTTP_X_FORWARDED_FOR);
>} else {
> $ipaddy = getenv(REMOTE_ADDR); }
>
>$referrer = $HTTP_REFERER;
>$date_stamp = date("Y-m-d H:i:s");
>
>/* Start session, and check registered variables */
>session_start();
>if (isset($HTTP_SESSION_VARS['ipaddy']) ||
>isset($HTTP_SESSION_VARS['referrer']) ||
>isset($HTTP_SESSION_VARS['date_stamp'])) {
> $main = "Video clips stored in Database";
>
>/* Insert client info to database table */
> require('/path/to/connection/class/con.inc');
> $table_sessions = "dev_sessions";
> $sql_insert = "INSERT INTO $table_sessions
>(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp')";
> $result = @mysql_query($sql_insert,$dbh) or die("Couldn't execute
>insert to database!");
>
>/* Pull video info from database table into an array */
> $table_content = "dev_videos";
> $sql_content = @mysql_query("SELECT * FROM $table_content",$dbh);
> while ($row = @mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
> $id = $row['id'];
> $file_name = $row['file_name'];
> $file_size = $row['file_size'];
> $file_properties = $row['file_properties'];
> $file_codec = $row['file_codec'];
> $file_author = $row['file_author'];
> $result = @mysql_query($sql_content,$dbh) or die("Couldn't execute
>query on database!");
>
>/* loop through records and print results of array into table */
> $current .=
>"$id$file_name$file_size$file_properties<
>/td>$file_codec$file_author"; }
> } else {
> /* Start a new session and register variables */
> session_start();
> session_register('ipaddy');
> session_register('referrer');
> session_register('date_stamp'); }
>?>
>Just to make sure everything is working correctly I have echoed every
>statement to the screen so I may see what's going on like so:
>
>
> // my results should be here but instead I get
>Resource ID #2 printed wtf?
>
>
>
>
>
>
>
>
>
>
>Any help would be great!
>Jas
>
>
>
>--
>PHP Database Mailing List (http://www.php.net/)
>To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP-DB] resource id#2 - ????
The quick answer is that that is what it's supposed to return... that's all
the result is. A nicer, longer answer is to give you some of my code so you
can see one way that you actually get the data out (I use sybase_fetch_row,
but you can also use _fetch_array which returns an associated array or
something - check the manual):
$dataResult = sybase_query($sql);
while ($array_ref = sybase_fetch_row($dataResult)) {
if ($array_ref[4] == 1) {
if ($array_ref[0] == $male ) {
$left[$i] = $array_ref[2];
$names[$i] = $array_ref[3];
$ages[$i] = $array_ref[1];
$i++;
}
elseif ($array_ref[0] == $female ) {
$right[$j] = $array_ref[2];
$j++;
}
}
}//while
Last note - the index of the array ref refers to something in your SQL
starting with 0 (so my sql started with select sex...)
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 1:42 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] resource id#2 -
Not sure how to over come this, the result of a database query keeps giving
me this: $id$file_name$file_size$file_properties<
/td>$file_codec$file_author"; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?>
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:// my
results should be here but instead I get Resource ID #2 printed wtf?
Any help would be great!
Jas
--
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[PHP-DB] resource id#2 - ????
Not sure how to over come this, the result of a database query keeps giving
me this:
$id$file_name$file_size$file_properties<
/td>$file_codec$file_author"; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?>
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
// my results should be here but instead I get
Resource ID #2 printed wtf?
Any help would be great!
Jas
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource ID#2 ????
How can I get the form below to list the single record in a db that matches
the request of $user_id?
$table = "auth_users";
$record = @mysql_query("SELECT * FROM $table WHERE user_id =
'$user_id'",$dbh);
while ($row = mysql_fetch_row($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= "
Edit Account
$user_id
First Name:i.e. John
Last Name:i.e. Doe
Email:i.e.
[EMAIL PROTECTED]
User Name:i.e. j-doe
Password:(password must be alpha-numeric, i.e.
pAs5w0rd)
Confirm Password:please confirm password
entered
";
echo $record;
Thanks in advance,
Jas
"Natalie Leotta" <[EMAIL PROTECTED]> wrote in message
7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG">news:7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG...
> I could be missing something, but it looks like you are using the result
of
> the mysql_query as the actual result. It actually returns some weird
> identifier. To access the real info you'd have to use something like
> mysql_fetch_array to get it.
