Hi, all
First of all, I apologize if this is a dump question but I can't find the
answer. I'm trying to use sessions to keep the information about user and
password in a database driven web application. I am trying with a simple web
page but I can't make it work correctly. Help me, please.
Hi Balazs,
The situation: I have 2 tables in an adverising site. In one table, i have
the cities used by advertisers, and in the other there are the ads.This ad
table has a field which refers to the id of the city in that table.
The problem: what mysql method should i use if i want to
I'm trying to connect to a Oracle database but i'm receiving the following error
message in the browser:
Fatal error: Call to undefined function: ora_logon() in
/home/httpd/html/restrito/testes/teste_bd.php on line 17
I'm starting to learn PHP now, so I don't know what does it means or what I
Hey did you get it to work yet?!?!
Dan
On Sunday, February 3, 2002, at 09:08 PM, [EMAIL PROTECTED] wrote:
Umm..
SET parameter is for the UPDATE query
Adam Royle [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
You missed the SET parameter in your
Morten,
ODBC stands for Open Database Connectivity. It's a connection layer (API)
that is non-database-specific.
This removes application dependence on one particular database, e.g. using
ODBC instead of native functions allows you to write your application
against MySQL and then deploy on
Since you gave no real details of your table structure, I'll conjecture.
Assume city table and advert table is defined as:
CREATE TABLE city (
city_id int not null, auto_increment,
city_name varchar(50) default
)
CREATE TABLE adverts (
ad_id int not null auto_increment,
ad_text varchar(200)
Make sure the path name is correct
Let's say you compiled apache with this prefix /web
you must have the full path name to work...like this
exec(cp '$ResumeUp' '/web/htdocs/resumes/$ResumeUp2');
Also check permissions...
I think that was my problem at first too!!!
You can also do
Hi,
Is there an easy way to see the data in a table?
I made a table that contains information about people. How do I see/change
the contenst of the table?
Thanks,
Morten
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PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Easiest way is to download and install phpMyAdmin.
http://phpmyadmin.sourceforge.net/
HTH
Richy
-Original Message-
From: Morten Nielsen [SMTP:[EMAIL PROTECTED]]
Sent: 04 February 2002 16:34
To: [EMAIL PROTECTED]
Subject:[PHP-DB] MySQL
Hi,
Is there an easy way to see
Go download PHPMyAdmin.
- Original Message -
From: Morten Nielsen [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, February 04, 2002 8:34 AM
Subject: [PHP-DB] MySQL
Hi,
Is there an easy way to see the data in a table?
I made a table that contains information about people. How
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www.sunmaia.net
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tel. 0121-242-1473
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-Original Message-
From: Peter
I would like to know how to have a php script loop through two different sql
statements and use one according to a yes or no answer. I am still kinda
new to php and I already have my sql statements, but I dont know how to use
php to tell it to use one of the other. My sql statements are as
Unfortunetly I'm stuck with the Oracle setup I've been dealt. It's an already
deployed production-level server and is not going to change any time soon.
So, I need to find some other way of getting around this using PHP and maybe
some other magic.
If anyone has any ideas, please let me know.
Hi all,
Ok I am learning this on my own but I could use some help.
I have started these two pages test.php and config.php here is the code:
test.php
?
// include the database configuration
include config.php;
// connect to database
I'm just learning MySQL and PHP so this fresh from my last snoop through the
documentation, hope it will help.
From the command line you can make a global query like - SELECT * FROM pets;
and all the info stored in the table will be shown.
To update a record use the update query:
UPDATE pet
On Monday 04 February 2002 10:54 am, you wrote:
Hi all,
Ok I am learning this on my own but I could use some help.
I have started these two pages test.php and config.php here is the code:
test.php
?
// include the database configuration
include config.php;
In looking at my include page
This just showed up, where was it stuck?
Essentially you do a SELECT into your target table, if the target key
exists, which you determine from row count, then you do an UPDATE,
otherwise an INSERT.
I don't know what else you have wrapped around this, in terms of control
structure, so here's
on 2/4/02 12:54 PM, J Leonard at [EMAIL PROTECTED] appended the following
bits to my mbox:
mysql_connect($db[host],$db[user],$db[password],$db[database]) or
The above function call only has three parameters. The database isn't
specified in the connect call but selected later. Try:
mysql_connect() function only takes 3 arguments (servername, username, pass)
.
Select your database by using mysql_select_db() function...
Try to rewrite the code as:
mysql_connect($db[host],$db[user],$db[password]);
mysql_select_db($db[database]);
Refer to:
RE Could not execute f query : Is it possibly a typo, such that INSERT
INTO Canidate should read Candidate (ie with the correct number of d's for
the english speling ;-)
Just a thought.
