[PHP-DB] mssql problems
Dear All, I hope somebody can help me. I am new to PHP (I've used ASP for three years:-( ) and I'm loving PHP and trying to get my work to convert but I've coming across a error with mssql_query() explorer its just crashes with a php.exe 'memory could not be read error'. I'm sure it a school boy error from me but I'm just stuck, please help me code is.. === $link = mssql_connect (laptop, sa, elmwood); print (brLINK was $link); $back = mssql_select_db ( elmback,$link); print(brback was $back); $query = SELECT * FROM tblperson; print(brquery was $query); $result = mssql_query($query,$link); //JUST CRASHES ON THIS LINE $r = mssql_rows_affected ( $link); print(brrows affected was $r); $close = mssql_close ($link); print(brclose was $close); = I run IIS4, NT4 (SP6a) and SQL 7 on the same machine as its my development one, out of date maybe but you should see my clothes but PHP was test downloaded a month ago - any ideas from a white knight? best regards Rob -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Connection error
Bisexuality doubles the chance of getting a date.
Hi All!
In order to make my job easy, a choose the following approach:
settings.php
?
$dbserver = 'server';
$dbname = 'name';
$dbpass = 'password';
?
main_file.php
?
include settings.php;
mysql_connect($dbserver,$dbname,$dbpass) or die (Error);
?
On the test machine (Windows 98, PHP Triad 221 - PHP 4.1.0) it works;
when I put the project online, it doesn't (the Error message shows up). If I
don't use this approach and, in main_file.php, I use the line:
mysql_connect('server','name','pass') or die (Error)
the sistem works.
What's the problem ?
--
Ciprian
Confucius Says: Man who fight with wife all day, get no piece at night!
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Re: [PHP-DB] Connection error
On Friday 19 April 2002 18:29, Ciprian Trofin wrote:
Bisexuality doubles the chance of getting a date.
Hi All!
In order to make my job easy, a choose the following approach:
settings.php
?
$dbserver = 'server';
$dbname = 'name';
$dbpass = 'password';
?
main_file.php
?
include settings.php;
mysql_connect($dbserver,$dbname,$dbpass) or die (Error);
?
On the test machine (Windows 98, PHP Triad 221 - PHP 4.1.0) it
works; when I put the project online, it doesn't (the Error message shows
up). If I don't use this approach and, in main_file.php, I use the line:
mysql_connect('server','name','pass') or die (Error)
the sistem works.
What's the problem ?
Always check for errors. Use mysql_error() to see what the error is. Also try
connecting to mysql from the shell/command-line.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
/*
It isn't easy being the parent of a six-year-old. However, it's a pretty
small
price to pay for having somebody around the house who understands computers.
*/
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[PHP-DB] Re:[PHP-DB] Connection error
Ciprian Trofin [EMAIL PROTECTED] wrote on 4/19/2002
5:29:00 PM:
Bisexuality doubles the chance of getting a date.
Hi All!
In order to make my job easy, a choose the following approach:
settings.php
?
$dbserver = 'server';
$dbname = 'name';
$dbpass = 'password';
?
main_file.php
?
include settings.php;
@mysql_connect($dbserver,$dbname,$dbpass) or die (Error);
?
On the test machine (Windows 98, PHP Triad 221 - PHP 4.1.0) it
works;
when I put the project online, it doesn't (the Error message shows
up). If I don't use this approach and, in main_file.php, I use the
line:
@mysql_connect('server','name','pass') or die (Error)
the sistem works.
What's the problem ?
try double quote instead single quote.
$dbserver = server;
_
adi.baliroomfinder.net
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Re: [PHP-DB] Re:[PHP-DB] Connection error
Monday is an awful way to spend 1/7th of your life. a try double quote instead single quote. a $dbserver = server; Tried that, no effect. -- Ciprian Oxymoron: Government budget -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Connection error
Instead of suppressing the error message using the @ sign, why not
remove it and your die() statement and see what error is produced on
your live server?
