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
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 get
: 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 have very little experience with PHP and SQL, but I learn
quickly. :) Our resident PHP
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 and see if it passes:
$SelectQuery=SELECT SUM(GiftSubscriptions
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 have very little experience with PHP and SQL, but I
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
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 resource
What does
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_data
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?
$result = mysql_query('select * from aannh_towns;');
echo 'info stored';
while ($query_data = mysql_fetch_array($result)) {
echo
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 you checked
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 stuffed up
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
-
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 snippet of code is
something I use all
the time
;
}
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
here is the sql
$parent_sql = SELECT * FROM categories
Regards,
Andrey Hristov
- Original Message -
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
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
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
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
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
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