>
> Check out this and see if it helps:
>
> http://www.php.net/manual/en/function.mysql-query.php
>
> -Natalie
>
> -Original Message-
> From: Jas [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, May 29, 2002 12:18 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Resource ID#2
>
>
> Ok here is my problem, I set this up so a user selects a name form a
select
> box and that name or $user_id is then passed to this page so the user can
> edit the contact info etc. However it does not pull the selected $user_id
> and place each field into my form boxes, all I get is a Resource ID #2.
Not
> sure how I can over come this...
>
> $table = "auth_users";
> $record = @mysql_query("SELECT * FROM $table WHERE user_id =
> '$user_id'",$dbh);
> $var_form .= " name=\"$user_id\" method=\"post\" action=\"del_account.php\">
> Edit Account
> $user_id
> First Name: type=\"text\" name=\"$f_name\" size=\"30\" maxlength=\"30\"
> value=\"$f_name\">i.e. John
> Last Name: type=\"text\" name=\"$l_name\" size=\"30\" maxlength=\"30\"
> value=\"$l_name\">i.e. Doe
> Email: type=\"text\" name=\"$email_addy\" size=\"30\" maxlength=\"30\"
> value=\"$email_addy\">i.e.
> [EMAIL PROTECTED]
> User Name: type=\"text\" name=\"$un\" size=\"30\" maxlength=\"30\"
value=\"$un\"> class=\"copyright\">i.e. j-doe
> Password: type=\"password\" name=\"$pw\" size=\"30\" maxlength=\"30\"> class=\"copyright\">(password must be alpha-numeric, i.e.
> pAs5w0rd)
> Confirm Password: width=\"80%\"> maxlength=\"30\">please confirm password
> entered
> type=\"submit\" name=\"add\" value=\"edit user\"> type=\"reset\" name=\"reset\" value=\"reset\">
>";
> echo $record;
> } else {
> blah blah
> }
> Thanks in advance,
> Jas
>
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP-DB] Resource ID#2 ????
On Thursday 30 May 2002 00:41, Jas wrote:
> If you look at the previously posted code at the bottom of the form there
> is a echo for the sql select statement that is echoing "Resource id #2" on
> the page. Now that error is the correct field id number in the database, I
> am just not sure how to itemize the data from that table, at least I think.
> Any help would be great!
You are confused -- #2 just happens to coincide with your field id.
You're not RFTM.
mysql_query() returns a result_id ("Resource id #2").
You feed this into a mysql_fetch_row() (or any of the other similar functions)
to actually retrieve records.
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
/*
1: No code table for op: ++post
*/
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RE: [PHP-DB] Resource ID#2 ????
I could be missing something, but it looks like you are using the result of
the mysql_query as the actual result. It actually returns some weird
identifier. To access the real info you'd have to use something like
mysql_fetch_array to get it.
Check out this and see if it helps:
http://www.php.net/manual/en/function.mysql-query.php
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, May 29, 2002 12:18 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource ID#2
Ok here is my problem, I set this up so a user selects a name form a select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2. Not
sure how I can over come this...
$table = "auth_users";
$record = @mysql_query("SELECT * FROM $table WHERE user_id =
'$user_id'",$dbh);
$var_form .= "
Edit Account
$user_id
First Name:i.e. John
Last Name:i.e. Doe
Email:i.e.
[EMAIL PROTECTED]
User Name:i.e. j-doe
Password:(password must be alpha-numeric, i.e.
pAs5w0rd)
Confirm Password:please confirm password
entered
";
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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Re: [PHP-DB] Resource ID#2 ????
If you look at the previously posted code at the bottom of the form there is
a echo for the sql select statement that is echoing "Resource id #2" on the
page. Now that error is the correct field id number in the database, I am
just not sure how to itemize the data from that table, at least I think.
Any help would be great!
Jas
"Jason Wong" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> On Thursday 30 May 2002 00:17, Jas wrote:
> > Ok here is my problem, I set this up so a user selects a name form a
select
> > box and that name or $user_id is then passed to this page so the user
can
> > edit the contact info etc. However it does not pull the selected
$user_id
> > and place each field into my form boxes, all I get is a Resource ID #2.
> > Not sure how I can over come this...
> >
> > $table = "auth_users";
> > $record = @mysql_query("SELECT * FROM $table WHERE user_id =
> > '$user_id'",$dbh);
>
> Make liberal use of error checking (see manual) and mysql_error().
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
> Open Source Software Systems Integrators
> * Web Design & Hosting * Internet & Intranet Applications Development *
>
>
> /*
> grep me no patterns and I'll tell you no lines.
> */
>
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Re: [PHP-DB] Resource ID#2 ????
