. - . mysql_error());
Thank you for any help !
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Hi,
I created a form asking username, password, country, etc.
On the submit of this form I make a sql connection and update the database,
add the user.
The problem is that whenever the field 'password' is filled in,
it (I don't know what) is asking to confirm the change of the password.
I
On Wed, August 29, 2007 2:03 pm, debussy007 wrote:
I created a form asking username, password, country, etc.
On the submit of this form I make a sql connection and update the
database,
add the user.
The problem is that whenever the field 'password' is filled in,
it (I don't know what) is
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Good morning friends,
I have a script that collects data from a form and puts together a mysql
query to search a database.
Now, everything worked fine until I added a few new form fields... now the
$_POST['var'] don't reach the script...
I have about 20 to 25 form fields which are all taken
2007. 05. 21, hétfő keltezéssel 09.26-kor Christian Haensel ezt írta:
Good morning friends,
I have a script that collects data from a form and puts together a mysql
query to search a database.
Now, everything worked fine until I added a few new form fields... now the
$_POST['var'] don't
Németh [EMAIL PROTECTED]
To: Christian Haensel [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, May 21, 2007 9:36 AM
Subject: Re: [PHP] PHP MySQL Problem
2007. 05. 21, hétfő keltezéssel 09.26-kor Christian Haensel ezt írta:
Good morning friends,
I have a script that collects data
MySQL Problem
2007. 05. 21, hétfő keltezéssel 09.26-kor Christian Haensel ezt írta:
Good morning friends,
I have a script that collects data from a form and puts together a mysql
query to search a database.
Now, everything worked fine until I added a few new form fields... now
] PHP MySQL Problem
2007. 05. 21, hétfő keltezéssel 09.43-kor Christian Haensel ezt írta:
Right, here we go... this page has about 1000 lines of code, so here the
relevant stuff.
$s_query = SELECT * FROM database WHERE
kategorie LIKE '$s_cat' AND
marke LIKE '$s_marke' AND
leistung_kw
Sent: Monday, May 21, 2007 9:58 AM
Subject: Re: [PHP] PHP MySQL Problem
2007. 05. 21, hétfő keltezéssel 09.43-kor Christian Haensel ezt írta:
Right, here we go... this page has about 1000 lines of code, so here the
relevant stuff.
$s_query = SELECT * FROM database WHERE
kategorie
10:16 AM
Subject: Re: [PHP] PHP MySQL Problem
2007. 05. 21, hétfő keltezéssel 10.04-kor Christian Haensel ezt írta:
I thought so, too. But the part where it receives the data is this:
$s_marke = $_POST['marke'];
$s_modell = $_POST['modell'];
$s_preis_von = $_POST['preis_von'];
$s_preis_bis
On Mon, May 21, 2007 2:26 am, Christian Haensel wrote:
Good morning friends,
I have a script that collects data from a form and puts together a
mysql
query to search a database.
Now, everything worked fine until I added a few new form fields... now
the
$_POST['var'] don't reach the
a great coding-day :o))
Chris
- Original Message -
From: Richard Lynch [EMAIL PROTECTED]
To: Christian Haensel [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, May 22, 2007 2:42 AM
Subject: Re: [PHP] PHP MySQL Problem
On Mon, May 21, 2007 2:26 am, Christian Haensel wrote
hi
i need help regarding a sql query in my php app.
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
i want to sort this query by the number of the repeated EMail counts.
can anyone help me with that please ?
Me2resh Lists wrote:
hi
i need help regarding a sql query in my php app.
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
i want to sort this query by the number of the repeated EMail counts.
can anyone help me
$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 help
Me2resh Lists wrote:
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
the only php I see it $SQL,$startingID,$items_numbers_list. This is a
mysql question.
--
PHP General Mailing List (http://www.php.net/)
To
i need help regarding a sql query in my php app.
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE
Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
i want to sort this query by the number of the repeated EMail counts.
can anyone help me with that please ?
A)
clive wrote:
Me2resh Lists wrote:
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
the only php I see it $SQL,$startingID,$items_numbers_list. This is a
mysql question.
so, you don`t know the answer, right?
cajbecu wrote:
clive wrote:
Me2resh Lists wrote:
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
the only php I see it $SQL,$startingID,$items_numbers_list. This is a
mysql question.
so, you don`t know the
cajbecu wrote:
clive wrote:
Me2resh Lists wrote:
the query is :
$SQL = SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list;
the only php I see it $SQL,$startingID,$items_numbers_list. This is a
mysql question.
so, you don`t know the answer,
if I asked you whether your wife/girlfriend enjoyed having sex with
your best friend while you are at work then you may or may not know the
answer but it is hopefully clear that this is not the list to ask such a
question (but no doubt that there is a list somewhere that caters to that
kind of
$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] wrote
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 id
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 items are
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 items
down
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, 3,
7. I
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.
