On Thu, May 5, 2011 at 10:29, Grega Leskovšek wrote:
> Can smbd please look at this sentence - I got an error and do not
> know how to fix it - I am still very unfamiliar with MYSQL:
>
> CREATE TABLE log ( idlog int auto_increment not null, imepriimek
> varchar(50), clock timestamp, action va
Mysql Username or password error.please testing connect setting
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"NADARAJAH SIVASUTHAN NADARAJAH" , haber iletisinde
sunlari yazdi:col106-w49cf8a0029c9582f1c0161d5...@phx.gbl...
Dear all,
When I try to retrive data from two tables using JOIN OR INNER JOIN
it display error me
You're very welcome, I understand that you need a solution which allows you
to keep using the connection this way.
I doubt that your php.ini is the cause so I'll suggest the following:
Try executing your MySQL queries by passing the $link variable to the
function as the connection resource:
mysq
[snip]
There is a tool call "idera" (SQL diagnostic manager). Basically it is a
performance monitoring and diagnostics tool.
It has a feature;
Identifying of worst-performing codes -
Identifies performance bottlenecks such as the worst-performing stored
procedures, long-running queries, mos
You can use something like
$articleDate= date ("Y-m-d 00:00:00",strtotime($day ." " . $month . " "
. $year));
If $day = 10 and $month = "January" and $year = 2004
then $articleDate will be set to "2004-01-10 00:00:00"
Then you can insert $articleDate into a datetime field in mysql.
On Wed, 28 Jul 2004 02:36, Redhat wrote:
> I am getting a weird error on one of my servers that says something
> about "query requires full tablescan". I was not getting this error
> until today that I noticed. I Googled for a minute and saw where a ton
> of other sites had the same problem but
So given the same database diagram ("MovieMain", "Actors", "Genre", then a
table named "MovieActor" to allow multiple actors to be associated with a
movie, and "MovieGenre" to allow multiple genre's to be associated with a
movie), can you steer me in the right direction on writnig the PHP to
displa
Alexey Kupershtokh wrote:
Hi.
In trying to learn PHP and MySQL, I'm trying to create a web-based
database
application that will keep track of my movies. I have five tables (shown
in
the graphic). I'm trying to be able to display the results of the tables
with one movie title, all of the actors a
Hi.
> In trying to learn PHP and MySQL, I'm trying to create a web-based
database
> application that will keep track of my movies. I have five tables (shown
in
> the graphic). I'm trying to be able to display the results of the tables
> with one movie title, all of the actors associated with it,
Excellent! That's exactly what I needed to know! Thanks a million!
"Justin Patrin" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> News.Php.Net wrote:
> > Hi everyone!
> >
> > In trying to learn PHP and MySQL, I'm trying to create a web-based
database
> > application that will kee
News.Php.Net wrote:
Hi everyone!
In trying to learn PHP and MySQL, I'm trying to create a web-based database
application that will keep track of my movies. I have five tables (shown in
the graphic). I'm trying to be able to display the results of the tables
with one movie title, all of the actors
I've tried playing around with it a bit but I'm getting quite confused now.
It still doesn't like my declaration of variables and I'm not sure how to
declare 'id' either as it isn't used in the form as a field of entry - just
an index in the database.
Any help would greatly be appreciated cause I'
On Wed, 18 Feb 2004 09:35:22 -, Slippyaah wrote:
>Well basically I think it's a problem with the declaration of variables
>and/or the way they are being processed by the form.
>
>Here is the individual addNews function:
>
>function addNews() {
> global $db;
>
> /* declare variables
Hey David,
Well basically I think it's a problem with the declaration of variables
and/or the way they are being processed by the form.
Here is the individual addNews function:
function addNews() {
global $db;
/* declare variables here? */
/* insert the new news entry */
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
> hey,
>
> I've been working on a PHP/MYSQL weblog and I've got most of it working
> apart from the adding new news function and a login section. I've set the
> form up but it is coming up with a few errors and the actual submitting of
>
Hey, Also the webhost only allows us a direct database connection
using phpMyadmin, I did notice that on the table that stores the info, it
keep getting an error after someone is posting the form. The error says
something about Overhead: 275 bytes, Is this just an MySQL limitation that
is set by th
I have it running under Windows NT 4.0..no problems...
Shadow
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lt;[EMAIL PROTECTED]>
Sent: Sunday, 22 June, 2003 10:13
Subject: [PHP-DB] Re PHP, MySql
I have a web site interfacing to a mysql databases which eventually may have
a large number of users. Should I use persistent coneconnection mode or let
each screen make and drop a connection ?
Where will I fi
I have a web site interfacing to a mysql databases which eventually may have
a large number of users. Should I use persistent coneconnection mode or let
each screen make and drop a connection ?
Where will I find guide lines on this?
Is this the correct list for this question?
Regards
Peter Gog
[EMAIL PROTECTED] (Julie Keywell) writes:
> I am querying our database for 11,000 different zip codes and it seems
> to be overloading MySQL. However, common sense says it shouldn't.
>
> My question is:
> When needing to query 11,000 things, is it best to make 1 SQL statement
> that has 11,000 or
-Original Message-
From: Jason Wong [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, December 04, 2002 11:48 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Re: PHP/MySQL not available remotely...
On Thursday 05 December 2002 00:44, Chase wrote:
> Yes I am.