HTH
--Sam
-Original Message-
From: Ciprian Trofin [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 6:29 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Connection error
Bisexuality doubles the chance of getting a date.
Hi All!
In order to make my job easy, a choose the following approach:
settings.php
?
$dbserver = 'server';
$dbname = 'name';
$dbpass = 'password';
?
main_file.php
?
include settings.php;
@mysql_connect($dbserver,$dbname,$dbpass) or die (Error);
?
On the test machine (Windows 98, PHP Triad 221 - PHP 4.1.0) it
works;
when I put the project online, it doesn't (the Error message shows
up). If I
don't use this approach and, in main_file.php, I use the line:
@mysql_connect('server','name','pass') or die (Error)
the sistem works.
What's the problem ?
--
Ciprian
Confucius Says: Man who fight with wife all day, get no piece at
night!
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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[PHP-DB] PHP Compiling issues
Background: Server version: Apache/1.3.24 (Darwin) MySQL: Ver 8.23 Distrib 3.23.49, for apple-darwin5.2 on powerpc PHP version 4.1.2 from the Darwin CVS repository -- After much hassle and grief, I was able to make and install PHP 4.1.2 from the Darwin CVS source repository using this configure syntax: ./configure --with-mysql=/usr/local/mysql \ --with-apxs=/usr/sbin/apxs \ --with-curl=/sw \ --with-zlib-dir=/usr/local \ --with-pear However, when I ran 'apachectl configtest' I got a bunch of errors dealing with Undefined symbols, all of them starting with _mysql_* So, after doing some searching investigating, I re-ran the configure script with the following flags: ./configure --with-mysql=shared,/usr/local/mysql \ --with-apxs=/usr/sbin/apxs \ --with-curl=/sw \ --with-zlib-dir=/usr/local \ --with-pear but when I ran make, the following errors showed: *** Warning: This library needs some functionality provided by -lmysqlclient. *** I have the capability to make that library automatically link in when *** you link to this library. But I can only do this if you have a *** shared version of the library, which you do not appear to have. *** Warning: libtool could not satisfy all declared inter-library *** dependencies of module mysql. Therefore, libtool will create *** a static module, that should work as long as the dlopening *** application is linked with the -dlopen flag. but since the make wsa able to finish, I decided to try make install and see what happens. Upon running 'apachectl configtest' again, I got Syntax OK, as expected. What I want to know is, will the issues mentioned in the errors cause me any troubles down the road? Or can these errors be safely ignored. The intended use for this server is as a test platform, and it will not come under any type of heavy load. Please cc me, as I am in digest mode. TIA chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] RE: troubles with select * from... to a ms-sql-server
Have you tried removing the [] brackets? And what is ReferencesComplete - a standard table, a view or a stored procedure? Ollie -Original Message- From: Hermann Otteneder [mailto:[EMAIL PROTECTED]] Sent: 17 April 2002 16:30 To: [EMAIL PROTECTED] Subject: troubles with select * from... to a ms-sql-server hi, i try to get some data from our ms-sql-server the following statement: $SQL = SELECT * FROM [ReferencesComplete]; causes an error of the php.exe! i tried many various of this statement but nothing helped. ether it came no error and the [$result = mssql_query($SQL,$IDconnection);] variable was false or the php.exe crashed! with mysql it workes ok but with the ms-sql-server not! please give me help - this so what simple - but does not work - i lost so much time... hermann -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] RE: mssql problems
