All:
I'm running PHP in ISAPI mode with IIS accessing (or attempting to
access) an Oracle 9i database on a RedHat box. Here are the versions of
everything:
Web Server: IIS v5.1
PHP: 5.2.5
Database: ORACLE v9i-R2
I am attempting to load a page with the following code:
//Database credentials
To: Fortuno, Adam
Cc: php-db@lists.php.net
Subject: Re: [PHP-DB] Unable to Login to Oracle 9i-R2 Database Using PHP
5.2.5
//Return the database connection
OCILogon($username, $passwd, $db);
I receive an error stating, Fatal error: Call to undefined function
OCILogon() in C:\Documents
]
Sent: Monday, November 24, 2008 5:54 PM
To: Fortuno, Adam
Cc: php-db@lists.php.net
Subject: Re: [PHP-DB] Unable to Login to Oracle 9i-R2 Database Using PHP
5.2.5
Rick, when I delve into the php_info() shmaz, I see this snipet - see
below. I'm not sure if I should be looking for something else
Mr. Froasty,
From your note, it sounds like you want to use foreign keys; as Daniel
pointed out. I think an example would be helpful here. The subject of
foreign keys is bigger than a bread box so I'll just touch on the pieces
I think you'll find helpful. There is all sorts of literature
Fred,
If you're using integrated security (e.g., a domain account such as
domain\bob), you won't supply a password. However, the sa account is
a SQL login (v.s. a domain login). In that case, it would make sense
that you supply a password. See the following page for some code to
connect using
Dee,
If all you're trying to code is that a value of NULL equates to FALSE
and a date value (whatever date value) equates to true, you can use
something like this:
If ($MyVariable) {
//... true path blah...
} else {
//... false path blah...
}
You can use this because NULL
All,
Chris makes a good point i.e., unknown compared to anything is
unknown. His comment made me want to clarify my earlier note. In my
earlier example, I wasn't suggesting that-that logic be coded in the
database. I was assuming it would be in the page:
?php
$TestNullValue = NULL;
If
Mika,
Put the dollar sign (i.e., $) outside the curly brace.
$query=SELECT * FROM pic_upload WHERE band_id='${band_id}';
A-
-Original Message-
From: Mika Jaaksi [mailto:mika.jaa...@gmail.com]
Sent: Thursday, February 12, 2009 12:27 PM
To: php-db@lists.php.net
Subject: [PHP-DB] Re:
Sashi,
This (likely) means you have a some generic page (i.e., picture.php)
that displays some picture. The picture it displays depends on the
parameter passed when the page is called (i.e., 123).
html
head
titleSashi's Test Page/title
/head
body
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = SELECT path FROM Image WHERE ID = %s;
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of the escaped
$value to the %s, lets say
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