I should've checked the assumption.
A little strange though for a scripting language. I'd say, since
default behaviour of mysql_* functions is to assume the last opened
link when a parameter is not passed, and since default behaviour is to
evaluate null, false and 0 the same, mysql_* functions sho
> [new code]
> if (!mysql_connect($a, $b, $c)) return;
>
> if (!mysql_select_db($dbname)) return;
>
> $res = mysql_query("SELECT * FROM manual;", $link);
> [/new code]
Isn't going to work because $link is not defined and definitely not a
resource like mysql_query expects.
> OR, optionally, to
Eric,
You gessed right. Both of the files are in my "docroot" directory.
I tried this. Unfortunely, it did not work ( perhaps, I did not use it
right), although, I followed your instructions.
Isaak gave me an idea which solved my problem (see his replies).
Thank you
Eric-275 wrote:
>
>
> --
It's great that you got your problem solved :).
On Thu, Jun 19, 2008 at 12:30 PM, bateivan <[EMAIL PROTECTED]> wrote:
>
> Hello Isaak,
>
> You've scored the bulleye! It works!
> Also, I tested by including "common.php" instead of "database.php" and it
> worked too.
> This solution is perfect for
Hello Isaak,
You've scored the bulleye! It works!
Also, I tested by including "common.php" instead of "database.php" and it
worked too.
This solution is perfect for me as I can keep my code compact and tidy.
I would, definetely, recommend this site to my fellow developers.
Thank you very much,
What you said wasn't wrong - you'd just produce a couple of warnings.
By taking away the assignment of the $link variable you're unsetting
it, meaning it will have a value equal to NULL. All of the following
should work.
[old code]
if (!($link = mysql_connect($a, $b, $c))) return;
if (!mysql_sele
My mistake then, it's been a while that I used it that way so some things do
fade away from my mind.
On Thu, Jun 19, 2008 at 1:57 AM, Chris <[EMAIL PROTECTED]> wrote:
> Isaak Malik wrote:
> > Because then the connection resource isn't stored in the $link variable
> > and you will be able to use t
Isaak Malik wrote:
> Because then the connection resource isn't stored in the $link variable
> and you will be able to use the mysql functions without passing that
> variable to each function.
RTM again.
The link parameter is completely optional.
If you don't specify it, it uses the last connect
To Evert:
4. I am posting some of my code to compare. Module "carmade.php" works v.s.
"login.php" which does not. Also, I am posting my "database.php" which is
establishing the connection and is included in every page through
"layout.php"->"common.php"->"database.php".
To bateivan:
I can only gu
Pastebin works lots better when you're posting code: www.pastebin.com
I don't see your database.php included in both of these modules. Where
do you include it?
EVert
On Wed, Jun 18, 2008 at 9:38 PM, bateivan <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> To answer these qwestions:
> 1. I am using "in
Hello,
To answer these qwestions:
1. I am using "include" in my code.
2. About the "$link = mysql_connect('localhost', 'root', 'testing');"
variable. I can just include this line before mysql_query function and it
works even without entering into mysql_query the "$link" variable. (Like the
connec
Thank you, Chris,
I tried your suggestion using regular connection. Unfortunately, I got the
same result.
See my reply to Isaak,s reply where I am going to post portion of my code to
visualize what I am talking about.
Thanx again
chris smith-9 wrote:
>
>
>> It means that either your mysql co
Because then the connection resource isn't stored in the $link variable and
you will be able to use the mysql functions without passing that variable to
each function.
On Wed, Jun 18, 2008 at 1:51 AM, Chris <[EMAIL PROTECTED]> wrote:
>
> > It means that either your mysql conenction details are no
excuseme but, you can access by terminal? [shell]
ODBC is the user? or is a ODBC database?
Please answer me and i can help you.
**Excuse me my english is bad. :(
El 17/06/2008, a las 06:51 p.m., Chris escribió:
It means that either your mysql conenction details are not
correctly set or
th
> It means that either your mysql conenction details are not correctly set or
> the connection resource isn't accessible for your mysql functions. I suggest
> you first try by replacing:
>
> $link = mysql_pconnect('localhost', 'root', 'testing');
>
> into:
>
> mysql_pconnect('localhost', 'root'
http://myprojects.srhost.info
eric{at}myprojects{dot}srhost{dot}info
- Original Message -
From: "bateivan" <[EMAIL PROTECTED]>
To:
Sent: Tuesday, June 17, 2008 11:19 PM
Subject: [PHP-DB] PHP-MySQL connection for particular module
:
If you get an error of this kind:
"Warning: mysql_query() [function.mysql-query]: Access denied for user
'ODBC'@'localhost' (using password: NO) in D:\Program Files\Apache Software
Foundation\Apache2.2\htdocs\login.php on line 17
Warning: mysql_query() [function.mysql-query]: A link to the server
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