Hi PHP Is quite cool isn't it As for your problem you might want to try a few things: - Running PHP as the ISAPI module make sure you have the latest version (you should get the latest version anyway). - Removing this line : $r = mssql_rows_affected ( $link); - as there is a mysql function for this but not mssql - it should have read - mssql_rows_affected($result) anyway I believe - it works on the result rather than the connection). Can't really think of any thing else without more info, although I've had bad experiences with trying to run MSSQL / IIS on the same NT4 box (although on 2000 and XP it's fine) - one of my colleages solution was to run Apache for win32 instead of IIS as there appeared to be clashes between SQL and IIS (something we have since rectified). Good luck Ollie -Original Message- From: Rob Fraser [mailto:[EMAIL PROTECTED]] Sent: 19 April 2002 08:44 To: [EMAIL PROTECTED] Subject: mssql problems Dear All, I hope somebody can help me. I am new to PHP (I've used ASP for three years:-( ) and I'm loving PHP and trying to get my work to convert but I've coming across a error with mssql_query() explorer its just crashes with a php.exe 'memory could not be read error'. I'm sure it a school boy error from me but I'm just stuck, please help me code is.. === $link = mssql_connect (laptop, sa, elmwood); print (brLINK was $link); $back = mssql_select_db ( elmback,$link); print(brback was $back); $query = SELECT * FROM tblperson; print(brquery was $query); $result = mssql_query($query,$link); //JUST CRASHES ON THIS LINE $r = mssql_rows_affected ( $link); print(brrows affected was $r); $close = mssql_close ($link); print(brclose was $close); = I run IIS4, NT4 (SP6a) and SQL 7 on the same machine as its my development one, out of date maybe but you should see my clothes but PHP was test downloaded a month ago - any ideas from a white knight? best regards Rob -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] page expires
How about this incorporated somewhere: $s = (SERVER_PROTOCOL == HTTP/1.0) ? Pragma: no-cache : Cache-Control: no-cache, must-revalidate; header($s); This code must be at the start of the script BEFORE anything is output to the browser - as it sends an additional HTTP header (hence the function header()). Hope that helps Ollie -Original Message- From: Jonathan Hilgeman [mailto:[EMAIL PROTECTED]] Sent: 18 April 2002 19:37 To: 'Natividad Castro'; Subject: RE: [PHP-DB] page expires http://php.sitecreative.com/faq.php -Original Message- From: Natividad Castro [mailto:[EMAIL PROTECTED]] Sent: Wednesday, April 17, 2002 6:12 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] page expires Hi to all, I have form that users fill out and submit. What I'm trying to do is not to let users to go back to the form after they have submmited, or if they go bak, make the page expires. Can someone please tell me or give an idea on how to do it? Thanks in advanced Nato -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Date of Birth From Form
Hey all, I have a credit application form, and one of the required elements is the applicant's Date of Birth. I have creditapp table with a birthdate field and it is a DATE datatype. On the form, I want to be able to have them enter their Date of Birth as mm/dd/ and have it go into the database properly (-mm-dd). Is there a way to do it this way, or am I going about it wrong? Any help would be greatly appreciated. Thanks! Brandon -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date of Birth From Form
On Fri, 19 Apr 2002, Brandon Paul wrote: going about it wrong? Any help would be greatly appreciated. 1) Validate that the input is correct. 2) Tokenize the string based on the /s 3) Re-format the date. Or, use the date functions - I'd swear one of them can re-write dates. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date of Birth From Form
A combination of strtotime() and date() can make this easy. date(Y-m-d, strtotime($birthday)) Or see docs and comments at http://www.php.net/manual/en/function.strtotime.php -Steve On Friday, April 19, 2002, at 01:29 PM, Brandon Paul wrote: Hey all, I have a credit application form, and one of the required elements is the applicant's Date of Birth. I have creditapp table with a birthdate field and it is a DATE datatype. On the form, I want to be able to have them enter their Date of Birth as mm/dd/ and have it go into the database properly (-mm-dd). Is there a way to do it this way, or am I going about it wrong? Any help would be greatly appreciated. Thanks! Brandon -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] IP Address?
How can I find out the IP address of the client requesting a given php file? I know that you can use $HTTP_SERVER_VARS[REMOTE_ADDR] in more recent versions of PHP. Unfortunately, we only have PHP3.0.14 and I don't think that this option is available in our ancient version. I also can't use server side includes. An upgrade to PHP4 is in the works, but I can't wait that long. How can I get a client's IP address with PHP3.014 and/or JavaScript? Thanks in advance. -Rob -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] IP Address?
