On Sun, 2002-11-03 at 01:44, Jason Wong wrote:
On Sunday 03 November 2002 10:23, John Coder wrote:
I seem to somehow get an boolean instead of an array from a
mysql_fetch_row function and I have no idea how.
Here's the code;
Try adding some error checking into your code (see manual
It's hard to tell which output is from which loop, you might be getting
confused (if anything, you might be confusing anyone that might help)...
the Boolean was probabally due to a bad result, for whatever reason
(maybe your connect code was wrong? I'm not sure but I think you get a
valid
Hi everyone,
on my site I created a login which is supposed to be secure. I'm not
familiar with how to surpass signups, but was wondering if people can see if
they can get my page to view without signing up. The page that is supposed
to be secured is the about me index. (the rest is still open).
Howdy --
How do I dynamically generate the value for radio buttons off the DB
(MySQL) backend? Here's some code that almost works
TIA,
David
- Almost works
?php
$header = mysql_query (SELECT * FROM chart
ORDER BY acct );
if ($row =
change your query to this:
select count(distinct itemid) from business where name like 'word1 word2
word3%' or description like 'word1 word2 word3%';
Peter
On 4 Nov 2002, Chris Barnes wrote:
Hi,
I've got a dilly of a problem. I'm probably doing something wrong but I
don't know what. I'm
Gonna need some additional information, such as:
1. The first three rows your query returns
2. The expected output
3. The actual problem and/or error.
Peter
On Sun, 3 Nov 2002, David Jackson wrote:
Howdy --
How do I dynamically generate the value for radio buttons off the DB
(MySQL)
Chris Barnes wrote:
Hi,
I've got a dilly of a problem. I'm probably doing something wrong but I
don't know what. I'm trying to use the LIKE statement in a query where
more than one word is used in with LIKE..e.g.
select count(distinct itemid) from business where name or description
like 'word1
Peter Beckman wrote:
Peter,
Thanks for your prompt reply.
Gonna need some additional information, such as:
1. The first three rows your query returns
2. The expected output
3. The actual problem and/or error.
What I'm trying todo is build a radio box form
for the selection of ledger
On Sunday 03 November 2002 20:58, John Coder wrote:
Try adding some error checking into your code (see manual examples) and
using mysql_error() to find out what's going.
What error checking to insert into this code I'm clueless. I found out
Again, have a look at the examples in the manual
Thanks!
I was able to get native mysql support by adding
extension=mysql.so
to php.ini. So now mysql_connect() works
John Coder wrote:
While Micah is correct, you do have mysql support on php but as a dbs
not the built in php fuctions. Therefore you must use the dbx functions
instead.
Here's your problem:
acct[] doesn't equal anything.
$acct[4] might.
So:
echo input type='radio' name='gl_acct' value='{$acct[4]}';
might give you want you want.
The name='acct[]' is used for PHP form processing, putting the correct
value in the variable. For example, $acct[Yes] might be set
if you want to search for multiple words, u have to use multiple like
operators:
select count(distinct itemid) from business where name like 'word1' or
name like 'word2' or name like 'word3';
or the IN statement with wildcards:
select count(distinct itemid) from business where name IN
Hey all, I've had a bizarre experience. When I uploaded a php script it
worked fine and when I was looking through my code I found that it
shouldn't have worked because two variables had been used instead of
one. What I'm meaning is, I used $Year and $year where I was just
supposed to use $Year,
Yeah I really need to search for multiple words. Can anyone confirm if
the IN statement will work for me in this situation?
On Mon, 2002-11-04 at 09:31, [EMAIL PROTECTED] wrote:
if you want to search for multiple words, u have to use multiple like
operators:
select count(distinct itemid) from
You can't use wildcards with IN, only with LIKE or regular expressions.
---John Holmes...
-Original Message-
From: [EMAIL PROTECTED] [mailto:Amit_Wadhwa;Dell.com]
Sent: Sunday, November 03, 2002 5:31 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] LIKE statement or IN statement?
Chris Barnes wrote:
Yeah I really need to search for multiple words. Can anyone confirm if
the IN statement will work for me in this situation?
Chris --
Why not just try it you self and let's us know.
Also check to MySQL doc at http://mysql.org
David
On Mon, 2002-11-04 at 09:31, [EMAIL
Peter Beckman wrote:
Here's your problem:
acct[] doesn't equal anything.
$acct[4] might.
So:
echo input type='radio' name='gl_acct' value='{$acct[4]}';
Peter -- Thanks again for you help.
I wonder if it's not a quoting issue on my part?
How would I use the $row[column_name], in the above
ok so you would have to use :
--select count(distinct itemid) from business where name like 'word1' or
name like 'word2' or name like 'word3';
no other go.
..
You can't use wildcards with IN, only with LIKE or regular
On Sun, 2002-11-03 at 15:01, Jason Wong wrote:
On Sunday 03 November 2002 20:58, John Coder wrote:
Try adding some error checking into your code (see manual examples) and
using mysql_error() to find out what's going.
What error checking to insert into this code I'm clueless. I found out
ok so you would have to use :
--select count(distinct itemid) from business where name like 'word1'
or
name like 'word2' or name like 'word3';
no other go.
If you're not going to use wildcards, then you can use IN. The whole
idea of using LIKE is that you can use _ and % as wildcards when
I keep forgetting where clauses. It's my dumbest mistake, and I always
make it. I just lost the password of everyone on my site trying to
change my own - not good. Anyway, I'm trying to edit PhpMyAdmin to warn
if there's more than a certain number of affected rows for a query. Is
there a
You can suppress the error message on failure by prepending a
http://www.php.net/manual/en/language.operators.errorcontrol.php to
the function name..
How can you get the error string into a local variable?
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I have a php page on server #1 trying to access a database on server
#2. What do I put in the following blanks:
$hostname: ___ (have tried IP, http, localhost, nothing works)
$username: blah
$password: blah
$database: _
$link=mysql_connect($hostname, $user, $pass) or
Cassy M Rowe wrote:
I have a php page on server #1 trying to access a database on server
#2. What do I put in the following blanks:
$hostname: ___ (have tried IP, http, localhost, nothing works)
$username: blah
$password: blah
$database: _
$link=mysql_connect($hostname,
OK, this works but there has to be a pretty way?
David
- Works -
html
headtitleOperation Sticky Bun/title/head
body
h3 align=centerOperation Sticky Bun /h3
?php require('connect.php'); ?
?php
print 'form action=hello.php method=post';
$header = mysql_query
How do I check to see if a table exists?
On Monday 04 November 2002 12:10, Bob Lockie wrote:
You can suppress the error message on failure by prepending a
http://www.php.net/manual/en/language.operators.errorcontrol.php to
the function name..
How can you get the error string into a local variable?
Look at the track_errors setting
RTFM?
On Monday 04 November 2002 10:47, Leif K-Brooks wrote:
I keep forgetting where clauses. It's my dumbest mistake, and I always
make it. I just lost the password of everyone on my site trying to
change my own - not good. Anyway, I'm trying to edit PhpMyAdmin to warn
if there's more than a
On Monday 04 November 2002 07:25, Graeme McLaren wrote:
Hey all, I've had a bizarre experience. When I uploaded a php script it
worked fine and when I was looking through my code I found that it
shouldn't have worked because two variables had been used instead of
one. What I'm meaning is, I
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