Re: [PHP-DB] Re: php-db foreign key
On Aug 9, 2011, at 3:41 PM, Peter Lind wrote:
On 9 August 2011 22:38, Chris Stinemetz chrisstinem...@gmail.com
wrote:
Yes, debug your code and figure out why it's looping twice instead.
For instance, try the other query in the mysql console.
Thank you! It was the first query. I put a LIMIT 1 on it and now it
is
working correctly. I appreciate your help!
You're fixing the symptom, not the problem. Your query was returning
multiple values because something is wrong with the query or your
data. If you don't correct it, the problem will likely just grow
bigger.
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Hi Chris,
Is this what your looking for?
?php
//store.php
include_once 'includes/connect.php';
include_once 'includes/header.php';
//print(pre.print_r($_GET,true)./pre); //Not sure what this is
for, uncomment if needed.
if(!$_SESSION['signed_in']) {//Check to see if signed in first, then
do search, otherwise display and search nothing.
echo('table class=table1 border=1
tr
td colspan=2You must be a href=signin.phpsigned in/a to
reply. You can also a href=signup.phpsign up/a for an account./
td
/tr
/table');
}
else {
$sql = SELECT * FROM stores WHERE store_subject =
'.mysql_real_escape_string($_GET['id']).';
$result = mysql_query($sql);
//fetch the posts from the database
$posts_sql = SELECT
stores.store_subject,
stores.store_comments,
stores.store_date,
stores.store_tptest,
stores.store_by,
users.user_id,
users.user_name,
users.first_name,
users.last_name
FROM
stores
LEFT JOIN
users
ON
stores.store_by = users.user_id
WHERE
stores.store_subject = '.mysql_real_escape_string($_GET['id']).'
ORDER BY
stores.store_date
DESC LIMIT=30;// Set a number of records if you dont want to display
ALL records on one page and want to build in pagination.
if(!$result || mysql_num_rows($result) == 0) {
echo('This store visit doesnprime;t exist.');
}
else {
echo('table class=table1 border=1');
while($row = mysql_fetch_array($result)) {
//display post data
$posts_result = mysql_query($posts_sql);//This may be the culprit
you had. Do you mean to do the $posts_sql query for ever result in
$sql query?? If not, move out of while loop above. Also, can you get
the same result by just using the $posts_sql query and axe the $sql
query?? They look like you get the same results to me. May be redundant.
echo('trtd
colspan=2'.$row['store_subject'].'/td/tr');
if(!$posts_result || mysql_num_rows($posts_result) ==
0) {
echo ('trtdNo Post results./td/tr/table');
} else {
while($posts_row =
mysql_fetch_array($posts_result)) {
echo('tr class=topic-post
td class=user-post'.$posts_row['first_name'].'$nbsp;'.
$posts_row['last_name'].'br/'.date('m-d-Y h:iA',
strtotime($posts_row['store_date'])).'/td
/tr
trtd class=post-content'.$posts_row['store_tptest'].'br/
'.htmlentities(stripslashes($posts_row['store_comments'])).'/td
/tr');
}
}
//show reply box
echo('trtd colspan=2h2Reply:/h2br /
form method=post
action=reply.php?id='.$row['store_id'].'
textarea name=reply-content/textareabr /br /
input type=submit value=Submit reply /
/form/td/tr');
//finish the table
echo('/table');
}
}
}
include_once 'includes/footer.php';
?
