Hi
Yes it is possible
I'm using it currentlyon some of my major qry's
You'd prepare the qry and the instead of using the ibase_query just use the
ibase_execute
in the same way
Niel
Arne Borkowski (borko.net) [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
Hi,
I can use
Hi All,
I'm building a standard shopping cart style e-commerce site using
PHP and MySQL running on Apache.
I store my users' cart info in this table:
++--+--+-+-+---+ |
| Field | Type | Null | Key | Default | Extra | +
: 28 April 2001 23:53
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Select statements - A Quest !!!
I have two tables namely,
1. cpkaizen : The following is the desc
Field | Type | Null | Key | Default | Extra
Have you thought about normalizing your data model? It could make your
task a lot easier.
Doug
At 11:33 PM 4/13/01 +0530, Sharmad Naik wrote:
I have three table called table1, table2, table3 all having fields like:
table1 contain id and username and id referencing table4
table2 contains id and
On Fri, Apr 13, 2001 at 02:23:50PM -0400, Doug Semig wrote:
Have you thought about normalizing your data model? It could make your
task a lot easier.
I have normalized my table structure this is just a practice model i m trying
I want that the search should be on username whereby first i
on 4/11/01 7:48 PM, Mike Baerwolf at [EMAIL PROTECTED] wrote:
SELECT substring_index( body, ". ", 2) FROM news;
This works great from the mysql client but when I try it using php with
this:
$result = mysql_query("SELECT substring_index(body, "." ,2) FROM news"
The way you have this
* Julio Cuz, Jr. ([EMAIL PROTECTED]) wrote:
Ron,
Thanks for your help, but my problem still there even when I made the
following changes:
$sql = "SELECT * FROM \"Remodel\" WHERE Email=\"".$find."\"";
You don't have to quote the table name either. You can probably get
away with
Ron,
Thanks for your help, but my problem still there even when I made the
following changes:
$sql = "SELECT * FROM \"Remodel\" WHERE Email=\"".$find."\"";
By the way, you're correct when saying that "WO" and "Email" are column
names and '"$find" is what the user entered when searching for a
There are probably a million ways to do this... here is one :)
(note that the file should be named "fieldtest.php"
--
Regards,
Gary "SuperID" Huntress
===
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-Original Message-
From: David Balatero [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 29, 2001 4:33 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Select Database number show it
I believe this is what you want:
SELECT * FROM mydatabase LIMIT 0,5
Hi
I believe this is what you want:
SELECT * FROM mydatabase LIMIT 0,5
---
David Balatero
[EMAIL PROTECTED]
---
-Original Message-
From: Naga Sean [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 28, 2001 5:06 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select Database number show it
Information Systems Limited. Registered in England. No
2117205.
Registered Office: Park House Mews, 77 Back Lane, Off Broadway, Horsforth,
Leeds, LS18 4RF
-Original Message-
From: boclair [mailto:[EMAIL PROTECTED]]
Sent: 22 March 2001 12:45
To: [EMAIL PROTECTED]
Su
m: boclair [mailto:[EMAIL PROTECTED]]
Sent: 22 March 2001 12:45
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Select where date is in period
___Morris___
The database has a table of equipments with date fields for
date_online and date_offline.
The requirement is to find which equip
"boclair" [EMAIL PROTECTED] wrote in message
005101c0b2d2$4e6b3920$[EMAIL PROTECTED]">news:005101c0b2d2$4e6b3920$[EMAIL PROTECTED]...
|
| - Original Message -
| From: boclair [EMAIL PROTECTED]
| To: [EMAIL PROTECTED]
| Sent: Thursday, March 22, 2001 10:45 PM
| Su
On Wed, 21 Mar 2001, boclair wrote:
This is simple but I cannot see where I am going wrong
I have a table members with one of the fields
status, varchar(10)
The values may be active or retired or deceased or null
If I run the select
SELECT * FROM members WHERE status = 'deceased';
- Original Message -
From: boclair [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, March 21, 2001 12:02 AM
Subject: [PHP-DB] Select where
This is simple but I cannot see where I am going wrong
I have a table members with one of the fields
status, varchar(10)
The values
Thanks, I only gave the mySQL but the php scripting is
$deceased = mysql_query(SELECT * FROM members where
status=\'deceased\'");
and later
while ($myrow = mysql_fetch_row($deceased))
MySQL now says in relation to the *while* line
Warning: Supplied argument is not a
You probably want
this.options[this.selectedIndex].value
-Original Message-
From: Julio Cuz, Jr. [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 22, 2001 2:36 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] SELECT problem (2)
Hi--
Sorry, I forgot to finish the previous
I am having most of my problems because I am trying to do this walking
through the data a single time, with the assumption any records parentID
has an ID less than the child and therefore would already be populated in
the array...
On Sun, 21 Jan 2001, Stephen wrote:
Date: Sun, 21 Jan 2001
page. If I work it out Ill be sure to post it :)
On Sun, 21 Jan 2001 [EMAIL PROTECTED] wrote:
Date: Sun, 21 Jan 2001 15:50:16 -0500
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] select with parentID field into multidimensional
array?
Hi,
Maybe it would be better
PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] select with parentID field into multidimensional
array?
Hi,
Maybe it would be better not to use a multidimentionnal array in a db.
You should create to additionnal columns that stores the ParentID
of the row.
It's
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