[PHP-DB] Output a few tupels of 493033 tupels

[Haiko Etzel]

Hello,

I have got a little problem with my datas. In one table I have got 493033 tupels. A 
simple Output of all datas is to much for the Browsers (and the intranet-connection) 
and to ask for all and then only give out a few is also a big problem for the server.

Is there any way only to ask for some tupels in a postgres-database with PHP?

H. Etzel
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RE: [PHP-DB] Picture Upload

[Luke Woollard]

use the GD Library functions - look up on php.net

The GD library is available from http://www.boutell.com/
It is included by default with PHP 4.3.x

NOTE: This is a question for [EMAIL PROTECTED]


-Original Message-
From: Dallas Freeman [mailto:[EMAIL PROTECTED]
Sent: Tuesday, 11 March 2003 5:07 PM
To: 
Subject: [PHP-DB] Picture Upload


Is there a possibility that you can resize an image before uploading to
the server?
 
If so, how?
 
Thanks,
 
 
Dallas Freeman


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[PHP-DB] Picture Upload

[Dallas Freeman]

Is there a possibility that you can resize an image before uploading to
the server?
 
If so, how?
 
Thanks,
 
 
Dallas Freeman

[PHP-DB] Re: adding to an array variable using a loop...?

[Adam Royle]

I assume the field 'date' in your database is a date field, and in your code
you are trying to do a date + 1, but you are not specifying *what* you are
adding to the date (ie. 1 day, 1 week, 1 month), therefore the whole
variable is being replaced (with a 1). Try looking at the various date
functions (or string functions) to work out how to do this. Depending on how
your date is setup you would use strtotime() or mktime() (or both).

Adam

--- Original Message ---
I want to create an array from results returned from a mysql query. I want
it to go through each row in the returned result, and if the variable in the
array staff exists add 1 to the total, if it doesnt exist, set it as one and
carry on.

But when i run this and do var_dump, it just returns "1" in the array for
every date.

I don't think my logic for this is correct, please help.
Cheers,
Dave.
==
$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
INTERVAL 6 DAY) ";
 $result = mysql_query($query);
 while ($row = mysql_fetch_array($result)){
  $x = 1;
  $date = $row['Date'];
  if ( isset($staff[$date] ) ){
   $staff[$date] = $staff[$date] + $x ;
  }
  else{
   $staff[$date] = $x ;
  }

 }


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Re: [PHP-DB] Date in a dropbox box

[Chris Payne]

Hi there,

Thank you John, you are a lifesaver :-)

Chris

> > I have a form which displays 3 dropdowns, day, month and year, I have
> it
> > displaying the default of the dropdown for month no problem, but how
> can I
> > get it to select the current date (day) in the dropdown, but show 31
> days
> > nomatter how many days are in the month?  This is important even
> though it
> > sounds odd :-)
> > 
> > so if it was the 5th of a month, it would show 5 as a default value,
> but
> > allow you to select 1-4 before that and 6-31 after it.
> 
> Lots of ways you can do it.
> 
> $today = date('d');
> for($x=1;$x<=31;$x++)
> {
> echo ' '>$x';
> }
> 
> ---John W. Holmes...
> 
> PHP Architect - A monthly magazine for PHP Professionals. Get your copy
> today. http://www.phparch.com/
> 
> 
> 
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RE: [PHP-DB] Date in a dropbox box

[John W. Holmes]

> I have a form which displays 3 dropdowns, day, month and year, I have
it
> displaying the default of the dropdown for month no problem, but how
can I
> get it to select the current date (day) in the dropdown, but show 31
days
> nomatter how many days are in the month?  This is important even
though it
> sounds odd :-)
> 
> so if it was the 5th of a month, it would show 5 as a default value,
but
> allow you to select 1-4 before that and 6-31 after it.

Lots of ways you can do it.

$today = date('d');
for($x=1;$x<=31;$x++)
{
echo '$x';
}

---John W. Holmes...

PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/



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[PHP-DB] Date in a dropbox box

[Chris Payne]

Hi there everyone,

I have a form which displays 3 dropdowns, day, month and year, I have it displaying 
the default of the dropdown for month no problem, but how can I get it to select the 
current date (day) in the dropdown, but show 31 days nomatter how many days are in the 
month?  This is important even though it sounds odd :-)

so if it was the 5th of a month, it would show 5 as a default value, but allow you to 
select 1-4 before that and 6-31 after it.

