mysql_result(): supplied argument is not a valid.....

[bettina@coaster.ch]

The program works local ok. The data were found in the datenbank and
display by the program.
When I wanted to test my program online, I get the following message:
mysql_result(): supplied argument is not a valid MySQL result resource
in ....

this message several times for all the mysql_result... .

The connection to the Datenbank seems to work because I didn't receive
an error message..... that's the script for the connection:
$db = @MYSQL_PCONNECT ($db_server,$db _user,$db_passw ort)
or die ("Connection failed");
$db_select = @MYSQL_SELECT_D B($db_name);


Could it be that however the datenbank was not really connected?
Because I cannot see it on my Provider Control Panel. Where should I
put the Datenbank really? Under mysql/data... just like I have it local
and it works fine?

Any help will be welcomed.

[kalenj@gmail.com]

Re: mysql_result(): supplied argument is not a valid.....

might want to check to see what access priviledges your mysql user has
on the database you're accessing. i know that access can be determined
by the host. maybe your local computer has access but the live server
you're working with doesn't. long shot but maybe.

[Tony]

Re: mysql_result(): supplied argument is not a valid.....

bettina@coaster .ch wrote:[color=blue]
> The program works local ok. The data were found in the datenbank and
> display by the program.
> When I wanted to test my program online, I get the following message:
> mysql_result(): supplied argument is not a valid MySQL result resource
> in ....
>
> this message several times for all the mysql_result... .
>
> The connection to the Datenbank seems to work because I didn't receive
> an error message..... that's the script for the connection:
> $db = @MYSQL_PCONNECT ($db_server,$db _user,$db_passw ort)
> or die ("Connection failed");
> $db_select = @MYSQL_SELECT_D B($db_name);[/color]

That's the connection code, now what about the query code? That's where the
error probably is.

The error you're getting means that you're trying to access a query result
that hasn't been set. Be sure you're assigning the result of mysql_query()
to a variable, and double-check the variable spelling. My guess is that
you're repeating the exact same error each time.

--
Tony Garcia
Web Right! Development
Riverside, CA

https://www.WebRightDevelopment.com

[bettina@coaster.ch]

Re: mysql_result(): supplied argument is not a valid.....

here is my code:
line 39 $total_1 = mysql_query("SE LECT COUNT(*) FROM COASTERS");
line 40 $total_coasters = mysql_result($t otal_1,0,0);
line 41 $total_2 = mysql_query("SE LECT COUNT(*) FROM COUNTRIES");
line 42 $total_countrie s = mysql_result($t otal_2,0,0);
line 43 $GET_DATE = mysql_query("SE LECT DATE_FORMAT(DAT E,'%d.%m.%Y')
FROM COASTERS ORDER BY DATE DESC LIMIT 1");
line 44 $update = mysql_result($G ET_DATE,0,0);

I get errors on line 40, 42 and 44 and I'm only counting the records
and getting the latest date,,, And what's strange is that it works
local!!!!

[Andy Hassall]

Re: mysql_result(): supplied argument is not a valid.....

On 12 Jul 2005 11:45:33 -0700, bettina@coaster .ch wrote:
[color=blue]
>When I wanted to test my program online, I get the following message:
>mysql_result() : supplied argument is not a valid MySQL result resource
>in ....[/color]

This always means you didn't check a previous call for errors.

Check every mysql_* call's return value.

If false, use mysql_error() to print out an informative error, and stop.

--
Andy Hassall / <andy@andyh.co. uk> / <http://www.andyh.co.uk >
<http://www.andyhsoftwa re.co.uk/space> Space: disk usage analysis tool

[Jerry Stuckle]

Re: mysql_result(): supplied argument is not a valid.....

bettina@coaster .ch wrote:[color=blue]
> here is my code:
> line 39 $total_1 = mysql_query("SE LECT COUNT(*) FROM COASTERS");
> line 40 $total_coasters = mysql_result($t otal_1,0,0);
> line 41 $total_2 = mysql_query("SE LECT COUNT(*) FROM COUNTRIES");
> line 42 $total_countrie s = mysql_result($t otal_2,0,0);
> line 43 $GET_DATE = mysql_query("SE LECT DATE_FORMAT(DAT E,'%d.%m.%Y')
> FROM COASTERS ORDER BY DATE DESC LIMIT 1");
> line 44 $update = mysql_result($G ET_DATE,0,0);
>
> I get errors on line 40, 42 and 44 and I'm only counting the records
> and getting the latest date,,, And what's strange is that it works
> local!!!!
>[/color]

When a query fails, mysql_query returns false. At that time you need to check
the contents of mysql_error();

Note you cannot just place the MySQL database files in a directory. That won't
add it to MySQL's catalogs. You need to create the database and tables, then
insert your data.

phpMyAdMin will help you a lot with this process. It allows you to easily
export your local database then import it to the one on the server.

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[Tony]

Re: mysql_result(): supplied argument is not a valid.....

bettina@coaster .ch wrote:[color=blue]
> here is my code:
> line 39 $total_1 = mysql_query("SE LECT COUNT(*) FROM COASTERS");
> line 40 $total_coasters = mysql_result($t otal_1,0,0);
> line 41 $total_2 = mysql_query("SE LECT COUNT(*) FROM COUNTRIES");
> line 42 $total_countrie s = mysql_result($t otal_2,0,0);
> line 43 $GET_DATE = mysql_query("SE LECT DATE_FORMAT(DAT E,'%d.%m.%Y')
> FROM COASTERS ORDER BY DATE DESC LIMIT 1");
> line 44 $update = mysql_result($G ET_DATE,0,0);
>
> I get errors on line 40, 42 and 44 and I'm only counting the records
> and getting the latest date,,, And what's strange is that it works
> local!!!![/color]

As others have pointed out, you should be having some error checking:

if (!$total_1) { print mysql_error(); }

after line 39 (and one for $total_2 after line 41)- that would tell you why
you're not getting a proper result back. Otherwise, it doesn't seem to be a
code problem. Add the error checking & come back with the exact error
message you get.

--
Tony Garcia
Web Right! Development
Riverside, CA

https://www.WebRightDevelopment.com