Sean Greenslade wrote:
[MASSIVE
SNIP]
Well, from what I saw while wading through your
code, you allow
unsanitized
variables to be
concatenated to your queries. Big no-no! For ANY
client-generated variable, always sanitize with
mysql_real_escape_string.
In
fact, sanitize all your
[SNIP]
added and else clause.
while ($_parent != 0)
{
if
($num_rows 0)
{
perform some action
}
else
{
$_parent =
0;
}
}
and that solved the
problem.
Thank you, everyone for your help.
Curtis
A small remark:
I think it is good programming practice to
On 11 May 2011 at 19:25, Curtis Maurand cur...@maurand.com wrote:
$_cartTotal=$0.00;
Surely that should be:
$_cartTotal = 0.00;
tim
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Tim Streater wrote:
On 11 May 2011 at 19:25, Curtis
Maurand cur...@maurand.com wrote:
$_cartTotal=$0.00;
Surely that should
be:
$_cartTotal = 0.00;
Good
pickup. I missed that. I didn't write the code, I'm just
trying to figure out what's going on.
Thanks, I'll look at
that. --C
I'm running PHP 5.3.6, Mysql 5.1.51 and Apache 2.2.17
I have
code that does a simple mysql_query(); the query on the commandline
returns an empty set. when run via PHP and the web server, the page
hangs, it never gets to the if (mysql_num_rows($result) 0) {}. and
the queries per second on
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script for
best results.
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Marc Guay wrote:
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script
for
best results.
It looks like the site is
under attack, because I keep seeing the query, SELECT catagory_parent FROM
t_catagories where catagory_ID= .
$_currentCat
where
Marc Guay wrote:
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script
for
best results.
It looks like the site is
under attack, because I keep seeing the query, SELECT catagory_parent FROM
t_catagories where catagory_ID= .
$_currentCat
where
On Wed, May 11, 2011 at 2:25 PM, Curtis Maurand cur...@maurand.com wrote:
Marc Guay wrote:
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script
for
best results.
It looks like the site is
under attack, because I keep seeing the query,
Hi all -
I could use a lead on a problem. I just don't know where to start.
I have a PHP script that populates a database table. No big deal. It creates
mailing labels. However, a weird things keeps happening - every once in a
while, a query is run twice. It is the same query, same
Jed R. Brubaker wrote:
I could use a lead on a problem. I just don't know where to start.
I have a PHP script that populates a database table. No big deal. It
creates
mailing labels. However, a weird things keeps happening - every once in a
while, a query is run twice. It is the same query,
Jed R. Brubaker wrote:
Hi all -
I could use a lead on a problem. I just don't know where to start.
I have a PHP script that populates a database table. No big deal. It creates
mailing labels. However, a weird things keeps happening - every once in a
while, a query is run twice. It is the same
I'm filling a database with info, and I'm using id as identifiers in the
list, the id is auto increment, and I deleted one entry, now I have a hole
in the database, is there any way to fix this?
lets say I have
1
2
3
4
5
6
And I deleted 6
Now it looks like
1
2
3
4
5
7
8
9 and so on
is there
Hi Hawk,
(snip auto-incrementing PKs)
is there somewhere the next number is located ?
No. Why would you care, anyway? The thing about PKs is that they have to be
unique, not sequential.
If you're *really* bothered by it, you'll have to dump the contents of the
table to a file, drop the
Primary Key's, by nature, are designed to always be uniqiue, which
means that even if you delete row 6, the next row you insert will be
10 because there is no id 10. If you simply need to get the list of
items in a query, in the order they were inserted, I would suggest using
something like:
Hi on this code:
$link = mysql_connect(localhost, login, passwd);
mysql_select_db(table);
$result = mysql_query(select * from table);
while ($row = mysql_fetch_object($result)) {
echo $row-ID;
echo $row-Drank;
}
mysql_free_result($result);
mysql_close($link);
It resulst in
Warning:
: Tuesday, March 12, 2002 5:19 PM
Subject: [PHP] mysql problems
Hi on this code:
$link = mysql_connect(localhost, login, passwd);
mysql_select_db(table);
$result = mysql_query(select * from table);
while ($row = mysql_fetch_object($result)) {
echo $row-ID;
echo $row-Drank
: . mysql_error());
-Teresa
-Original Message-
From: hugh danaher [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, March 12, 2002 8:27 PM
To: Maarten Weyn; php
Subject: Re: [PHP] mysql problems
Maarten,
Perhaps table isn't the name of the table you want.
If mysql can't find the table (in line 13
Having a problem with this, I have a working login that supports multiple
users, and I'm trying to make it possible for the users to change their own
settings etc, but I get killed when trying to send the data to the MySQL
database.
To connect I use mysql_connect($host,$user,$pswd) and
The reason you aren't seeing any errors is because you used or
die() -- to the best of my knowledge, this replaces any standard errors
that PHP would normally generate.
It appears that you have omitted the second argument to mysql_query().
A common mistake that I've made many times myself.
-
From: Hawk [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, January 17, 2002 1:18 PM
Subject: [PHP] PHP MySQL problems, updating database..
Having a problem with this, I have a working login that supports multiple
users, and I'm trying to make it possible for the users to change their
own
Hawk,
If you have a working login, can we safely assume that there is information
in the database for each user? If so, then we won't bother discussing
insert statements, but concentrate on updates. We'll also assume that $user
has update privileges on the database.
The normal form of an
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