Thanks everyone

Chris

RE: [PHP-DB] Database design

[Beverly Steiner]

Shaun,

I took a quick look at your database layout and noticed that Practice_ID and
Clinical_Trial_ID are repeated in the Booking table.  This isn't necessary
because the Booking table links to the Project table which contains this
information.  In the Project table, Practice_ID and Clinical_Trial_ID should
probably just be foreign keys and not primary keys to that table.  Same
thing for User_ID and Project_ID in the Booking table.

To keep track of the data you will need to create a new table which is
linked to the Clinical_Trial table.  For instance you could create a table
called Clinical_Data which has the following fields:
* Clinical_Data_ID (PK)
* Clinical_Trial_ID (FK)
* Data_Description - field that describes a piece of collected data (e.g.
blood pressure before hypnosis)

Is the User table where the client information goes?  You also need to
create a joining table between the User (or whatever table contains the
client info.) table and the Clinical_Data table.

--
Beverly Steiner
[EMAIL PROTECTED]


-Original Message-
From: shaun [mailto:[EMAIL PROTECTED]
Sent: Monday, March 10, 2003 8:53 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Database design


Hi,

I am creating a database for a web application. The idea for the application
is for a company administrator to be able to log into the site and allocate
staff to a project which will be a clinical trial at a particular practice.
Staff will be able to log in and update the status of the project they are
working on and clients will be able to log in and book a member of staff.
Administrators will be able to add / edit / delete - staff / clients /
clinical trials.

Here is my database diagram:

http://www.mania.plus.com

My problem is each different clinical trail will need to have different data
collected for it. How could I handle this in the database model? The
administrator would like to be able to edit the data collected for each
trial via the site...

Any other comments on my database model would be appreciated as this is my
weakest area of web development by far!




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[PHP-DB] adding to an array variable using a loop...?

[David Rice]

I want to create an array from results returned from a mysql query. I want 
it to go through each row in the returned result, and if the variable in the 
array staff exists add 1 to the total, if it doesnt exist, set it as one and 
carry on.

But when i run this and do var_dump, it just returns "1" in the array for 
every date.

I don't think my logic for this is correct, please help.
Cheers,
Dave.
==
$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start + 
INTERVAL 6 DAY) ";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		$x = 1;
		$date = $row['Date'];
		if ( isset($staff[$date] ) ){
			$staff[$date] = $staff[$date] + $x ;
		}
		else{
			$staff[$date] = $x ;
		}

	}

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RE: [PHP-DB] Query Help...

[NIPP, SCOTT V (SBCSI)]

Thanks for all the advice, I thought it might have something to do
with being a reserved word.  I changed the table name from group to grps,
and everything works fine now.

-Original Message-
From: Rob Bryant [mailto:[EMAIL PROTECTED]
Sent: Monday, March 10, 2003 11:39 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Query Help...


- Original Message -
From: "NIPP, SCOTT V (SBCSI)" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, March 10, 2003 9:23 AM
Subject: [PHP-DB] Query Help...


> I am having a lot of trouble with a query that works fine from a
> basic SQL command line, but fails in my web script.  Here is the
portion of
> code including the query:
>
> mysql_select_db($database, $Prod);
> $query_groups = "SELECT name FROM group";
> $groups = mysql_query($query_groups, $Prod) or die(mysql_error());
>
> Here is the error I am receiving:
>
> You have an error in your SQL syntax near '"group"' at line 1
>

Have you tried using backticks in your query? E.g.,

$query_groups = "SELECT name FROM `group`";

--
rob


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Re: [PHP-DB] Query Help...

[Rob Bryant]

- Original Message -
From: "NIPP, SCOTT V (SBCSI)" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, March 10, 2003 9:23 AM
Subject: [PHP-DB] Query Help...


> I am having a lot of trouble with a query that works fine from a
> basic SQL command line, but fails in my web script.  Here is the
portion of
> code including the query:
>
> mysql_select_db($database, $Prod);
> $query_groups = "SELECT name FROM group";
> $groups = mysql_query($query_groups, $Prod) or die(mysql_error());
>
> Here is the error I am receiving:
>
> You have an error in your SQL syntax near '"group"' at line 1
>

Have you tried using backticks in your query? E.g.,

$query_groups = "SELECT name FROM `group`";

--
rob


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RE: [PHP-DB] Query Help...