How can I find out the IP address of the client requesting a
given php file?
I know that you can use $HTTP_SERVER_VARS[REMOTE_ADDR] in more recent
versions of PHP. Unfortunately, we only have PHP3.0.14 and I don't think
that this option is available in our ancient version. I also can't use
server side includes. An upgrade to PHP4 is in the works, but I can't wait
that long. How can I get a client's IP address with PHP3.014 and/or
JavaScript? Thanks in advance.
I believe
$ip = getenv('REMOTE_ADDR');
will do what you want.
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RE: [PHP-DB] IP Address?
Thanks, Craig. That was exactly what I was looking for!
-Rob
-Original Message-
From: Craig Vincent [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 2:35 PM
To: Rob Day; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] IP Address?
How can I find out the IP address of the client requesting a
given php file?
I know that you can use $HTTP_SERVER_VARS[REMOTE_ADDR] in more recent
versions of PHP. Unfortunately, we only have PHP3.0.14 and I don't think
that this option is available in our ancient version. I also can't use
server side includes. An upgrade to PHP4 is in the works, but I can't wait
that long. How can I get a client's IP address with PHP3.014 and/or
JavaScript? Thanks in advance.
I believe
$ip = getenv('REMOTE_ADDR');
will do what you want.
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RE: [PHP-DB] IP Address?
?
phpinfo();
?
should give you all of the info you need.
Sincerely
berber
Visit http://www.weberdev.com Today!!!
To see where PHP might take you tomorrow.
-Original Message-
From: Rob Day [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 9:29 PM
To: 'Craig Vincent'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] IP Address?
Thanks, Craig. That was exactly what I was looking for!
-Rob
-Original Message-
From: Craig Vincent [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 2:35 PM
To: Rob Day; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] IP Address?
How can I find out the IP address of the client requesting a
given php file?
I know that you can use $HTTP_SERVER_VARS[REMOTE_ADDR] in more
recent
versions of PHP. Unfortunately, we only have PHP3.0.14 and I don't
think
that this option is available in our ancient version. I also can't use
server side includes. An upgrade to PHP4 is in the works, but I can't
wait
that long. How can I get a client's IP address with PHP3.014 and/or
JavaScript? Thanks in advance.
I believe
$ip = getenv('REMOTE_ADDR');
will do what you want.
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To unsubscribe, visit: http://www.php.net/unsub.php
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[PHP-DB] Help with a JOIN statement
I have 2 tables, 'users' and 'files' (defined below) Can I construct a single SELECT statement so that it will return ONLY users that have files and ONLY the last file that the user has? So it would return 1 row per user containing their most recent file. users Id int Namevarchar Files Id int Userid int Filenamevarchar Stamp int (unix_timestamp) The way I'm doing it now, I use SELECT DISTINCT to get all the users with files, then a separate query for each user returned to retrieve their latest file. This method requires many, many individual queries to the database? Any help is appreciated. Thanks, Mikey -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Help with a JOIN statement
You could try: SELECT users.Name, Files.Filename FROM Files LEFT JOIN users ON Files.Userid = users.Id GROUP BY Files.Userid ORDER BY Files.Stamp DESC I'm not sure if this would work, depends on if MySQL ORDER's before GROUP. You could try it though. Capitolization may be incorrect, I tried to use your convention, but I may have missed something. Robert V. Zwink http://www.zwink.net/daid.php -Original Message- From: Michael Robbins [mailto:[EMAIL PROTECTED]] Sent: Friday, April 19, 2002 4:55 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Help with a JOIN statement I have 2 tables, 'users' and 'files' (defined below) Can I construct a single SELECT statement so that it will return ONLY users that have files and ONLY the last file that the user has? So it would return 1 row per user containing their most recent file. users Id int Namevarchar Files Id int Userid int Filenamevarchar Stamp int (unix_timestamp) The way I'm doing it now, I use SELECT DISTINCT to get all the users with files, then a separate query for each user returned to retrieve their latest file. This method requires many, many individual queries to the database? Any help is appreciated. Thanks, Mikey -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: mssql problems