>
> I was just asked to
The phpinfo() call works great on the server itself, but I cannot get the
information to display to anyone that connects to the server.
"Jason Wong" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> On Thursday 05 December 2002 00:44, Chase wrote:
> > Yes I am.
>
On Thursday 05 December 2002 00:44, Chase wrote:
> Yes I am.
>
> I was just asked to create a simple web page that would display the
> visitors IP address using:
>
>
So you're saying that the above works ...
> and when I call this up over the intranet or internet it works just fine.
> However, I
com) or username@'%' (which would work
for
> > that username from any host)
> >
> >
> > hth
> > jeff
> >
> >
> >
> > "Chase"
> > > biz>
> "Chase"
>
>
> biz> cc:
>
>
Subject: [PHP-DB] Re: PHP/MySQL not
available remotely...
12/04/2002
Okay, it was suggested that I may not have my "register_globals" flagged to
"on" but I do. I still can't seem to make any remote connect to the server.
Again, there are no errors, it just flat don't work.
Chase
"Chase" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTE
MessageAccording to http://www.mysql.com/doc/D/I/DISTINCT_optimisation.html,
DISTINCT is converted to a GROUP BY on all columns, DISTINCT combined with
ORDER BY will in many cases also need a temporary table.
Instead of using DISTINCT, consider adding GROUP BY for all the columns
except ID.
--
look at http://www.webmasterbase.com/article/228
--
Best regards,
George Nicolae
IT Manager
___
X-Playin - Professional Web Design
www.x-playin.f2s.com
"Sg" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi,
>
> I'm trying to do a sort of
The error indicates that the result resource is invalid. The result
resource in your example is $rs. Either this is not the variable that you
used when sending the query or something went wrong with the query. You
should always add "or die(mysql_error) to the end of your mysql_query
statements,
I agree with mr. Torrent here...
you are not logged on to your mysql server when you issue the query!
Check your password/username...
> -Opprinnelig melding-
> Fra: torrentUK [mailto:[EMAIL PROTECTED]]
> Sendt: 20. november 2001 18:35
> Til: [EMAIL PROTECTED]
> Emne: [
I think you can get this error when you sql query fails.
Try copy and pasting the $sql query to the mysql command line and see what
response you get (it looks as though this would work though unless you
haven't got a table called "test").
I guess it might be that you are not successfully logging o
LIKE is a very good option. You can use:
LIKE \n'$id'\n;
Or similar, where \n stands for the Mysql carriage return 'symbol'.
Also, you may try selecting the whole text field, explod()ing contents by
carriage returns and checking with in_array() for the id if the above
doesn't works.
--
Use sql keyword LIKE, unless you have access to the design of the database,
in which case design it properly (i.e. something approaching normal form).
Tim
--
From: Peter Brown [SMTP:[EMAIL PROTECTED]]
Sent: 12 November 2001 05:28
To: [EMAIL PROTECTED];
Found the solution:
I had to change row
$fpName = tempnam($dir, "tableData");
to
$fpName = tempnam("", "tableData");
So file will be created to a temporary directory.
Niklas
P.S. :)
-Original Message-
From: Niklas Lampén [mailto:[EMAIL PROTECTED]]
Sent: 6. marraskuuta 2001 14:5
> if ($row = mysql_fetch_array($result)) {
>
> /* Change all instances of db_table to match your db table structure */
> do {
> PRINT "Your search results: ";
> PRINT "db_table: ";
> print $row["db_table"];
> print ("");
> .
> [ NUMEROUS REPETITIONS WERE HERE ]
> .
> P
they have something you need
http://phpwizard.net
--
Hidulf
http://www.hidulf.com
"Andrius Jakutis" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hello all,
>
> I need your recomendations, where to download from a good (profesional -
> many options) banner ex
> The code should not be on the page which comes
> after clicking, but on the same page where the button is.
if this is the requirement, then look maybe use javascript
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For additional commands, e-mail:
Trick is to have a server side script which increments a counter and
redirects to the desired page.
So you'd create something like this:
/counter?uri=
and in your database you'll have fields for the uri and the number of hits
so far.
goodluck :)
"Pranot Kokate" <[EMAIL PROTECTED]>
Try http://www.aletiahosting.com/. I have my sites hosted by them.
Excellent support especially the ones via AIM/ICQ. The people are always
around to answer your questions.
Plans start at $9.95/month.
Hope this helps.
Best wishes,
Indri
>> -Original Message-
>> From: Ken Sommers [mail
You might want to try nomonthlyfees.com. I host two sites with them (that's
my only relation) and I've been happy so far. It's $200 up front then
$35/year which includes the domain name registration.
Aral :)
__
([EMAIL PROTECTED])
New Media Producer, Kismia, Inc.
(
I used Cedant.com to host my own site for a few months, and they were
excellent - no down time that I was aware of, and a handy web-based control
panel (and of course SSH shell access for anything not covered by the
control panel.) Although they didn't have a large tech support phone bank
(Usuall
"Jens Kisters" <[EMAIL PROTECTED]> wrote:
> Hi there, is there a possibility to use MySQL's MATCH ... AGAINST to
> look for a string like "foo bar" ?
>
> if i MATCH keyname AGAINST ('foo bar') i get resuts that match foo or
> bar, but not both of them separated by a space.
>
> My guess is that t
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