[George Pitcher]

Scott,

'Group' is a reserved word so you can't use it for a column name. try
changing it to 'groups' or soething else that makes sense but is safe.

George

> -Original Message-
> From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
> Sent: 10 March 2003 5:24 pm
> To: '[EMAIL PROTECTED]'
> Subject: [PHP-DB] Query Help...
>
>
>   I am having a lot of trouble with a query that works fine from a
> basic SQL command line, but fails in my web script.  Here is the
> portion of
> code including the query:
>
> mysql_select_db($database, $Prod);
> $query_groups = "SELECT name FROM group";
> $groups = mysql_query($query_groups, $Prod) or die(mysql_error());
>
>   Here is the error I am receiving:
>
>   You have an error in your SQL syntax near '"group"' at line 1
>
>   I am wondering if for some reason group is trying to be
> interperetted as the "GROUP BY" MySQL function?  I am running PHP 4.2.3
> against a MySQL DB on an Apache 1.3.27 server.  Thanks in advance for the
> help.
>
> Scott Nipp
> Phone:  (214) 858-1289
> E-mail:  [EMAIL PROTECTED]
> Web:  http:\\ldsa.sbcld.sbc.com
>
>
>
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[PHP-DB] Query Help...

[NIPP, SCOTT V (SBCSI)]

I am having a lot of trouble with a query that works fine from a
basic SQL command line, but fails in my web script.  Here is the portion of
code including the query:

mysql_select_db($database, $Prod);
$query_groups = "SELECT name FROM group";
$groups = mysql_query($query_groups, $Prod) or die(mysql_error());

Here is the error I am receiving:

You have an error in your SQL syntax near '"group"' at line 1

I am wondering if for some reason group is trying to be
interperetted as the "GROUP BY" MySQL function?  I am running PHP 4.2.3
against a MySQL DB on an Apache 1.3.27 server.  Thanks in advance for the
help.

Scott Nipp
Phone:  (214) 858-1289
E-mail:  [EMAIL PROTECTED]
Web:  http:\\ldsa.sbcld.sbc.com



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[PHP-DB] Prepare Sql statement

[Alejandro Michelin Salomon \( Adinet \)]

Hi:

I am working with php 4.1.1, in a linux i386.

How i prepare a sql statement with  odbc_prepare ?

How create parameters for a sql, to use in odbc_execute ?

Alejandro Michelin Salomon

PROBrax e-Tech, Brasil


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[PHP-DB] Database design

[shaun]

Hi,

I am creating a database for a web application. The idea for the application
is for a company administrator to be able to log into the site and allocate
staff to a project which will be a clinical trial at a particular practice.
Staff will be able to log in and update the status of the project they are
working on and clients will be able to log in and book a member of staff.
Administrators will be able to add / edit / delete - staff / clients /
clinical trials.

Here is my database diagram:

http://www.mania.plus.com

My problem is each different clinical trail will need to have different data
collected for it. How could I handle this in the database model? The
administrator would like to be able to edit the data collected for each
trial via the site...

Any other comments on my database model would be appreciated as this is my
weakest area of web development by far!




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[PHP-DB] Re:PHP + Interbase / Firebird extensions

[Philippe Makowski]

Any idea where I can download the interbase module??


Here :

--
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Firebird serveur SQL open-source en français http://firebird-fr.eu.org

Ma clé PGP : http://makowski.eu.org/pgpkey.html



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Re: [PHP-DB] table relationship

[Miles Thompson]

Three tables seems wasteful, and could mean that you have to do three 
queries when looking for someone. Why not approach it this way.

Users table - info on all users, regardless of category

Levels table - sets different access levels, e.g. clients, staff, 
administrators

User_levels table - assigns levels to user id's

This would give you more long-term flexibility as you would only have to 
extend the levels table to add granularity of access or control levels.

I'd also have a look at various network permission schemes, because there 
are subtleties that are not immediately apparent when working up an access 
scheme.

Cheers - Miles Thompson





At 09:35 AM 3/10/2003 +, shaun wrote:
Hi,

I am creating a web site which will have different types of users:
Administrators, clients and staff. Is it possible/good practice to have 3
tables related to one table i.e.
 USER
 user_id (PK)
  |
  |
  |   ||
  |   ||
AdministratorClient  Staff
admin_id(PK)   client_id(PK)   staff_id(PK)
user_id(FK)  user_id(FK)user_id(FK)
Thanks in advance for any advice offered.