- Original Message -
From: Ross [EMAIL PROTECTED]
To: php-general@lists.php.net
Sent: Tuesday, May 02, 2006 12:00 PM
Subject: [PHP] php mysql problem
Just say I have a db
CREATE TABLE `mytable` (
`id` int(4) NOT NULL auto_increment,
`fileName` varchar(50) NOT NULL default
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`
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
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,
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
At 11:00 AM +0100 5/2/06, Ross wrote:
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
Depending on your needs, here is another tack to think about (I know,
everyone else will be gasping and running for air...)
A few years ago I had to make a shopping cart. I had an archaic system
that would not share information with my web/db server, so I had to
write a query to dump the
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.
CREATE TABLE `board_papers` (
`id` int(4) NOT NULL auto_increment,
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.
CREATE TABLE `board_papers` (
`id` int(4) NOT NULL auto_increment,
On Tue, May 2, 2006 5:00 am, Ross wrote:
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
[snip]
$sql = SELECT * FROM `table1` LEFT JOIN `table2` USING `id` WHERE ...
GROUP BY `table1`.`id`;
mysql_query($sql);
the problem is, that, when in table2 is not matching data using that id,
i lose that id from output array. but i don`t want to.. is there any
posibility to keep that id?
thanks for your posting, but I have:
table1
id
name
value
table2
id
x
y
i want to
id name value x y
but there is no id from table 2 that is in table2.id and i want the
resource to bu something like this:
id(from table 1) name value (empty) (empty)
Jay Blanchard wrote:
[snip]
$sql =
[snip]
table1
id
name
value
table2
id
x
y
i want to
id name value x y
but there is no id from table 2 that is in table2.id and i want the
resource to bu something like this:
id(from table 1) name value (empty) (empty)
SELECT table2.*
FROM table1 LEFT OUTER JOIN table2
ON(table1.id =
[snip]
SELECT table1., *table2.*
FROM table1 LEFT OUTER JOIN table2
ON(table1.id = table2.id)
[/snip]
Oops, typo.
SELECT table1.*, table2.*
FROM table1 LEFT OUTER JOIN table2
ON(table1.id = table2.id)
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thanks a lot for your posting,
[cut]
Posted by Fred Mitchell on December 11 2004 2:47pm [Delete] [Edit]
Let's say you are doing a LEFT JOIN with a table that shares a column
name in common with another table, and that you are selecting for
instances where the join is missing, that is IS
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,
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_title`
[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
I have installed PHP5 and i get the following error from a script that was
working on PHP4
Fatal error: Call to undefined function mysql_pconnect() in
D:\htdocs\cdalex\Connections\listacon.php on line 9
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Hi,
I am generating a mysql statement and then printing it to a formatted field.
Here is my code:
$sql=SELECT * from tablename where name in ('Web','HTML','PHP') group by
name;
$mysql_rslt1 = mysql_query($sql, $mysql_bconn)
or die (Could not get data);
while ($rec1 = mysql_fetch_array
Jianping Zhu wrote:
Hi, thank you for your respinse.
It the /etc/php.ini i have something like:
;
; extension=modulename.extension
;
; For example:
;
extension=mysql.so
if you go to
http://coopunit.forestry.uga.edu:8080/test.php
you will find mysql is not installed as an extension.
but i do
I have redhat 9.0 and Server version: Apache/2.0.40.
i have installed rpms php-4.2.2-17.2.i386.rpm
php-mysql-4.2.2-17.2.i386.rpm
After i create a database called mydb and serveral tables in mysql,
I tried to run following testdb.php script
Hello,
[snip]
but i got error message with:
http://coopunit.forestry.uga.edu:8080/testdb.php
the error is:
Fatal error: Call to undefined function:
mysql_connect() in /var/www/html/testdb.php on line 13
How can Fix this problem? Thanks
[/snip]
Your PHP is compiled without
Hi, thank you for your respinse.
It the /etc/php.ini i have something like:
;
; extension=modulename.extension
;
; For example:
;
extension=mysql.so
if you go to
http://coopunit.forestry.uga.edu:8080/test.php
you will find mysql is not installed as an extension.
but i do not know how to
Thank you for your response.
I get the php rpm distributed with redhat 9.0.
Do I have to recompile phd instead from source code with mysql support?
Thanks
On Thu, May 13, 2004 at 06:15:23PM +0200, Oliver Hankeln wrote:
Hello,
[snip]
but i got error message with:
Jianping Zhu wrote:
Thank you for your response.
I get the php rpm distributed with redhat 9.0.