Neil.
At 02:13 04/02/2002 +, Todd Williamsen [EMAIL PROTECTED] wrote:
$query = INSERT INTO
[EMAIL PROTECTED] (Adv. Systems Design) writes:
I was simply going by the docs themselves:
http://www.mysql.com/doc/A/N/ANSI_diff_Sub-selects.html
(see bottom of page)
yes, using REPLACE may be a solution. But you have to make a copy of
the original table, because REPLACE cannot access the
GOOD
What was the problem!?!?
Dan
On Monday, February 4, 2002, at 01:44 PM, [EMAIL PROTECTED] wrote:
Ok,
I got it woohoo!!!
It works!
Thanks for all your help
-Original Message-
From: Dan Brunner [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 04, 2002 10:35 AM
To:
Hi,
I am having problems with this code. I want to update a members number of
days they have been a member can anyone help?
$query[days]=(UPDATE users set days_member VALUES = ('days_member' +
'1'));
$result=mysql_query($query[days]);
Thanks in advance
JD
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PHP Database Mailing List
Hello,
I am not sure why you want to do this. Wouldn't it be better to use
calculate the days from the date signed up, rather than through updating
your database every day. Try using datediff.
Jeremy
At 01:51 PM 2/4/2002 -0800, Jennifer Downey wrote:
Hi,
I am having problems with this
Try it without quotes around days_member and 1.
Gurhan
-Original Message-
From: Jennifer Downey [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 04, 2002 4:51 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Update
Hi,
I am having problems with this code. I want to update a members
Hi Jennifer,
I am having problems with this code. I want to update a members number of
days they have been a member can anyone help?
$query[days]=(UPDATE users set days_member VALUES = ('days_member' +
'1'));
$result=mysql_query($query[days]);
=1 RTFM 6.4.5 UPDATE Syntax
=2 Why would
Yeah I didn't notice that..
VALUES must not be there . it should be :
set days_member=days_member+1
thanks for pointing that out:)
Gurhan
-Original Message-
From: Lokesh Verma [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 04, 2002 5:12 PM
To: Gurhan Ozen
Subject: Re: [PHP-DB]
Why? There's absolutely no point -- record the date a person became a
member and calculate the number of days from that.
In addition, what happens when you fail to run the update routine? Or
someone runs it twice? Or it half runs, then dies because the database has
gotten so large that the
Why? Because I am learning. It seemed like the way to do it. I didn't know
all those things could happen. There is NO documentation, manual or tutorial
for that matter that gives clear, in English directions on how to do
something. People just assume that everyone knows techie talk. So they
Hello,
I have three tables:
CREATE TABLE event (
eid int(10) unsigned NOT NULL auto_increment,
tid smallint(5) unsigned NOT NULL default '0',
itimestamp int(10) unsigned NOT NULL default '0',
etimestamp int(10) unsigned NOT NULL default '0',
somethingelse tinyint(4) NOT NULL default
Jennifer,
I appreciate and I'm glad that you're learning. What you were proposing is
an easy trap to fall into. It's just that I've worked on systems where, for
no apparent reason, the designer had decided to store the number of
days/months someone had been a member / tenant / on-account
Miles is correct, this the standard way (without knowing your application
and if I could figure out another trick).
But if you absolutely don't want to use two SQL blocks you can set your SQL
block to something like this:
$sql =
DELETE FROM $table_name
WHERE our_serv = \$our_serv\
INSERT
Hans,
On 2/4/02 6:30 PM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
From: hz11 [EMAIL PROTECTED]
Date: Mon, 04 Feb 2002 21:17:15 -0500
To: [EMAIL PROTECTED]
Subject: Dealing with One-to-Many SELECTS (MySQL)
to get a single event. This all works fine, however this is the result
Does anyone have experience of using odbc_primarykeys???
I'm writing a script to let me port a database from Access to MySQL without
having to write all the table definitions by hand. But I can't get
odbc_primarykeys to work - it comes back with:
Warning: SQL error: , SQL state 0 in
Richard,
Simple example using MS SQL Server:
?php
$dsn=Northwind;
$usr=sa;
$pwd=;
$qual=Northwind;
$owner=dbo;
$table=Orders;
$conn_id = odbc_connect($dsn, $usr, $pwd) or die (odbc_error());
$result_id = odbc_primarykeys($conn_id, $qual, $owner, $table) or die
(odbc_error());
The oracle extension is not installed. This is why PHP doesn't recognize
ora_logon().
So you have to install PHP with oracle extension. BTW, you would prefer
installing oci8 extension instead of oracle. oci8 is the newest extension to
connect to an oracle 7 or 8 database (see ocilogon() and
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