2007. 03. 21, szerda keltezéssel 00.04-kor Richard Lynch ezt írta:
> On Tue, March 20, 2007 11:08 am, Ford, Mike wrote:
> >> what do you want with that '@' here?
> >> that operator can be used to suppress error messages when calling
> >> functions but not when using a variable
>
> This is most def
On Tue, March 20, 2007 11:08 am, Ford, Mike wrote:
>> what do you want with that '@' here?
>> that operator can be used to suppress error messages when calling
>> functions but not when using a variable
This is most definitely way wrong.
> What complete tosh! @ is a unary operator, so can be app
2007. 03. 20, kedd keltezéssel 16.08-kor Ford, Mike ezt írta:
> On 20 March 2007 13:26, Németh Zoltán wrote:
>
> > 2007. 03. 20, kedd keltezéssel 15.09-kor Pavel Kaznarskiy ezt írta:
> > > Hello !
> > > I have problem with access in mysql
> > >
> > > it is my code:
> > >
> > > SQL Query Sender
>
On 20 March 2007 13:26, Németh Zoltán wrote:
> 2007. 03. 20, kedd keltezéssel 15.09-kor Pavel Kaznarskiy ezt írta:
> > Hello !
> > I have problem with access in mysql
> >
> > it is my code:
> >
> > SQL Query Sender
> >
> > > $host="";
> > $user="";
> > $password="";
> > /* Section that execute
Németh Zoltán wrote:
> 2007. 03. 20, kedd keltezéssel 15.09-kor Pavel Kaznarskiy ezt írta:
>> Hello !
...
>
> what do you want with that '@' here?
> that operator can be used to suppress error messages when calling
> functions but not when using a variable
>
not true - although it's a lazy/bad
> >
> > Warning: mysql_select_db(): Access denied for user
> 'ODBC'@'localhost' (using password: NO) in
> z:\home\localhost\www\2.php on line 12
> >
> > Warning: mysql_select_db(): A link to the server could not
> be established in z:\home\localhost\www\2.php on line 12
> >
> > Warning: mysql_qu
On 3/20/07, Pavel Kaznarskiy <[EMAIL PROTECTED]> wrote:
Hello !
I have problem with access in mysql
it is my code:
SQL Query Sender
{$_POST['database']}
Query: $queryResults";
if($result == 0)
echo "Error ".mysql_errno().": ".mysql_error().
"";
elseif (@mysql_num_rows($result) == 0)
echo("Quer
2007. 03. 20, kedd keltezéssel 15.09-kor Pavel Kaznarskiy ezt írta:
> Hello !
> I have problem with access in mysql
>
> it is my code:
>
> SQL Query Sender
>
> $host="";
> $user="";
> $password="";
> /* Section that executes query */
> if(@$_GET['form'] == "yes")
what do you want with that '@'
IL PROTECTED]
Sent: Thursday, 24 July 2003 8:24 AM
To: David Nicholson
Cc: PHP List
Subject: Re: [PHP] Problem with MySQL Query
Tried NOT LIKE and that didnt exclude it either.
I am trying to exclude only 'Meal Plans'
Phil
- Original Message -
From: "David Nicholson" <
dnesday, July 23, 2003 6:19 PM
Subject: Re: [PHP] Problem with MySQL Query
> Hello,
>
> This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at
> 22:54, lines prefixed by '>' were originally written by you.
>
> > > $dbqueryshipping1 = "s
Hello,
This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at
22:54, lines prefixed by '>' were originally written by you.
> > $dbqueryshipping1 = "select * from tempuserpurchase
where
> > (usersessionid="$User_Shopping_Id") and day="1" and
> > type<>'Meal Plans'
> Tryed both me
> > $dbqueryshipping1 = "select * from tempuserpurchase where
> > (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and
> > type<>'Meal Plans'
> >
>
> Tryed both methods and it is still not excluding anything matching Meal
> Plans
Been a short while since I used SQL with my PHP, but try pu
>
> $dbqueryshipping1 = "select * from tempuserpurchase where
> (usersessionid=\"$User_Shopping_Id\") and day=\"1\" and
> type<>'Meal Plans'
>
Tryed both methods and it is still not excluding anything matching Meal
Plans
---
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Hello,
This is a reply to an e-mail that you wrote on Wed, 23 Jul 2003 at
22:28, lines prefixed by '>' were originally written by you.