Do I have to recompile phd instead from source code with mysql support?
Thanks
You need to install the php-mysql rpm. It will be on one of your Red
Hat Discs.
--
John C. Nichel
KegWorks.com
Hi all,
We just installed PHP 4.3.6 on FreeBSD 5.2.1 on some SMP machines and we are
experiencing some weird problems.
The PHP processes are running as FastCGI under Zeus with these compile
options:
./configure --with-curl \
--enable-sockets \
--enable-dbx \
--with-jpeg-dir \
--with-jpeg \
* Thus wrote Lasse Laursen ([EMAIL PROTECTED]):
Hi all,
We just installed PHP 4.3.6 on FreeBSD 5.2.1 on some SMP machines and we are
experiencing some weird problems.
...
The database backend is MySQL 4.0.18 and is run on a seperate dedicated
server.
What mysqlclient is php linked to?
- Original Message -
From: Curt Zirzow [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Sunday, April 25, 2004 5:52 PM
Subject: Re: [PHP] PHP /MySQL problem on SMP machines
* Thus wrote Lasse Laursen ([EMAIL PROTECTED]):
Hi all,
We just installed PHP 4.3.6 on FreeBSD 5.2.1 on some SMP
To: [EMAIL PROTECTED]
Subject: [PHP] PHP, MySQL problem
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 database. Nearly
everything is working, as it should except for the communication between
HTML form and PHP. If I try
Sorry, I should have done that from the beginning, but here it is. I have
looked further on the variables and it seams like the name attribute
sumbit (in the form) is not converted to the $submit variable when I
press the button. If I make a var_dump() in the beginning and end of the
code, $submit
Hello Nicolai,
Wednesday, January 14, 2004, 12:14:09 PM, you wrote:
NE Sorry, I should have done that from the beginning, but here it is. I have
NE looked further on the variables and it seams like the name attribute
NE sumbit (in the form) is not converted to the $submit variable when I
NE if
Hi
Add records with this code.
?php
$name=isset($_POST['name']) ? $_POST['name'] :'';
$address=isset($_POST['address']) ? $_POST['address'] :'';
if (!empty($name)) {
mysql_connect($server,$user,$pass) or die (Error conecting);
mysql_select_db($dbnamn,$conection) or die (no db .$dbnamn);
$query =
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 database. Nearly
everything is working, as it should except for the communication between
HTML form and PHP. If I try to add a record to my database by pushing a
submit the
Good Afternoon!!
I just installed RedHat Linux 9.0 with Apache/Php/MySql for a project. I
have Apache and PHP running. However, PHP does not have the Mysql module??
installed and this is the key to our project. Could someone point me in the
right direction to recompile PHP with proper
This is a good tutorial and it starts off with installiing php and mysql
http://hotwired.lycos.com/webmonkey/programming/php/tutorials/tutorial4.html
Robin Kopetzky wrote:
Good Afternoon!!
I just installed RedHat Linux 9.0 with Apache/Php/MySql for a project. I
have Apache and PHP
I just installed RedHat Linux 9.0 with Apache/Php/MySql for a project. I
have Apache and PHP running. However, PHP does not have the Mysql module??
installed and this is the key to our project. Could someone point me in the
right direction to recompile PHP with proper MySql module
Hi everyone,
first, sorry to all if my english is so poor.
second, i have the follow question: when i create a existing MySQL DBase,
what happend??
how can i avoid this problem??
i attached my create_table code. Thanks for all
html
head
titleCreación de una Base de Datos/title
/head
body
h2
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,
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/lib/php4. The php.ini file has this in it:
[daniel
Hi!
I'm trying to connect to my mysql database using something like
mysql_connect( 'localhost', 'root', 'thepassword' )
or die ( 'Unable to connect to server.' );
But I get the error message:
Fatal error: Call to undefined function: mysql_connect() in
PM
Subject: [PHP] MySQL problem with RedHat 8
Hi!
I'm trying to connect to my mysql database using something like
mysql_connect( 'localhost', 'root', 'thepassword' )
or die ( 'Unable to connect to server.' );
But I get the error message:
Fatal error: Call to undefined function
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 3:58 PM
Subject: [PHP] MySQL problem with RedHat 8
Hi!
I'm
Hi everyone! I can't figure out why this doesn't work. The call to the
CONNECT function works and it connects with no problem, and it does save the
Resource ID in $this-DBLink[]...however, when CHOOSEDB is called, the
Resource ID just saved is gone. $this-DBLink is just empty, and I can't
figure
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 commands like mysql_connect()
when trying to start phpMyAdmin I get: cannot load MySQL extension
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 commands
I am using PHP and MySQL on a 2k dev box to be uploaded to a linux box
I have a piece of SQL
INSERT INTO `contracts` (`key`, `content`) VALUES (1,'blah blah blah
character 39 is a single speach mark '+CHAR(39)+' blah blah blah')
Instead of inserting: 'blah blah blah character 39 is a single
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
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
Hello,
I am extremely new to MySQL and have never managed to get working
smoothly with PHP before. I am trying really hard to understand
how to work it, and am almost there.