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[PHP-DB] Help with file upload

[Ahmed Shams]

I am trying to upload a file but the problem i am having is that i get no 
value if i used $HTTP_POST_FILES['userfile']['name']  i have my register 
globals on and everything but i just can't see what am i doing wrong. here 
is my code:
/*
-$dir = ".\Apache\htdoc\ ";
-$dir .= $_POST['img1'];
-$temp1 = $_POST['img1'];

-$filename = $HTTP_POST_FILES['img1']['name'];
-echo "filename is $filename\n";
-$temp_file = $HTTP_POST_FILES["img1"]["tmp_name"];
-$file_type = $HTTP_POST_FILES["img1"]["type"];
-if($HTTP_POST_VARS['img1']['name'] != ""){
copy($temp1,$dir) or
die ("Couldn't copy the file");
-}else{
die("No input file specified");
-}
-?>

-
-
-Successful upload 
-
-You sent: , a 
- byte file of type . 
-
-
*/
please help me because i need this ASAP.. thank you all

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RE: [PHP-DB] Passing value in URL to Form

[Hutchins, Richard]

If I understand your question correctly, you are interested in finding a way
to pull values from the URL and insert them into a form using just HTML, no
PHP.

Going on that assumption, I'm pretty sure the answer is no. I think the only
way you can pull that off is with SOME kind of scripting language. PHP is
one option, but you may be able to use JavaScript to do it as well.

As a matter of fact, a quick search on Google for "Javascript GET values"
yielded many matches, of which, this one may be of particular interest to
you:

http://javascript.internet.com/forms/passing-values-source.html

You might also want to consider posting your question to a JS or general
HTML mailing list. You may get better/more responses since this list is
primarily concerned with the use of PHP with a database and your question
appears to be mostly HTML-based.

Good luck.

> -Original Message-
> From: Stephen K Knight [mailto:[EMAIL PROTECTED]
> Sent: Sunday, March 09, 2003 10:06 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Passing value in URL to Form
> 
> 
> I have managed to pass information through a URL and extract 
> it and insert
> it into a form on a php document.  Can this be done for an 
> html document?
> Passing the value through the URL to an html page with form 
> and then through
> a URL again into a php doc.  I tried the same method I used 
> in the php doc
> in the html doc but it didn't work.
> 
> Any ideas?
> 
> In kindness,
> Stephen
> 
> 
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Re: Fwd: Re: [PHP-DB] need help with foreach()

[Jason Wong]

On Monday 10 March 2003 17:56, David Rice wrote:
> Here is the complete function I am using.
> I returned, for testing i commented out the foreach loop and returned
> $staff, then $tips both arrays returned NULL when i did a
> var_dump(pointvalue($startdate));

I haven't been following this thread so I'm not sure what your objective is. 
But ...


> function pointvalue($start){
>   $query = "SELECT * FROM Tips WHERE date >= $start and date <= ($start +
> INTERVAL 6 DAY) ";
>   $result = mysql_query($query);
>   while($row = mysql_fetch_array($result)){
>   $date = $row[Date];

You should really be using:

   $date = $row['Date'];

Also what is $date supposed to be storing? Each iteration of the while-loop 
it's being overwritten with the 'latest' date.

>   $tips[$date] = $row[TotalTips];

As above you should put single-quotes around your array subscripts.

>   }
>   $query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
> INTERVAL 6 DAY) ";
>   $result = mysql_query($query);
>   while ($row = mysql_fetch_array($result)){
>   $date = $row[Date];

Again, see above comments.

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Re: [PHP-DB] table relationship

[Ignatius Reilly]

Yes. You have to use the "subtype" relational design:

 USER
 user_id (PK)
  |
  |
  |   ||
  |   ||
AdministratorClient  Staff
user_id(PFK)   user_id(PFK)   user_id(PFK)
admin_field1 user_field1   staff_field1
admin_field2
.
With a 1-1 relationship from Administrator to User, 0-1 relationship from
User to Administrator
(PFK : PRIMARY KEY and FOREIGN KEY)
This is particularly useful when the subtypes have different fields, so you
don't want to have only one table with many blanks for fields that do not
relate to the type at hand.

