I just set up a pc with mysql-3.23.51/php-4.2.1/apache-1.3.26. I have
created a database and tables and a web interface to insert data into
it. I can insert data from the command line just fine. From the web
interface there is nothing inserted and no error messages. I have tried
connection string
On Saturday 13 July 2002 14:49, Chip Wiegand wrote:
I just set up a pc with mysql-3.23.51/php-4.2.1/apache-1.3.26. I have
created a database and tables and a web interface to insert data into
it. I can insert data from the command line just fine. From the web
interface there is nothing
Richard Mail wrote:
hi,
i've tryed to install php 4.2.1 with apache 1.3.24
php works, but i can't pass any variables.
is it in php.ini that i have to change something ?
or is the installation of php not correct ??..
thanks in advance,
Richard Pijnenburg
The netherlands
I'm only running
Hi everybody,
I am just trying to find out how some db errors happened during a user was
browsing my web app.
This is out of the Apache log:
GET /subapp_profiles/act_upload_image.php HTTP/1.1 200 160
POST /subapp_profiles/act_upload_image.php HTTP/1.1 302 5
I do not have a clue where this GET
Does everyone have any experiences connecting to MySQL 4.0.2 through the
socket ? Here is the deal, I have just installed MySQL 4.0.2 as the second
mysqld server on my linux machine, and , to specify which mysql to connect
to , I am trying to connect to 4.0.2 via socket in PHP, i.e.
I dont know what causus it, but I have a kind of a solution for your
problem...
you probably have something like $_GET[myvar] in your query, try to do
something like this :
$_VARS=_$REQUEST_METHOD;
$_VARS=$$_VARS;
And then instead of using the _GET, you use the _VARS...
Andy [EMAIL PROTECTED]
Hi all!
I have managed to set up the connection to my MS-SQL Server, but I had
to change the User-Authentification modus from NT SQL
Authentification to just SQL Authentification in the SQL-Server
management, otherwise I always got the error message: Login failed.
I would like to know, if it
Hi there
I am having a problem querrying to my database.
I wish to check to see if a username is listed in the database so that I can
authenticate that individual. However, when I try to echo my $_result i'm
always getting 0 can someone please help. I think it should be either 1 or
0.
These are
On Sat, 2002-07-13 at 00:52, Jason Wong wrote:
On Saturday 13 July 2002 14:49, Chip Wiegand wrote:
I just set up a pc with mysql-3.23.51/php-4.2.1/apache-1.3.26. I have
created a database and tables and a web interface to insert data into
it. I can insert data from the command line just
Your _Name1 in your SELECT doesn't have a $ in front of it.
$_query = select userName from users where userName = '$_Name1' ;
or
$_query = select userName from users where userName = '.$_Name1.' ;
Also after you query your database you have to get the results from your
select.
$_result after
Use mysql_result or mysql_fetch_array to get your results.
Example:
$userName = mysql_reqult($result,0);
Just use mysql_num_rows($result) and count the amount of rows selected.
if(mysql_num_rows($result) == '0')
{
do_user_doesnt_exist_stuff();
}
else
{
do_user_exists_stuff();
}
First of all, you are not using the variable when you are passing it to
the sql string. Second, you are doing a pointless var check, and
thirdly, your condition statement is flawed.
1. SQL should be (if you have register globals turned on, otherwise you
must use $_GET['_Name1'] ):
$_query =
I have a database, all the data is numbers. I want to make a query that
will extract the data and then make it available in an array, so the
array is populated 'real-time'. I could just enter the number into an
array manually, but I want to automate the job. I don't know what to
start looking
I have a database, all the data is numbers. I want to make a query that
will extract the data and then make it available in an array, so the
array is populated 'real-time'. I could just enter the number into an
array manually, but I want to automate the job. I don't know what to
start
On Sunday 14 July 2002 03:43, Chip Wiegand wrote:
$sql = INSERT INTO legs
(squats,squats_reps,legpress,legpress_reps,legext,legext_reps,legcurls,
legcurls_reps,calfraise,calfraise_reps,hipabducter,hipabducter_reps,
hipadducter,hipadducter_reps,comments)
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