I have a problem which I do not know how to resolve and was
wondering if anybody could help me. I have no idea what is
Here's the example from the PHP manual:
The tutorial here are very helpfull:
http://www.melonfire.com/community/columns/trog/
-- David
?php
// Connecting, selecting database
$link = mysql_connect(mysql_host, mysql_login, mysql_password)
or die(Could not connect);
print Connected
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
?php
* 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 in
: 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
?php
// Connecting, selecting database
$link = mysql_connect(mysql_host, mysql_login, mysql_password)
or die
- 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:
?php
// database connect script
: Wednesday, December 05, 2001 4:59 PM
To: Javier Muniz; '[EMAIL PROTECTED]'
Subject: Re: [PHP] PHP + MySQL problem (strange behavior)
On Thu, 6 Dec 2001 08:32, Javier Muniz wrote:
Hello,
I'm having trouble determining what's going wrong with a MySQL query
that I'm doing from PHP. Now before
Hello,
I'm having trouble determining what's going wrong with a MySQL query that
I'm doing from PHP. Now before you go blaming MySQL read on :)
I have a table with the following columns:
id (int)
name (varchar 20)
starttime (int)
duration (int)
now, i have a row that has a starttime of 60,
On Thu, 6 Dec 2001 08:32, Javier Muniz wrote:
Hello,
I'm having trouble determining what's going wrong with a MySQL query
that I'm doing from PHP. Now before you go blaming MySQL read on :)
I have a table with the following columns:
id (int)
name (varchar 20)
starttime (int)
duration
ject: [PHP] mySQL problem
I'm having a wierd problem with mySQL query.
$Query = "SELECT UCASE(Company), Icons, ID, LogoD FROM feComps";
returns right amount of rows, but field Company is empty.
$Query = "SELECT Company, Icons, ID, LogoD FROM feComps."
ject: [PHP] mySQL problem
I'm having a wierd problem with mySQL query.
$Query = "SELECT UCASE(Company), Icons, ID, LogoD FROM feComps";
returns right amount of rows, but field Company is empty.
$Query = "SELECT Company, Icons, ID, LogoD FROM feComps."
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 = SELECT UCASE(Company
Message-
From: Niklas Lampén [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 02, 2001 2:22 AM
To: Php-General
Subject: [PHP] mySQL problem
I'm having a wierd problem with mySQL query.
$Query = SELECT UCASE(Company), Icons, ID, LogoD FROM feComps;
returns right amount of rows, but field
I'm having a wierd problem with mySQL query.
$Query = SELECT UCASE(Company), Icons, ID, LogoD FROM feComps;
returns right amount of rows, but field Company is empty.
$Query = SELECT Company, Icons, ID, LogoD FROM feComps.;
works fine.
First query works great when I run it in
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
- weekending (this is a -MM-DD format date)
- (and others)
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
- weekending
] MySQL problem
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
]]
Sent: Thursday, April 26, 2001 10:28 AM
To: [EMAIL PROTECTED]
Subject: [PHP] MySQL problem...
MySQL doesn't like me
Sometimes, my pages that connect to the database get the error Warning:
Supplied argument is not a valid MySQL result resource... repeated over
and
over again (someth
]]
Sent: Thursday, April 26, 2001 10:28 AM
To: [EMAIL PROTECTED]
Subject: [PHP] MySQL problem...
MySQL doesn't like me
Sometimes, my pages that connect to the database get the error Warning:
Supplied argument is not a valid MySQL result resource... repeated over and
over again
MySQL doesn't like me
Sometimes, my pages that connect to the database get the error Warning:
Supplied argument is not a valid MySQL result resource... repeated over and
over again (something like 1000 times...)
What's causing this error? Obviously, PHP isn't getting a result back from
o:[EMAIL PROTECTED]]
Sent: Thursday, April 26, 2001 10:28 AM
To: [EMAIL PROTECTED]
Subject: [PHP] MySQL problem...
MySQL doesn't like me
Sometimes, my pages that connect to the database get the error Warning:
Supplied argument is not a valid MySQL result resource... repeated over
and
ov
[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...
$db = mysql_connect(localhost,user,pass);
mysql_select_db(db,$db);
$gmdquery=SELECT * FROM game_of_the_day;
$the_info = mysql_query
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
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