Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt [EMAIL PROTECTED] wrote:
Umm, he is putting them into an array, I quote:
They don't do anything, but my point was, he said that what he pulled
from the DB needed to be put into an array, and I was pointing out, it
already was.
On Wed, 2002-11-27 at 12:23, Mark wrote:
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a
Add: print_r($row)
In your while loop, that will show you everything that is being returned
with both it's numeric and text based position.
On Tue, 2002-11-26 at 13:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search
Try this...
either...
while ($row = mysql_fetch_array($result)) {
$title = $row['Books.Title'];
$author = $row['Books.Author'];
...
print $title;
}
or...
while ($row = mysql_fetch_array($result)) {
print $row['Title'];
...
}
-Original
On Wednesday 27 November 2002 02:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array()
The quick answer is that that is what it's supposed to return... that's all
the result is. A nicer, longer answer is to give you some of my code so you
can see one way that you actually get the data out (I use sybase_fetch_row,
but you can also use db_fetch_array which returns an associated
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps giving
me this:
?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy =
; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] resource id#2 -
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps
giving me this: ?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy
Thanks a ton, I was really looking right past that variable.
Jas
Paul Dubois [EMAIL PROTECTED] wrote in message
news:p05111763b92d3addd3b4@[192.168.0.33]...
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps
giving
me this:
?php
/* Get
On Thursday 30 May 2002 00:17, Jas wrote:
Ok here is my problem, I set this up so a user selects a name form a select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into
If you look at the previously posted code at the bottom of the form there is
a echo for the sql select statement that is echoing Resource id #2 on the
page. Now that error is the correct field id number in the database, I am
just not sure how to itemize the data from that table, at least I
I could be missing something, but it looks like you are using the result of
the mysql_query as the actual result. It actually returns some weird
identifier. To access the real info you'd have to use something like
mysql_fetch_array to get it.
Check out this and see if it helps:
On Thursday 30 May 2002 00:41, Jas wrote:
If you look at the previously posted code at the bottom of the form there
is a echo for the sql select statement that is echoing Resource id #2 on
the page. Now that error is the correct field id number in the database, I
am just not sure how to
How can I get the form below to list the single record in a db that matches
the request of $user_id?
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
while ($row = mysql_fetch_row($record)) {
$user_id = $row['user_id'];
$f_name =
: [EMAIL PROTECTED]
Sent: Wednesday, January 23, 2002 12:21 PM
Subject: Re: [PHP-DB] Resource Id #2
Use something like this...
...
...
...
$rows = mysql_num_rows($result);
for ($y = 0; $y $rows; $y++){
$data = mysql_fetch_object($result);
echo $data-Field_Name
AHAHAHHAHA
AHHAHAHAH
Dan
On Wednesday, January 23, 2002, at 02:19 PM,
[EMAIL PROTECTED] wrote:
Dan,
Good, I'm glad it worked!!!
I see a lot of people using mysql_fetch_array().
Which is fine...But I'm more into Object Oriented programming...and
that's why I always use
On Thursday 29 November 2001 01:29, Kevin Ruiz wrote:
I've come across yet another problem.
[snip]
$ci = select contactid from users where username='$username' and
password='$password';
$cir = mysql_query($ci)
or die(Couldn't execute);
$query = select
I am trying to select a message from a table in a database. the message will will be
in column 'msg' and will be
in same row as id='0'.
I need that msg put to the screen. With the code below I get the error: Resource id #2
Can someone please explain how I can fix this..
Thankyou in advance.
?
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