Ramil,
Got that. thanks. Yeah, I mis-named a field in the seond table. Needed to name title
for district
$sql2 = "select * from ".$db.".".$table_GLOfficers." where title='" .
$mydata->district. "'";
Thanks for that. Getting late, i guess.
John
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PHP General Mailing List (http://www.php.net/
On Sat, 25 Sep 2004 01:01:13 -0400, John Taylor-Johnston
<[EMAIL PROTECTED]> wrote:
> Hi,
Hi, the error on the line
line 101> while ($mydata2 = mysql_fetch_object($news2))
says that $news2 is null. The line
$news2 = mysql_query($sql2); //desc => z-a
probably didn't work. How about trying;
$n
Hi,
My question: can I open $table2 within $table1 - I thought so.
I thought I had this set up correctly. Sorry to throw all these lines of code at you.
But I need to situate the error for someone to see plainly what I'm doing.
I'm getting this error:
Warning: mysql_fetch_object(): supplied arg
From: "Kurosh Burris" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 02, 2003 7:54 PM
Subject: [PHP] Supplied argument is not a valid MySQL result resource
> Hi Everyone,
>
> I'm trying to get a very simple "sum" of the values in an
t use
the "new Query" part.
Kurosh
-Original Message-
From: Hugh Danaher [mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 03, 2003 12:53 AM
To: Kurosh Burris
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Supplied argument is not a valid MySQL result resource
try something simpler an
lt;[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 02, 2003 7:54 PM
Subject: [PHP] Supplied argument is not a valid MySQL result resource
> Hi Everyone,
>
> I'm trying to get a very simple "sum" of the values in an MySQL database
> field column.
> I'm trying to get a very simple "sum" of the values in an MySQL
database
> field column. I have very little experience with PHP and SQL, but I
learn
> quickly. :) Our resident PHP expert who built the full script is
really
> busy, so I've been trying to do this myself. The farthest I could ge
Hi Everyone,
I'm trying to get a very simple "sum" of the values in an MySQL database
field column. I have very little experience with PHP and SQL, but I learn
quickly. :) Our resident PHP expert who built the full script is really
busy, so I've been trying to do this myself. The farthest I co
On 28 Apr 2002 at 15:14, Dan McCullough wrote:
> What does that error mean?
>
> $result = mysql_query('select * from aannh_towns;');
> echo 'info stored';
> while ($query_data = mysql_fetch_array($result)) {
> echo "", $query_data['town'], "", $query_data['town_id'], ""; }
> ?>
Means you
On Sun, 28 Apr 2002, Dan McCullough wrote:
> What does that error mean?
>
> $result = mysql_query('select * from aannh_towns;');
> echo 'info stored';
> while ($query_data = mysql_fetch_array($result)) {
> echo "", $query_data['town'], "", $query_data['town_id'], ""; }
> ?>
You'd know if y
than
- Original Message -
From: "Dan McCullough" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, April 28, 2002 4:14 PM
Subject: [PHP] Supplied argument is not a valid MySQL result resource
What does that error mean?
$result = mysql_query('sele
Probably means your query failed. What line did the error occur on? You
should be able to track down your problem with that.
On Sunday 28 April 2002 15:14 pm, you wrote:
> What does that error mean?
>
> $result = mysql_query('select * from aannh_towns;');
> echo 'info stored';
> while ($query_
It means your query failed.
www.php.net/mysql_error
---John Holmes...
> -Original Message-
> From: Dan McCullough [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, April 28, 2002 3:15 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Supplied argument is not a valid MySQL result reso
What does that error mean?