> As per your suggestion
> $dbqueryshipping1 = "select * from tempuserpurchase where
> (usersessionid="$User_Shopping_Id") and day="1" and
type!='Meal
> Plan
As per your suggestion
$dbqueryshipping1 = "select * from tempuserpurchase where
(usersessionid=\"$User_Shopping_Id\") and day=\"1\" and type!='Meal Plans'
";
$resultshipping1 = mysql_db_query($dbname,$dbqueryshipping1);
if(mysql_error()!=""){echo mysql_error();}
$result1 = my
Take out the plus sign... type != 'Meal Plans'
And using single quotes in your query might make things easier (no
escaping...)
$dbqueryshipping1 = "select * from tempuserpurchase where
usersessionid='$User_Shopping_Id' and day='1' and type!='Meal Plans'";
You don't need quotes around '1' since i
source versions of MySQL, in that
case, use
--with-mysql=/usr/local
- Original Message -
From: "Jon Haworth" <[EMAIL PROTECTED]>
Newsgroups: php.general
To: "Ryan Vennell" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Monday, March 31, 2003 5:0
;
Sent: Monday, March 31, 2003 5:00 PM
Subject: RE: [PHP] problem with mysql.
> Hi Ryan,
>
> > when configuring php i use --with-mysql and it configures just
> > fine. i've even added an =/path/to/php after it to no avail.
>
> Try --with-mysql=/path/to/mysql inste
Sorry about that. i did use /path/to/mysql. i wasnt paying attention when writing my
pervious post.
-Ryan
>>> Jon Haworth<[EMAIL PROTECTED]> 03/31/03 04:00PM >>>
Hi Ryan,
> when configuring php i use --with-mysql and it configures just
> fine. i've even added an =/path/to/php after it to no
Hi Ryan,
> when configuring php i use --with-mysql and it configures just
> fine. i've even added an =/path/to/php after it to no avail.
Try --with-mysql=/path/to/mysql instead of --with-mysql=/path/to/php.
Cheers
Jon
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit:
if you check on the FIRST insert (the primary table that holds "master"
usernames) to ensure the uname doesn't exist, THEN do the other queries,
there shouldn't be a problem.
I think the key here is that you're using the username as the unique key,
and putting that username in multiple tables...
MySql should insert a value into that column when you update if you are using an
auto_increment field. this value will always be unique. period.
-Chris
-Original Message-
From: Chad Day [mailto:[EMAIL PROTECTED]]
Sent: Wed 2/19/2003 1:16 PM
To
Ok. Go to this link...
http://www.magidesign.com/movielist.php and select "12 Monkeys" You will see
that only picked one actor...
It should have pick four
mysql> SELECT concat_ws(" ", fname, lname) as actor FROM actormovie WHERE
title = "12 Monkeys" ORDER BY lname;
+-+
| act
you need to put your $myrow in a while loop:
while ($myrow = mysql_fetch_array($result)) {
$title = $myrow[title];
$videoid = $myrow[videoid];
$catergory = $myrow[catergory];
$appraisal = $myrow[appraisal];
// blah blah blah everything else
}
Tyler Longren
Captain Jack Communications
www.capta
On Thursday 11 April 2002 20:40, Roman Duriancik wrote:
> Please send me info : how to set path for *.pid file and ho to set path
> where will be mysql.sock in mysql database on linux ?
> i want : mysql.sock and localhost.pid in directory /var/lib/mysql
>
> i install mysql with this command...
>
On Fri, 22 Jun 2001 19:52, Null wrote:
> In a script I have an update query adding on to a LONGTEXT field in my
> database. Strangely it will no longer work after seemingly random
> string lengths. So far, one row stopped adding at 440 bytes and another
> at 1049 bytes.
>
> mysql_query("UPDATE dod
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