$sql = "SELECT count(Email) as numEmails, Email FROM mena_guests WHERE
Voted='yes' GROUP BY Email ORDER BY numEmails DESC LIMIT $startingID,
$items_numbers_list";
I answered this morning, I don't know why it got deleted
--
itoctopus - http://www.itoctopus.com
""Me2resh Lists"" <[EMAIL PROTECTED]>
$sql = "SELECT count(Email) as numEmails, Email FROM mena_guests WHERE
Voted='yes' GROUP BY Email ORDER BY numEmails DESC LIMIT $startingID,
$items_numbers_list";
--
itoctopus - http://www.itoctopus.com
""Me2resh Lists"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> hi
> i need hel
On 5/3/06, Richard Lynch <[EMAIL PROTECTED]> wrote:
On Tue, May 2, 2006 7:22 am, chris smith wrote:
> On 5/2/06, Ross <[EMAIL PROTECTED]> wrote:
>> This is my database now...I will use the item_id for the order but
>> what if I
>> want to change item_id 3 to item id 1? How can I push all the item
On Tue, May 2, 2006 7:22 am, chris smith wrote:
> On 5/2/06, Ross <[EMAIL PROTECTED]> wrote:
>> This is my database now...I will use the item_id for the order but
>> what if I
>> want to change item_id 3 to item id 1? How can I push all the items
>> down one
>> place? How can I delete any gaps when
On Tue, May 2, 2006 7:05 am, Ross wrote:
> This is my database now...I will use the item_id for the order but
> what if I
> want to change item_id 3 to item id 1? How can I push all the items
> down one
> place? How can I delete any gaps when items are deleted.
Change item_id 3 to 1.
... select i
Exactly - I don't think you really understand how a relational database
works. The ids are retained as they may relate to records in another table.
Internal sorting order is of no relevance at the application level. I think
you need to rethink your design a little.
On 02/05/06, T.Lensselink <[EMA
> This is my database now...I will use the item_id for the order but what if
> I
> want to change item_id 3 to item id 1? How can I push all the items down
> one
> place? How can I delete any gaps when items are deleted.
>
>
> CREATE TABLE `board_papers` (
> `id` int(4) NOT NULL auto_increment,
>
On 5/2/06, Ross <[EMAIL PROTECTED]> wrote:
This is my database now...I will use the item_id for the order but what if I
want to change item_id 3 to item id 1? How can I push all the items down one
place? How can I delete any gaps when items are deleted.
Why do you want to do that? There's no be
This is my database now...I will use the item_id for the order but what if I
want to change item_id 3 to item id 1? How can I push all the items down one
place? How can I delete any gaps when items are deleted.
CREATE TABLE `board_papers` (
`id` int(4) NOT NULL auto_increment,
`doc_date` va
Ross schrieb:
Just say I have a db
CREATE TABLE `mytable` (
`id` int(4) NOT NULL auto_increment,
`fileName` varchar(50) NOT NULL default '',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;
when I add items they go id 1,2,3 etc. Whn I delete them gaps appear. 1
Ross wrote:
Hi all,
I am trying to create a table on the remote server but it never seems to
work
CREATE TABLE `sheet1` (
`id` int(10) NOT NULL auto_increment,
`title` varchar(255) NOT NULL default '',
`fname` varchar(255) NOT NULL default '',
`sname` varchar(255) default NULL,
`job_tit
[snip]
1064 - You have an error in your SQL syntax. Check the manual that
corresponds to your MySQL server version for the right syntax to use
near
'DEFAULT CHARSET=latin1 AUTO_INCREMENT=303' at line 18
and this is what the manual says (not very helpful)
a.. Error: 1064 SQLSTATE: 42000 (ER_PA
Hi
Add records with this code.
name
address
Hope this helps.
Jan
"Nicolai Elmqvist" <[EMAIL PROTECTED]> skrev i meddelandet
news:[EMAIL PROTECTED]
> Hi
>
> I have just started working with PHP and MySQL and have gone through 3
> tutorials on how to add and delete records from a databa
Felipe,
I'm replying in spanish so you can understand better.
El problema que tienes es que la variable $nueva_base esta vacia. Si
deseas especificar el nombre con esa variable debes asignarle algun
valor antes de llamarla, si lo que quieres es crear una base de datos
que se llame nueva_base,
> Dimension Networks, Inc.
> (727) 723-8388
>
> -Original Message-
> From: Daniel Elenius [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, January 14, 2003 5:22 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] MySQL problem with RedHat 8
>
> Yes, mysql.so is in /usr
Yes, mysql.so is in /usr/lib/php4. The php.ini file has this in it:
[daniel@p85 etc]$ grep mysql php.ini
;extension=php_mysql.dll
extension=mysql.so
mysql.allow_persistent = On
mysql.max_persistent = -1
mysql.max_links = -1
; Default port number for mysql_connect(). If unset, mysql_connect()
will
Make sure that the shared module is in the correct directory.