$result = mysql_query('select * from aannh_towns;');
echo 'info stored';
while ($query_data = mysql_fetch_array($result)) {
echo "", $query_data['town'], "", $query_data['town_id'], ""; }
?>
=
dan mccullough
---
Hi Dan,
> $parent_sql = "SELECT * FROM categories ORDER BY sort_order ASC";
> $parent_result = mysql_query($parent_sql);
change the line to this one and run it again to see the error:
$parent_result = mysql_query($parent_sql) or die(mysql_error());
--
Jimmy
Brain of
: "Dan McCullough" <[EMAIL PROTECTED]>
> To: "PHP General List" <[EMAIL PROTECTED]>
> Sent: Tuesday, January 08, 2002 6:47 PM
> Subject: [PHP] Supplied argument is not a valid MySQL result resource
>
>
ame']."";
}
Regards,
Andrey
- Original Message -
From: "Dan McCullough" <[EMAIL PROTECTED]>
To: "Andrey Hristov" <[EMAIL PROTECTED]>
Sent: Tuesday, January 08, 2002 7:00 PM
Subject: Re: [PHP] Supplied argument is not a valid MySQL result resource
> he
-
From: "Dan McCullough" <[EMAIL PROTECTED]>
To: "PHP General List" <[EMAIL PROTECTED]>
Sent: Tuesday, January 08, 2002 6:47 PM
Subject: [PHP] Supplied argument is not a valid MySQL result resource
> I know this is a common error, but what does it mean. This snipp
I know this is a common error, but what does it mean. This snippet of code is
something I use all
the time.
help please.
dan
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PHP General Mailing List
Hi James,
> I've been getting the following error:
> Warning: Supplied argument is not a valid PostgreSQL
> result resource in dbasefunctions.php on line 87
that's because your SQL query return nothing, but you
try to fetch a row.
--
Jimmy
Live for Love, for without
Hi.
I've been getting the following error:
Warning: Supplied argument is not a valid PostgreSQL
result resource in dbasefunctions.php on line 87
The error occurs in the following function:
function dbFetchRow($result,$num) {
global $wireddb_h;
$r = pg_fetch_row($result,$num);
Because you are enclosing your $sql string within () brackets, $sql is
being set a boolean value, not the value of the string.
Change it to $sql = "SELECT * ... '$group_id'";
-Mike
Dan McCullough wrote:
>Warning: Supplied argument is not a valid MySQL result resource in
>/home/sites/projects.
Warning: Supplied argument is not a valid MySQL result resource in
/home/sites/projects.heathermccullough.com/web/copeland/product.php on line 61
Here is the code, I fixed it once, and I cant figure out why it keeps returning.
if ($submit) {
$sql = ("SELECT * FROM store WHERE state = '$State' AN
Hi davek,
@ 1:50:14 PM on 4/14/2001, davek wrote:
> Does anybody know how to fix this error? "Supplied argument is not a
> valid MS SQL-Link resource" I am running:
...
> $Conn=mssql_connect('192.168.1.20','username','password')
or die("Could not connect to the mssql server");
> $ThisConnect=
http://us.php.net/manual/en/function.mssql-select-db.php
*_select_db does not return a connection id, it returns true or false.
Just change your script like this:
$Conn=mssql_connect('192.168.1.20','username','password');
mssql_select_db('My_Database',$Conn);
$QueryString="select id, name from
Does anybody know how to fix this error? "Supplied argument is not a valid
MS SQL-Link resource" I am running:
NT4
MSSQL7.0
IIS4
PHP 4.0.4pl1
I've been running php3 for some time now without any problems but want to
upgrade to 4. This error seems to be occuring at the mssql_query()
function.
This happened to me when I used _affected_rows instead of _count
make sure you're using the right function.
that drove me nuts before I got it working, AND I followed the durn directions.
chris
On Tue, 10 Apr 2001 09:02:32 -0700, elias wrote:
>can you show me some code?
>basically when you
That means that your mysql_query probably returned an error.
It should do that only when you screw up your syntax.
echo/print out the syntax used in your query, and post it here. That is
probably the problem you are having.
--
Plutarck
Should be working on something...
...but forgot what it wa
can you show me some code?
basically when you provide an invalid link to any MySql function this error
occur.
-elias
http://www.kameelah.org/eassoft
""Nathan Roberts"" <[EMAIL PROTECTED]> wrote in message
9atdps$kgv$[EMAIL PROTECTED]">news:9atdps$kgv$[EMAIL PROTECTED]...
> I am trying to get to
I am trying to get to grips with managing data in mysql, am am currentky
working on deleting records
I have worked through a turorial - no problems, but now I am trying to apply
it to my own database, I am getting the error
Warning: Supplied argument is not a valid MySQL result resource in o
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