HTH,
Karl DeSaulniers
Design Drumm
http://designdrumm.com
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[PHP-DB] foreign key
Okay. I am pretty new to mysql so this may seem like a ridiculous question to some people. I am trying to use a LEFT JOIN query, but the results I am finding unusual. For every record occurrence there is for the query, the results are duplicated by that amount. So if there are 3 records from the query results, then the output is 3 times what I expect.. if that makes sense. From what I have researched so far. I believe I may need to add a foreign key to build the relations between the two tables. Based on the query can any tell me the correct way of adding the foreign key if ,in fact, that is what I need? I can provide table structures and information if necessary. The query is: $posts_sql = SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = ' . mysql_real_escape_string($_GET['id']).' ORDER BY stores.store_date DESC ; The query dump is: SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Noland Park Plaza 3509 S. Noland Rd' ORDER BY stores.store_date DESC -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] foreign key
On 9 August 2011 20:31, Chris Stinemetz chrisstinem...@gmail.com wrote: Okay. I am pretty new to mysql so this may seem like a ridiculous question to some people. I am trying to use a LEFT JOIN query, but the results I am finding unusual. For every record occurrence there is for the query, the results are duplicated by that amount. So if there are 3 records from the query results, then the output is 3 times what I expect.. if that makes sense. From what I have researched so far. I believe I may need to add a foreign key to build the relations between the two tables. Based on the query can any tell me the correct way of adding the foreign key if ,in fact, that is what I need? I can provide table structures and information if necessary. The query is: $posts_sql = SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = ' . mysql_real_escape_string($_GET['id']).' ORDER BY stores.store_date DESC ; The query dump is: SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Noland Park Plaza 3509 S. Noland Rd' ORDER BY stores.store_date DESC Is users.user_id unique? -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] foreign key
Is users.user_id unique? yes it is. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] foreign key
On 9 August 2011 20:49, Chris Stinemetz chrisstinem...@gmail.com wrote: Is users.user_id unique? yes it is. What does your result look like? Hard to say what the problem is without seeing the result. -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] foreign key
What does your result look like? Hard to say what the problem is without seeing the result. I am echoing the query and printing the get array just for debugging purposes, but below you can see how it is repeating its' self. Thank you Array ( [id] = Bella Roe 4980 Roe Blvd ) Bella Roe 4980 Roe BlvdSELECT stores.store_subject, stores.store_comments, stores.store_date, stores.store_tptest, stores.store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Bella Roe 4980 Roe Blvd' ORDER BY stores.store_date DESC Chris Stinemetz 08-09-2011 02:08PM600kbps-3.8mbps testChris Stinemetz 08-09-2011 02:07PM0-250kbps test1Reply: Bella Roe 4980 Roe BlvdSELECT stores.store_subject, stores.store_comments, stores.store_date, stores.store_tptest, stores.store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Bella Roe 4980 Roe Blvd' ORDER BY stores.store_date DESC Chris Stinemetz 08-09-2011 02:08PM600kbps-3.8mbps testChris Stinemetz 08-09-2011 02:07PM0-250kbps test1Reply:
Re: [PHP-DB] foreign key
On 9 August 2011 21:23, Chris Stinemetz chrisstinem...@gmail.com wrote: What does your result look like? Hard to say what the problem is without seeing the result. I am echoing the query and printing the get array just for debugging purposes, but below you can see how it is repeating its' self. Thank you So you're saying that SELECT stores.store_subject, stores.store_comments, stores.store_date, stores.store_tptest, stores.store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Bella Roe 4980 Roe Blvd' ORDER BY stores.store_date DESC returns Chris Stinemetz 08-09-2011 02:08PM 600kbps-3.8mbps test Chris Stinemetz 08-09-2011 02:08PM 600kbps-3.8mbps test Chris Stinemetz 08-09-2011 02:07PM 0-250kbps test1 Chris Stinemetz 08-09-2011 02:07PM 0-250kbps test1 From the above, can't see where your problem is but something in your join is obviously not unique - whether it's the first or second table. -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] foreign key
Okay. I just validated it is not in the left join query, but in my php. I
was able to build the correct relations.
Does anyone see any think in the php code below that may cause duplicate
table rows?
Thank you all.
?php
//store.php
include_once 'includes/connect.php';
include_once 'includes/header.php';
print(pre.print_r($_GET,true)./pre);
$sql = SELECT store_id, store_subject, store_comments, store_date,
store_tptest, store_by
FROM stores
WHERE store_subject = ' . mysql_real_escape_string($_GET['id']).';
$result = mysql_query($sql);
if(!$result)
{
echo 'The latest post could not be displayed, please try again later.';
//echo mysql_error(); //debugging purposes, uncomment when needed
}
else
{
if(mysql_num_rows($result) == 0)
{
echo 'This store visit doesnprime;t exist.';
}
else
{
while($row = mysql_fetch_assoc($result))
{
//display post data
echo 'table class=table1 border=1
tr
th colspan=2' . $row['store_subject'] . '/th
/tr';
//fetch the posts from the database
$posts_sql = SELECT
stores.store_subject,
stores.store_comments,
stores.store_date,
stores.store_tptest,
stores.store_by,
users.user_id,
users.user_name,
users.first_name,
users.last_name
FROM
stores
LEFT JOIN
users
ON
stores.store_by = users.user_id
WHERE
stores.store_subject = ' . mysql_real_escape_string($_GET['id']).'
ORDER BY
stores.store_date DESC ;
//echo $posts_sql;
$posts_result = mysql_query($posts_sql);
if(!$posts_result)
{
echo 'trtdThe posts could not be displayed, please try again
later./tr/td/table';
//echo mysql_error(); //debugging purposes, uncomment when needed
}
else
{
while($posts_row = mysql_fetch_assoc($posts_result))
{
echo 'tr class=topic-post
td class=user-post' . $posts_row['first_name'] . ' ' .