I had this exact same problem and after a lot of time I switched to an ODBC connection (no mssql_connect... Just use the ODBC commands) and everything has worked flawlessly since. There's probably some right combination and way to use the mssql specific commands but life it too short and the ODBC stuff will do everything you need anyway. It worked first time and every time since for me. Good Luck, Frank On 4/19/02 1:54 PM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: -Original Message- From: Rob Fraser [mailto:[EMAIL PROTECTED]] Sent: 19 April 2002 08:44 To: [EMAIL PROTECTED] Subject: mssql problems Dear All, I hope somebody can help me. I am new to PHP (I've used ASP for three years:-( ) and I'm loving PHP and trying to get my work to convert but I've coming across a error with mssql_query() explorer its just crashes with a php.exe 'memory could not be read error'. I'm sure it a school boy error from me but I'm just stuck, please help me code is.. === $link = mssql_connect (laptop, sa, elmwood); print (brLINK was $link); $back = mssql_select_db ( elmback,$link); print(brback was $back); $query = SELECT * FROM tblperson; print(brquery was $query); $result = mssql_query($query,$link); //JUST CRASHES ON THIS LINE $r = mssql_rows_affected ( $link); print(brrows affected was $r); $close = mssql_close ($link); print(brclose was $close); = I run IIS4, NT4 (SP6a) and SQL 7 on the same machine as its my development one, out of date maybe but you should see my clothes but PHP was test downloaded a month ago - any ideas from a white knight? best regards Rob -- Frank Flynn Poet, Artist Mystic -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Complex (for me) table join - help please?
I'm trying to perform a join that seems like it should be simple, but I'm just ending up in a morass. Hopefully someone who's standing a step or two back from the problem can help me solve this one... What I have: Table A (Category): with a 3-tier category structure: -Nature -Category -Sub-Category Table B (Responsibility): with a reference to Table A, and a client id, and the id of a person responsible for that Sub-category for that client. -TableA-id -Client-id -Person-id So, what I want is a simple select which will fetch for me all rows in (Category) with the relevant person from (Responsibility) if there is one for a specific Client, but with a blank in that column if there isn't one. ie. (Category) has 10 entries (let's call them Ca-Cj), and (Responsibility) has 5 entries for Client 1 (Ca,Cb,Cc,Cd,Ce) and 3 entries for Client 2 (Cd,Ce,Cf). When I run the select for Client 1, I'd like to see... Ca - person Cb - person Cc - person Cd - person Ce - person Cf - Cg - Ch - Ci - Cj - And when I run the same select for Client 2 I'd like to see: Ca - Cb - Cc - Cd - person Ce - person Cf - person Cg - Ch - Ci - Cj - I tried a basic outer join type select, and of course for Client 1 I miss Cf, and when I use it for Client 2 I miss Ca-Cc. Anyone able to sort me out here? I need to use a basic Select (instead of building a temp table or using variables in a PHP script) because another database process that I don't control needs to do the same thing. Think it's possible, or am I going to have to go and insert records into Responsibility for every entry in Category for every Client, resulting in (#Categories x #Clients) rows? (The latter idea sure doesn't make me happy, but it would be 'easier' to build, if not to maintain. (And, as you can appreciate, since I'm not the one maintaining it, I'm like THAT CLOSE from just going that way and saying 'screw em'!) Hopefully, -Peter Westergaard -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Complex (for me) table join - help please?