On Thursday 30 May 2002 00:17, Jas wrote:
> Ok here is my problem, I set this up so a user selects a name form a select
> box and that name or $user_id is then passed to this page so the user can
> edit the contact info etc. However it does not pull the selected $user_id
> and place each field into my form boxes, all I get is a Resource ID #2.
> Not sure how I can over come this...
>
> $table = "auth_users";
> $record = @mysql_query("SELECT * FROM $table WHERE user_id =
> '$user_id'",$dbh);
Make liberal use of error checking (see manual) and mysql_error().
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
/*
grep me no patterns and I'll tell you no lines.
*/
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Resource ID#2 ????
Ok here is my problem, I set this up so a user selects a name form a select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2. Not
sure how I can over come this...
$table = "auth_users";
$record = @mysql_query("SELECT * FROM $table WHERE user_id =
'$user_id'",$dbh);
$var_form .= "
Edit Account
$user_id
First Name:i.e. John
Last Name:i.e. Doe
Email:i.e.
[EMAIL PROTECTED]
User Name:i.e. j-doe
Password:(password must be alpha-numeric, i.e.
pAs5w0rd)
Confirm Password:please confirm password
entered
";
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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Re: [PHP-DB] Resource Id #2
AHAHAHHAHA AHHAHAHAH Dan On Wednesday, January 23, 2002, at 02:19 PM, [EMAIL PROTECTED] wrote: > Dan, > >> Good, I'm glad it worked!!! >> >> I see a lot of people using mysql_fetch_array(). >> >> Which is fine...But I'm more into Object Oriented programming...and >> that's why I always use mysql_fetch_object() > > > hey this is a family show - but I guess 'what you do in the privacy > of' ... > > - and I'm just plain object-ionable, so I favor mysql_fetch_assoc() -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Resource Id #2
Dan, > Good, I'm glad it worked!!! > > I see a lot of people using mysql_fetch_array(). > > Which is fine...But I'm more into Object Oriented programming...and > that's why I always use mysql_fetch_object() hey this is a family show - but I guess 'what you do in the privacy of' ... - and I'm just plain object-ionable, so I favor mysql_fetch_assoc() =dn -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Resource Id #2
Good, I'm glad it worked!!!
I see a lot of people using mysql_fetch_array().
Which is fine...But I'm more into Object Oriented programming...and
that's why I always use mysql_fetch_object()
Good luck!!!
Dan
On Wednesday, January 23, 2002, at 11:45 AM, [EMAIL PROTECTED] wrote:
> My thanks for your quick response!
>
> Here's the problem I'm dealing with:
>
> I am inserting info with a unique identifier that auto-increments, then
> take
> that ID and place it into another table for later reference.
>
> I was trying to use something like:
>
> $query="select var1 from table1 where var2='$var3'
> and var4='$var5'";
>
> $somevar=mysql_query($query) or die.. yada yada
> echo $somevar;
>
> I tried your code, and it works wonderfully! Thanks so much. :)
>
> J. Wharton
> [EMAIL PROTECTED]
>
> - Original Message -
> From: "Dan Brunner" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Wednesday, January 23, 2002 12:21 PM
> Subject: Re: [PHP-DB] Resource Id #2
>
>
>> Use something like this...
>>
>> ...
>> ...
>> ...
>> $rows = mysql_num_rows($result);
>> for ($y = 0; $y < $rows; $y++){
>> $data = mysql_fetch_object($result);
>>
>> echo $data->Field_Name;
>> ...
>> ...
>> ...
>>
>> }
>>
>>
>> A Little more code would Help!!!
>>
>>
>>
>>
>> On Wednesday, January 23, 2002, at 10:56 AM, [EMAIL PROTECTED] wrote:
>>
>>> I'm having a problem with retrieving data from MySQL databases. I can
>>> input
>>> with no problems, but when I try to pull the data back out (single
>>> field)
>>> and echo it I get something showing up saying Resource Id #2. Any
>>> ideas?
>>>
>>>
>>>
>>>
>>>
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>>> To contact the list administrators, e-mail: php-list-
>>> [EMAIL PROTECTED]
>>>
>>
>>
>
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Re: [PHP-DB] Resource id #2 ?
You have to fetch the row that contains the count in it. Try this:
$resource = mysql_query("SELECT COUNT(*) cnt FROM catalogs WHERE PROCESSED
IS NULL");
$countreq = mysql_fetch_array($resource);
echo $countreq[cnt];
(note as well that if no row is returned by the query then $countreq will be
set to false, so checking that before using it as an array is a good idea)
- Patrick
- Original Message -
From: "James Kupernik" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, January 18, 2002 12:05 PM
Subject: [PHP-DB] Resource id #2 ?