HTH
Ignatius

- Original Message -
From: "shaun" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, March 10, 2003 10:35 AM
Subject: [PHP-DB] table relationship


> Hi,
>
> I am creating a web site which will have different types of users:
> Administrators, clients and staff. Is it possible/good practice to have 3
> tables related to one table i.e.
>
>
>  USER
>  user_id (PK)
>   |
>   |
>   |   ||
>   |   ||
> AdministratorClient  Staff
> admin_id(PK)   client_id(PK)   staff_id(PK)
> user_id(FK)  user_id(FK)user_id(FK)
>
>
> Thanks in advance for any advice offered.
>
>
>
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>
>


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Fwd: Re: [PHP-DB] need help with foreach()

[David Rice]

Here is the complete function I am using.
I returned, for testing i commented out the foreach loop and returned 
$staff, then $tips both arrays returned NULL when i did a 
var_dump(pointvalue($startdate));

can anyone see how this could be solved?

Cheers,
dave
==
function pointvalue($start){
	$query = "SELECT * FROM Tips WHERE date >= $start and date <= ($start + 
INTERVAL 6 DAY) ";
	$result = mysql_query($query);
	while($row = mysql_fetch_array($result)){
	$date = $row[Date];
		$tips[$date] = $row[TotalTips];
	}
	$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start + 
INTERVAL 6 DAY) ";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		$date = $row[Date];
		if (isset($staff[$date])){

			$staff[$date] = $staff[$date] + 1;

}
else{
$staff[$date] = 1;
}
}
return $staff;
}
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[PHP-DB] content management

[shaun]

I would really appreciate some advice from anyone who has worked with or
developed their own content management system.

This is my scenario, when i have finished creating a site, i want to be able
to add in the CMS with a minimum amount of fuss. I want to be able to get
the CMS to recognize all the tables and somehow allow me to set the tables
and fields that the client can update safely (i.e. if it is an employment
recruitment site then they will be able to add jobs but not job_id). This
will save me so much time rather than having to handcode the CMS for every
site.

I think i have an answer to my problem, and would be interested to hear your
opinion. When i install the CMS it will read the existing tables and create
2 new tables:

CMS_TABLES
cms_table_id(PK)
cms_table_name
cms_table_is_editable

CMS_FIELDS
cms_field_id(PK)
cms_table_id(FK)
cms_field_name
cms_field_is_editable
cms_field_type
cms_field_size
cms_field_is_primary_key

As an administrator I will be able to set fields and tables which are
editable. Now when i go to the database management page i can do 'SELECT *
FROM CMS_FIELDS WHERE cms_table_id = '$_GET[table_id]' AND
cms_field_editable = TRUE

Also does anyone have any suggestions for editing static content?

Any comments here would be greatly appreciated.

Thanks





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[PHP-DB] table relationship

[shaun]

Hi,

I am creating a web site which will have different types of users:
Administrators, clients and staff. Is it possible/good practice to have 3
tables related to one table i.e.


 USER
 user_id (PK)
  |
  |
  |   ||
  |   ||
AdministratorClient  Staff
admin_id(PK)   client_id(PK)   staff_id(PK)
user_id(FK)  user_id(FK)user_id(FK)


Thanks in advance for any advice offered.



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[PHP-DB] Re: Cannot connect PostgreSQL 7.3.1. Using PHP thru...

[H. Etzel]

-BEGIN PGP SIGNED MESSAGE-

I think  the problem is the username. you have a user "administrator"? At 
postresql the administratzor is the user postgres.

Perhaps you try first with 'reateuser' to create a new user, peraps with all 
rights, and then you can try to connect with  this username an password.

Hope this can help

H.Etzel



Patrick Lok wrote:

> tcpip_socket is enabled!
> access to postgresql is granted; ip-address is added to pg_hba.conf -
> METHOD = trust
> 
> ./pl
> 
> 
> 
> "Patrick Lok" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
>> I have the following setups:
>> *PostgreSql 7.3.1 and MSSQL on Win2k server (A)
>> *Apache 1.3.x + PHP 4.3.1.1(using CGI & loaded w/ php_pgsql.dll) +  ODBC
> for
>> PostgreSql v7.02.00.05 on Win2K server (B)
>>
>> I tried to use phppgadmin 2.4.2  and odbc for postgresql to connect
>> PostgreSql in (A) but both connection failed!
>>
>> Can somebody give me some advices?
>> Can PHP access remote DB server?
>> Can ODBC for PostgreSql access remote DB server?
>>
>> Best regards
>> ./pl
>>
>>
>>
>> Using  PHPPGADMIN 2.4.2 to administer but it says "Wrong
> username/password.
>> Access denied".
>>
>> The following is logged on psqlodbc_.log:
>> DSN info:
>>
> 
DSN='PostgreSQL',server='192.168.1.16',port='5432',dbase='tet',user='adminis
>> trator',passwd=''
>>
>> onlyread='0',protocol='6.4',showoid='0',fakeoidindex='0',showsystable='0'
>>   conn_settings='',conn_encoding='OTHER'
>>   translation_dll='',translation_option=''
>>
>>
>> I wrote a simple php program in (B) to connect to server (A) using odbc
> for
>> postgresql but the connection failed!
>>
>> > $rtn = odbc_connect("PostgreSql","administrator","");
>> $cur=odbc_exec($rtn,"select d_base, high_jrnal from salfmsc");
>> while(odbc_fetch_row($cur)){
>> $fld1=odbc_result($cur,1);
>> $fld2=odbc_result($cur,2);
>> echo $fld1;
>> echo ("-");
>> echo $fld2;
>> echo ("");
>> }
>> echo odbc_close($rtn);
>> ?>
>>
>> BUT Warning is returned.
>> Warning: SQL error: Could not connect to the server; Could not connect to
>> remote socket., SQL state 08001 in SQLConnect in
>> d:\tmp\webcodetest\mstest.php on line 3
>>
>> The following is logged on psqlodbc_.log:
>> DSN info:
>>
> 
DSN='PostgreSql',server='192.168.1.16',port='5432',dbase='tet',user='adminis
>> trator',passwd=''
>>
>> onlyread='0',protocol='6.4',showoid='0',fakeoidindex='0',showsystable='0'
>>   conn_settings='',conn_encoding='OTHER'
>>   translation_dll='',translation_option=''
>> conn = 15742608, PGAPI_Connect(DSN='PostgreSql', UID='teter', PWD='')
>> Global Options: Version='07.02.0005', fetch=100, socket=4096,
>> unknown_sizes=0, max_varchar_size=254, max_longvarchar_size=8190
>> disable_optimizer=1, ksqo=1, unique_index=1,
>> use_declarefetch=0
>> text_as_longvarchar=1, unknowns_as_longvarchar=0,
>> bools_as_char=1 NAMEDATALEN=64
>> extra_systable_prefixes='dd_;', conn_settings=''
>> conn_encoding='OTHER'
>> CONN ERROR: func=PGAPI_Connect, desc='Error on CC_connect', errnum=101,
>> errmsg='Could not connect to the server'
>> 
>> henv=15742552, conn=15742608, status=0, num_stmts=16
>> sock=15753592, stmts=15761872, lobj_type=-999
>>  Socket Info ---
>> socket=-1, reverse=0, errornumber=4, errormsg='Could not
> connect
>> to remote socket.'
>> buffer_in=15753664, buffer_out=15757768
>> buffer_filled_in=0, buffer_filled_out=0, buffer_read_in=0
>>
>>

- -- 


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Re: [PHP-DB] PHP + Interbase / Firebird extensions

[Koleszár Tibor]

Dear Andrew

You can download the source from:
http://packages.debian.org/unstable/web/php4-interbase.html


Cheers

Tibor

- Original Message -
From: "Andrew Neillans" <[EMAIL PROTECTED]>
To: "'Koleszár Tibor'" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Saturday, March 08, 2003 4:41 PM
Subject: RE: [PHP-DB] PHP + Interbase / Firebird extensions


> Tibor,
>
> Thanks for your response!
>
> I thought it would be fairly simple to find the RPM for Mandrake / Redhat
> and install it, but I can't seem to find one anywhere for php4-interbase.
>
> Apache and PHP are compiled for module support, as I have a couple of
> modules already installed and running.
>
> Any idea where I can download the interbase module??
>
> Andy
>
> > Dear Andy,
> >
> > Sorry if i cant answer for your question, I'm a Debian user and I don't
> > know
> > the apache and php packages of Readhat and Mandrake. In Debian simply
> > you have to apt-get the php4-interbase module for php and you have to
> > restart apache.
> > You should check your system about:
> >  - is apache configured and compiled for modules?
> >  - is php compiled for modules?
> >  - if so, compile or download interbase module for you version of php
and
> > restart apache.
> >
> > This should work.
> >
> > Tibor
>
>
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> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
>



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