Peter, I've you tried doing a inner join, left outer join, right outer join? What db program are you using? Could you post what your current select statement looks like? Mike - Original Message - From: Peter Westergaard [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, April 19, 2002 5:46 PM Subject: [PHP-DB] Complex (for me) table join - help please? I'm trying to perform a join that seems like it should be simple, but I'm just ending up in a morass. Hopefully someone who's standing a step or two back from the problem can help me solve this one... What I have: Table A (Category): with a 3-tier category structure: -Nature -Category -Sub-Category Table B (Responsibility): with a reference to Table A, and a client id, and the id of a person responsible for that Sub-category for that client. -TableA-id -Client-id -Person-id So, what I want is a simple select which will fetch for me all rows in (Category) with the relevant person from (Responsibility) if there is one for a specific Client, but with a blank in that column if there isn't one. ie. (Category) has 10 entries (let's call them Ca-Cj), and (Responsibility) has 5 entries for Client 1 (Ca,Cb,Cc,Cd,Ce) and 3 entries for Client 2 (Cd,Ce,Cf). When I run the select for Client 1, I'd like to see... Ca - person Cb - person Cc - person Cd - person Ce - person Cf - Cg - Ch - Ci - Cj - And when I run the same select for Client 2 I'd like to see: Ca - Cb - Cc - Cd - person Ce - person Cf - person Cg - Ch - Ci - Cj - I tried a basic outer join type select, and of course for Client 1 I miss Cf, and when I use it for Client 2 I miss Ca-Cc. Anyone able to sort me out here? I need to use a basic Select (instead of building a temp table or using variables in a PHP script) because another database process that I don't control needs to do the same thing. Think it's possible, or am I going to have to go and insert records into Responsibility for every entry in Category for every Client, resulting in (#Categories x #Clients) rows? (The latter idea sure doesn't make me happy, but it would be 'easier' to build, if not to maintain. (And, as you can appreciate, since I'm not the one maintaining it, I'm like THAT CLOSE from just going that way and saying 'screw em'!) Hopefully, -Peter Westergaard -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.330 / Virus Database: 184 - Release Date: 2/28/02 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Complex (for me) table join - help please?
Try SELECT TableA.Category, TableB.PersonID FROM TableA LEFT JOIN TableB ON (TableA.CategoryID = TableB.TableAID) WHERE TableB.ClientID = $cid ORDER BY TableA.Category ASC LEFT JOIN states that you should return all TableA, inserting NULL for TableB if there is no match. 'Luck -Szii - Original Message - From: Peter Westergaard [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, April 19, 2002 2:46 PM Subject: [PHP-DB] Complex (for me) table join - help please? I'm trying to perform a join that seems like it should be simple, but I'm just ending up in a morass. Hopefully someone who's standing a step or two back from the problem can help me solve this one... What I have: Table A (Category): with a 3-tier category structure: -Nature -Category -Sub-Category Table B (Responsibility): with a reference to Table A, and a client id, and the id of a person responsible for that Sub-category for that client. -TableA-id -Client-id -Person-id So, what I want is a simple select which will fetch for me all rows in (Category) with the relevant person from (Responsibility) if there is one for a specific Client, but with a blank in that column if there isn't one. ie. (Category) has 10 entries (let's call them Ca-Cj), and (Responsibility) has 5 entries for Client 1 (Ca,Cb,Cc,Cd,Ce) and 3 entries for Client 2 (Cd,Ce,Cf). When I run the select for Client 1, I'd like to see... Ca - person Cb - person Cc - person Cd - person Ce - person Cf - Cg - Ch - Ci - Cj - And when I run the same select for Client 2 I'd like to see: Ca - Cb - Cc - Cd - person Ce - person Cf - person Cg - Ch - Ci - Cj - I tried a basic outer join type select, and of course for Client 1 I miss Cf, and when I use it for Client 2 I miss Ca-Cc. Anyone able to sort me out here? I need to use a basic Select (instead of building a temp table or using variables in a PHP script) because another database process that I don't control needs to do the same thing. Think it's possible, or am I going to have to go and insert records into Responsibility for every entry in Category for every Client, resulting in (#Categories x #Clients) rows? (The latter idea sure doesn't make me happy, but it would be 'easier' to build, if not to maintain. (And, as you can appreciate, since I'm not the one maintaining it, I'm like THAT CLOSE from just going that way and saying 'screw em'!) Hopefully, -Peter Westergaard -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