Check your php.ini file to make sure but it is most likely at /usr/lib/php4
make sure that you have mysql.so
- Original Message -
From: "Daniel Elenius" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, January 14, 2003
Look into the logs, they should be more verbose. How did you install the
three.
BB wrote:
>I seem to have a php-mysql problem on my new Sun Qube3 (RH Linux).
>php works fine, mysql works fine, apache works fine, only the combination of
>the three seems troublesome.
>php does not recognize comma
I understand about the concat function, but that doesn't really fit into my
scheme of things
I run all text for the web through a function SafeSQL so that values from
the web don't make SQL error or potential hacks occur.
All SafeSQL was doing (for mssql, access and just about any other db) was
> INSERT INTO `contracts` (`key`, `content`) VALUES (1,'blah blah blah
> character 39 is a single speach mark '+CHAR(39)+' blah blah blah')
That's because you are adding strings together, not concatenating them
(this isn't javascript!)
Use CONCAT() in MySQL to join strings together.
mysql> sel
- Original Message -
From: David Jackson
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED] ; [EMAIL PROTECTED]
Sent: Monday, December 31, 2001 2:41 AM
Subject: Re: [PHP] MySQL problem
I almost forgot add a or mysql_error() for each line like this:
This will return "
> From: David Jackson
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Sent: Monday, December 31, 2001 1:48 AM
> Subject: Re: [PHP] MySQL problem
>
>
> Here's the example from the PHP manual:
> The tutorial here are very helpfull:
>
* GoodFella ([EMAIL PROTECTED]) [Dec 30. 2001 21:10]:
> Thanks for the quick reply. I used the PHP manual example and it connects
> to the database successfully but cannot select the database.
So you are using this line:
> > mysql_select_db("booktest");
Correct? What does the server "say" i
rom: David Jackson
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Monday, December 31, 2001 1:48 AM
Subject: Re: [PHP] MySQL problem
Here's the example from the PHP manual:
The tutorial here are very helpfull:
http://www.melonfire.com/community/columns/trog/
-- David
Here's the example from the PHP manual:
The tutorial here are very helpfull:
http://www.melonfire.com/community/columns/trog/
-- David
\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
print "\t\n";
foreach ($line as $col_value) {
print "\t\t$col_value\n";
}
The issue here is that you aren't getting an index of "Company" from that
query. It is probably stored under the index of "UCASE(Company)". Try:
$Query = "SELECT UCASE(Company) as ucCompany, Icons, ID,...";
Then the ucCompany field will contain your capitalized company data.
Jon
-Origina
"SELECT UCASE(Company) AS Company" works great, thanks!
Niklas
-Original Message-
From: Dimitris Kossikidis [mailto:[EMAIL PROTECTED]]
Sent: 2. marraskuuta 2001 11:11
To: 'Niklas Lamp¨¦n'
Cc: PHP General
Subject: RE: [PHP] mySQL problem
Try this
$Query = &
Try this
$Query = "SELECT UCASE(Company) as company, Icons, ID, LogoD FROM
feComps";
List ( $company, $icons, etc ) = mysql_fetch_row( );
> -Original Message-
> From: Niklas Lamp&igr;n [mailto:[EMAIL PROTECTED]]
> Sent: Friday, November 02, 2001 10:22 AM
> To: Php-General
> Subj
Try this
$Query = "SELECT UCASE(Company) as company, Icons, ID, LogoD FROM
feComps";
List ( $company, $icons, etc ) = mysql_fetch_row( );
> -Original Message-
> From: Niklas Lamp&igr;n [mailto:[EMAIL PROTECTED]]
> Sent: Friday, November 02, 2001 10:22 AM
> To: Php-General
> Subj
Sorry!!! I'm stupid! I forgot to mention that the list of causes has to be
for a specified accident_report.weekending
Cheers
Simon
> -Original Message-
> From: Don Read [mailto:[EMAIL PROTECTED]]
> Sent: 04 July 2001 23:21
> To: Simon Kimber
> Cc: [EMAIL PROTECTED]
On 04-Jul-01 Simon Kimber wrote:
> Hi All,
>
> Does anyone know if this can be done with one query?
>
> I have to create a chart based on info in two tables that are four tables
> apart.
>
> Here are the relevant tables and just the most relevant fields...
>
> accident_report
> - ID
> - weeke
Generally spoken, echo the SQL-Statement, and paste it in your local MySql
Client (e.g. MySql-Font). These Frontends give you a better error, and
you´ll find the problem in seconds. (hope so)
""Peter Houchin"" <[EMAIL PROTECTED]> schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]..
[EMAIL PROTECTED]
www.phpbeginner.com
-Original Message-
From: Brian Rue [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 26, 2001 12:24 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] MySQL problem...
Here's all the code that uses MySQL...
