Hey there, okay I ran into these and need some tips, pointers etc...
First I was getting the Resource ID#5 error with this query:
$query=SELECT * FROM importimages WHERE Category='Obits';
$result = mysql_query($query);
so then I read how mysql_query returns a resource, so I tried this:
$query =
On Fri, Dec 12, 2008 at 16:54, Terion Miller webdev.ter...@gmail.com 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:
?php
$sql =
On Fri, Dec 12, 2008 at 4:02 PM, Daniel P. Brown
daniel.br...@parasane.netwrote:
On Fri, Dec 12, 2008 at 16:54, Terion Miller webdev.ter...@gmail.com
wrote:
$query = SELECT * FROM importimages WHERE Category='Obits' ;
$result = mysql_query ($query);
$arr = mysql_fetch_row($result);
On Fri, Dec 12, 2008 at 4:52 PM, Terion Miller webdev.ter...@gmail.comwrote:
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 webdev.ter...@gmail.com
wrote:
$query = SELECT * FROM importimages WHERE
On Fri, Dec 12, 2008 at 18:03, Terion Miller webdev.ter...@gmail.com 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 =
On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown
daniel.br...@parasane.netwrote:
On Fri, Dec 12, 2008 at 18:03, Terion Miller webdev.ter...@gmail.com
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
Terion Miller wrote:
On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown
daniel.br...@parasane.netwrote:
On Fri, Dec 12, 2008 at 18:03, Terion Miller webdev.ter...@gmail.com
wrote:
Well I did some changes and I must be learning because although I have
the
same error I don't have new ones...
When I try to insert a field into my database it shows as Resource id#21?
I must be doing something dim.
R.
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When I try to insert a field into my database it shows as Resource id#21?
I must be doing something dim.
R.
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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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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()
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John C. Nichel
ÜberGeek
KegWorks.com
716.856.9675
[EMAIL PROTECTED]
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I just want to make sure of my understanding on something. I was pretty
sure I
had this down but now I'm beginning to question myself and I'm hoping
someone
could confirm.
* User A accesses page X, which makes a connection to the database.
Echoing
out the result of the mssql_pconnect() function
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
database.
I print variable to the screen and get the result Resource id #1. Any
ideas waht Resource id #1 is?
Thanks
_
Plan your next US getaway to one of the super destinations here.
http://special.msn.com/local/hotdestinations.armx
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PHP
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 screen
When trying to do this query:
$rsum =mysql_query(SELECT SUM(rating) FROM ratings where threadid = $ratevar)or
die (mysql_error());
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
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
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;
because
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;
~kurth
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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 [EMAIL
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 $result;
~kurth
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this is because mysql_query() returns a result identifier for select
statements NOT a string or whatever you expected...
if ($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
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 what I am doing wrong and guide
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
$query=mysql_query("Select pass from members where uname='$username'");
$result = mysql_query($query)
or die("You are not authorized to be here.");
?
$query=mysql_query("Select pass from members where uname='$username'");
$result = mysql_query($query);
// If the number of rows is less than
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 what I
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