> I run this query $countreq = mysql_query("SELECT COUNT(*) FROM catalogs
> WHERE PROCESSED IS NULL"); then try to echo $countreq but end up getting
> Resource id #2 instead of the count total. What would cause this?
>
> Thanks for any advise!
>
> James
>
>
>
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>
>
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[PHP-DB] Resource id #2 ?
I run this query $countreq = mysql_query("SELECT COUNT(*) FROM catalogs
WHERE PROCESSED IS NULL"); then try to echo $countreq but end up getting
Resource id #2 instead of the count total. What would cause this?
Thanks for any advise!
James
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Re: [PHP-DB] resource id #2, #3, #4......
On Thursday 29 November 2001 01:29, Kevin Ruiz wrote:
> I've come across yet another problem.
[snip]
> $ci = "select contactid from users where username='$username' and
> password='$password'";
> $cir = mysql_query($ci)
> or die("Couldn't execute");
> $query = "select * from my_contacts where contactid='$cir'";
>
> $result = mysql_query($query) or
> die( mysql_error() );
>
> When I try to print $cir to see if anything's getting passed I keep getting
> something that reads "resource id #2".
[snip]
> Does anyone know what this means and how I can work around it.
$cir does NOT contain the actual results of your query. It is only a
*pointer* to your results. To get at the actual result(s):
print "\n";
while ($line = mysql_fetch_array($cir)) {
print "\t\n";
while(list($col_name, $col_value) = each($line)) {
print "\t\t$col_value\n";
}
print "\t\n";
}
print "\n";
Read the manual for full details.
hth
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/*
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[PHP-DB] resource id #2, #3, #4......
I've come across yet another problem.
I have a table set up that houses four things, a person's real name,
username, pass, and id. The id is used to join another table which houses
that persons contacts. I've been validating the user and pass by comparing
the number of rows that the sql statement returns.
Once the user submits their user/pass they're sent to the validation page
where the user and pass are compared and if they're valid another sql
statement takes place:
$ci = "select contactid from users where username='$username' and
password='$password'";
$cir = mysql_query($ci)or die("Couldn't execute");
echo("");
I've registered the variables $username, $password, & $cir and if the user
is validated they're sent to the contacts page.
On the contacts page I'm trying to select which contacts to show based on
the contact id. This is what the code looks like on the page where the
contacts are going to be displayed...
mysql_connect("localhost") or
die("couldn't connect");
mysql_select_db("act") or
die("couldn't connect to the database");
$ci = "select contactid from users where username='$username' and
password='$password'";
$cir = mysql_query($ci)
or die("Couldn't execute");
$query = "select * from my_contacts where contactid='$cir'";
$result = mysql_query($query) or
die( mysql_error() );
When I try to print $cir to see if anything's getting passed I keep getting
something that reads "resource id #2".
If I were rename the variables in the $ci, $cir, and $query lines, run the
code, and try to print $cir again I'd get "resource id #3" and so on.
Does anyone know what this means and how I can work around it.
Thanks for your time.
Kevin
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RE: [PHP-DB] Resource id #2
Change code as follows: ".$row['msg'].""; ?> -Original Message- From: Andrew Duck [mailto:[EMAIL PROTECTED]] Sent: Sunday, October 28, 2001 7:00 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Resource id #2 I am trying to select a message from a table in a database. the message will will be in column 'msg' and will be in same row as id='0'. I need that msg put to the screen. With the code below I get the error: Resource id #2 Can someone please explain how I can fix this.. Thankyou in advance. $msg"; ?> -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Resource id #2
I am trying to select a message from a table in a database. the message will will be in column 'msg' and will be in same row as id='0'. I need that msg put to the screen. With the code below I get the error: Resource id #2 Can someone please explain how I can fix this.. Thankyou in advance. $msg"; ?> Andrew, Although the assignment statement looks as though the LHS (left-hand side) should contain the result of the query, it does not (as you have discovered). Instead of being an integer or string, it is something called a 'resource'. You have already 'met' a resource in connection with the db connect process. After issuing a query, you must then use one of the mysql_fetch_... functions to extract the data, one field, or one row at a time, from the 'resource'. Please check out the manual at http://uk2.php.net/manual/en/ref.mysql.php. Also you might consider working through one of the many excellent texts and/or online articles discussing the PHP/MySQL combination. Regards, =dn -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] Resource id #2
I am trying to select a message from a table in a database. the message will will be in column 'msg' and will be in same row as id='0'. I need that msg put to the screen. With the code below I get the error: Resource id #2 Can someone please explain how I can fix this.. Thankyou in advance. $msg"; ?>