$db = mysql_connect("localhost",&
Here's all the code that uses MySQL...
$db = mysql_connect("localhost","user","pass");
mysql_select_db("db",$db);
$gmdquery="SELECT * FROM game_of_the_day";
$the_info = mysql_query($gmdquery,$db);
while ($myrow = mysql_fetch_row($the_info)) {
(get info from the result)
}
... (dec
it might also be the characters escaping.
Sincerely,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
-Original Message-
From: Brian Rue [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 26, 2001 9:28 AM
To: [EMAIL P
some code would be nice to have a look at :)
Other than that, check table names, database names, also your result lines, I've found
i get that error by not calling a result or calling the incorrect table/database
Peter
-Original Message-
From: Brian Rue [mailto:[EMAIL PROTECTED]
identifier. Grr..
Thanks again. Your a deadline saver :)
- John Vanderbeck
- Admin, GameDesign
> -Original Message-
> From: Christian Reiniger [mailto:[EMAIL PROTECTED]]
> Sent: Saturday, March 10, 2001 2:32 PM
> To: PHP User Group
> Subject: Re: [PHP] MySQL problem - stum
that
call.
- John Vanderbeck
- Admin, GameDesign
> -Original Message-
> From: Julian Wood [mailto:[EMAIL PROTECTED]]
> Sent: Saturday, March 10, 2001 2:28 PM
> To: John Vanderbeck; PHP User Group
> Subject: Re: [PHP] MySQL problem - stumped
>
>
>
> Well, PHP
On Saturday 10 March 2001 19:58, you wrote:
> The following code is giving an me problems, I can't figure it out to
> save my soul. The last line gives:
>
> Here is the code:
>
> $link = db_connect();
> $query = "UPDATE Users SET firstname='$firstname', lastname='$lastname'
> WHERE username='$use
> > >The following code is giving an me problems, I can't figure it
> > out to save
> > >my soul. The last line gives:
> > >
> > >Here is the code:
> > >
> > >$link = db_connect();
> > >$query = "UPDATE Users SET firstname='$firstname', lastname='$lastname'
> > >WHERE username='$user' && password
&. But the
> problem remains. Same error.
> Any other ideas?
>
> - John Vanderbeck
> - Admin, GameDesign
>
>> -Original Message-
>> From: Rick St Jean [mailto:[EMAIL PROTECTED]]
>> Sent: Saturday, March 10, 2001 2:09 PM
>> To: PHP User Grou
> From: Rick St Jean [mailto:[EMAIL PROTECTED]]
> Sent: Saturday, March 10, 2001 2:09 PM
> To: PHP User Group
> Subject: Re: [PHP] MySQL problem - stumped
>
>
> At 01:58 PM 3/10/01 -0500, John Vanderbeck wrote:
>
>
> You are using && in your statement ... should b
At 19:58 10.03.2001, John Vanderbeck said:
[snip]
>$query = "UPDATE Users SET firstname='$firstname', lastname='$lastname'
>WHERE username='$user' && password='$password'";
[snip]
Try "AND" instead of "&&", this shou
At 01:58 PM 3/10/01 -0500, John Vanderbeck wrote:
You are using && in your statement ... should be "AND"
.
>The following code is giving an me problems, I can't figure it out to save
>my soul. The last line gives:
>
>Here is the code:
>
>$link = db_connect();
>$query = "UPDATE Users SET firs
I think if you echo the error (echo mysql_error();) you will see the exact
error you would get if connected directly to the db.
You can do something like:
$result = @mysql_query($sql_query); // hide any errors
If (!$result) {
echo "Error: " . Mysql_error() . "";
}
On 2/20/01 2:57 AM this wa
TECTED]]
> Sent: Monday, February 19, 2001 17:00
> To: PHP User Group
> Subject: Re: [PHP] mysql problem
>
>
> Ok, I've fixed the problem, it seems there's some major differences
> between the linux / win32 ports of the client.
>
> instead of
>
> blah in
AIL PROTECTED]>
Sent: Tuesday, February 20, 2001 11:42 AM
Subject: Re: [PHP] mysql problem
> Nope, I've been using autoincrement on that box for a year or so.
>
> It's not a copy/paste thing, cause I'm getting it when I type the lines
> in by hand, too...
>
> Gfu
hangovers, and hangovers to... suffering.
- Original Message -
From: "David Robley" <[EMAIL PROTECTED]>
To: "Josh G" <[EMAIL PROTECTED]>; "PHP User Group"
<[EMAIL PROTECTED]>
Sent: Tuesday, February 20, 2001 11:41 AM
Subject: Re: [PHP] my
On Tue, 20 Feb 2001 10:57, Josh G wrote:
> Hi, sorry to post this here, but it's driving me crazy. On my local
> machine, the following works no furys:
>
> create table category (category_id integer primary key
> auto_increment,name varchar(255) );
>
> But on the production machine, I get:
>
> ERR
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