Terion Miller wrote:
> On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown
> wrote:
>
>> On Fri, Dec 12, 2008 at 18:03, Terion Miller
>> wrote:
>>> Well I did some changes and I must be learning because although I have
>> the
>>> same error I don't have new ones...
>>> so now the code is like this:
On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown
wrote:
> On Fri, Dec 12, 2008 at 18:03, Terion Miller
> wrote:
> >
> > Well I did some changes and I must be learning because although I have
> the
> > same error I don't have new ones...
> > so now the code is like this:
> > $sql = "SELECT * FROM
On Fri, Dec 12, 2008 at 18:03, Terion Miller wrote:
>
> Well I did some changes and I must be learning because although I have the
> same error I don't have new ones...
> so now the code is like this:
> $sql = "SELECT * FROM `importimages` WHERE `Category` = 'Obits'";
> $result = mysql_query($sql)
On Fri, Dec 12, 2008 at 4:52 PM, Terion Miller wrote:
>
>
> On Fri, Dec 12, 2008 at 4:02 PM, Daniel P. Brown <
> daniel.br...@parasane.net> wrote:
>
>> On Fri, Dec 12, 2008 at 16:54, Terion Miller
>> wrote:
>> >
>> > $query = "SELECT * FROM importimages WHERE Category='Obits' ";
>> > $result = my
On Fri, Dec 12, 2008 at 4:02 PM, Daniel P. Brown
wrote:
> On Fri, Dec 12, 2008 at 16:54, Terion Miller
> wrote:
> >
> > $query = "SELECT * FROM importimages WHERE Category='Obits' ";
> > $result = mysql_query ($query);
> >
> > $arr = mysql_fetch_row($result);
> > $result2 = $arr[0];
> > echo ($re
On Fri, Dec 12, 2008 at 16:54, Terion Miller wrote:
>
> $query = "SELECT * FROM importimages WHERE Category='Obits' ";
> $result = mysql_query ($query);
>
> $arr = mysql_fetch_row($result);
> $result2 = $arr[0];
> echo ($result2);
Try this to get yourself started:
$v) {
echo stripsl
Ross wrote:
When I try to insert a field into my database it shows as Resource id#21?
I must be doing something dim.
Right after you try to do the insert, echo out mysql_error()
--
John C. Nichel
ÜberGeek
KegWorks.com
716.856.9675
[EMAIL PROTECTED]
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PHP General Mailing List (http://www.php
When I try to insert a field into my database it shows as Resource id#21?
I must be doing something dim.
Some could would definetely help here...
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On Thu, August 18, 2005 1:00 am, Chris Boget wrote:
> * User A accesses page X, which makes a connection to the database.
> Echoing
> out the result of the mssql_pconnect() function shows it's using
> 'Resource id #10'.
> * User B accesses the same page, X, making another connection to the
> databa
It's probably a database result set or a connection object or something
similar.
Shaunak
> -Original Message-
> From: Mike Mapsnac [mailto:[EMAIL PROTECTED]
> Sent: Monday, February 16, 2004 11:53 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Resource id #1
>
>
> I print variable to the scr
John,
This is the output:
Resource id #15
or some other seemingly arbitrary Resource ID number?
First of all what is a resource ID and second how do I get it to actually
show what I am trying to get it to show!
=When MySQL returns data to PHP, the information is put into a variable
called a "re
Kurth Bemis <[EMAIL PROTECTED]> wrote:
> i get this:
>
> Resource id #2
>
> when i run this code.whats resource id 2 mean? i just want to
> know if the query was ok or not
>
> $result = mysql_query("SELECT authcode FROM users WHERE email='$email'",$db);
> echo $result;
this is because mysql_query() returns a result identifier for select
statements NOT a string or whatever you expected...
if ($result) {
my query was syntactically ok, but I still don't know anything about the
result
}
else {
my query was semantically invalid.
}
have a look at mysql_resul
heh, that basically means that query was ok at least techincally.
When you echo that $result and get nothing - then something went wrong.
The right way to check if query was ok is:
if(is_resource($result))
echo "Query OK";
or just:
if($result)
echo "Query OK";
lenar.
"Kurth Bemis" <[E
Sorry...
Read morgans topic aswell...
Didn´t see the duplicate mysql_query(). :)
// Tobias
""Tobias Talltorp"" <[EMAIL PROTECTED]> wrote in message
9betpl$nb0$[EMAIL PROTECTED]">news:9betpl$nb0$[EMAIL PROTECTED]...
> > $query=mysql_query("Select pass from members where uname='$username'");
> > $
Greg K schrieb:
> I am trying to run a query and in my log I am getting a message the message
> resource id #2.
>
> $query=mysql_query("Select pass from members where uname='$username'");
> $result = mysql_query($query)
> or die("You are not authorized to be here.");
>
> Can someone tell me w
> $query=mysql_query("Select pass from members where uname='$username'");
> $result = mysql_query($query)
> or die("You are not authorized to be here.");
What you are doing is checking if the query is valid. Your die() would print
only if you had a faulty query (try changing pass to pass2), but
try
$query="Select pass from members where uname='$username'";
$result = mysql_query($query) or die("You are not authorized to be here.");
the mysql_query command is executing the statement
morgan
At 10:40 AM 4/16/2001, Greg K wrote:
>I am trying to run a query and in my log I am getting a me
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