ument is not a valid MySQL result resource
try something simpler and see if it passes:
$SelectQuery="SELECT SUM(GiftSubscriptions) AS GiftSubs FROM Registration";
$Result = mysql_query($SelectQuery) or die("Something died in here!
".mysql_error());
- Original Message -
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
try something simpler and see if it passes:
$SelectQuery="SELECT SUM(GiftSubscriptions) AS GiftSubs FROM Registration";
$Result = mysql_query($SelectQuery) or die("Something died in here!
".mysql_error());
- Original Message -
From: "Kurosh Burris" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTE
> 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
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
I am betting the problem is simply your semicolon... never use a semicolon in a php
mysql query. It
doesn't need one. In general, though, you should write your query such that it will
tell you exactly
what went wrong... Re-write this like so:
$sql = 'select * from aannh_towns';
$result = mysql_
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 resource
>
> What
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
Here is the code
$parent_select = "Parent Category:";
$parent_sql = "SELECT * FROM categories ORDER BY sort_order ASC";
$parent_result = mysql_query($parent_sql);
while ($parent = mysql_fetch_array($parent_result) ) { <-- the problem line
$parent_select .= "".$parent['cat_name']."";
}
$parent_sel
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
Possibilities:
1)There was an error when mysql_query(). Possibly the SQL statement is not correct.
Trace with var_dump()/mysql_errno().
2)You give as parameter to some of mysql_* functions a variable not returned from
mysql_query().
HTH
Regards,
Andrey Hristov
- Original Message -
Fr
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.
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
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