$posts_row['last_name'] . 'br/' . date('m-d-Y h:iA',
strtotime($posts_row['store_date'])) . '/td
td class=post-content' . $posts_row['store_tptest'] . 'br/' .
htmlentities(stripslashes($posts_row['store_comments'])) . '/td
/tr';
}
}
if(!$_SESSION['signed_in'])
{
echo 'trtd colspan=2You must be a href=signin.phpsigned in/a
to reply. You can also a href=signup.phpsign up/a for an account.';
}
else
{
//show reply box
echo 'trtd colspan=2h2Reply:/h2br /
form method=post action=reply.php?id=' . $row['store_id'] . '
textarea name=reply-content/textareabr /br /
input type=submit value=Submit reply /
/form/td/tr';
}
//finish the table
echo '/table';
}
}
}
include_once 'includes/footer.php';
?
[PHP-DB] Re: php-db foreign key
You already have a foreign key, that is stores.store_by references users.user_id. You might not have declared it (which is OK) but if that is the key you want that is fine. I suspect you are seeing an inadvertent Cartesian product. The way you have written this query you will get one row from the stores table for each row in the users table where a store_by = user_id - and because you said LEFT JOIN you will get one row from the stores table even if there are no matching rows in the users table. So you say you get 3 x the rows you're expecting; are there 3 users that match that store_by? Good Luck, Frank On Aug 9, 2011, at 11:31 AM, php-db-digest-h...@lists.php.net wrote: From: Chris Stinemetz chrisstinem...@gmail.com Subject: foreign key Date: August 9, 2011 11:31:51 AM PDT To: php-db@lists.php.net Okay. I am pretty new to mysql so this may seem like a ridiculous question to some people. I am trying to use a LEFT JOIN query, but the results I am finding unusual. For every record occurrence there is for the query, the results are duplicated by that amount. So if there are 3 records from the query results, then the output is 3 times what I expect.. if that makes sense. From what I have researched so far. I believe I may need to add a foreign key to build the relations between the two tables. Based on the query can any tell me the correct way of adding the foreign key if ,in fact, that is what I need? I can provide table structures and information if necessary. The query is: $posts_sql = SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = ' . mysql_real_escape_string($_GET['id']).' ORDER BY stores.store_date DESC ; The query dump is: SELECT store_subject, store_comments, store_date, store_tptest, store_by, users.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.user_id WHERE stores.store_subject = 'Noland Park Plaza 3509 S. Noland Rd' ORDER BY stores.store_date DESC
[PHP-DB] Re: php-db foreign key
On Tue, Aug 9, 2011 at 2:55 PM, Frank Flynn fr...@declan.com wrote: You already have a foreign key, that is stores.store_by references users.user_id. You might not have declared it (which is OK) but if that is the key you want that is fine. I suspect you are seeing an inadvertent Cartesian product. The way you have written this query you will get one row from the stores table for each row in the users table where a store_by = user_id - and because you said LEFT JOIN you will get one row from the stores table even if there are no matching rows in the users table. So you say you get 3 x the rows you're expecting; are there 3 users that match that store_by? No there is only 1 user that match the sotre_by. If the query is not written correctly do you mind trying to show me how to correctly right it? When I dump the query and run it in console I get the results I want. Not sure what I am doing wrong. mysql SELECT stores.store_subject, stores.store_comments, stores.store_date, stores.store_tptest, stores.store_by, user s.user_id, users.user_name, users.first_name, users.last_name FROM stores LEFT JOIN users ON stores.store_by = users.use r_id WHERE stores.store_subject = 'Bella Roe 4980 Roe Blvd' ORDER BY stores.store_date DESC ; +-++-+-+--+-+--- -++---+ | store_subject | store_comments | store_date | store_tptest| store_by | user_id | user_name | first_name | last_name | +-++-+-+--+-+--- -++---+ | Bella Roe 4980 Roe Blvd | test | 2011-08-09 14:08:05 | 600kbps-3.8mbps |1 | 1 | chrisstinemetz | Chris | Stinemetz | | Bella Roe 4980 Roe Blvd | test1 | 2011-08-09 14:07:49 | 0-250kbps |1 | 1 | chrisstinemetz | Chris | Stinemetz | +-++-+-+--+-+--- -++---+ 2 rows in set (0.00 sec)
Re: [PHP-DB] Re: php-db foreign key
When I dump the query and run it in console I get the results I want. Not sure what I am doing wrong. Your php code had more than one query running (one inside the other). It's the outer query that runs twice, not the inner one returning double the results -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: php-db foreign key
Your php code had more than one query running (one inside the other). It's the outer query that runs twice, not the inner one returning double the results Thanks Peter, Do you have any suggestions on how to fix this? Thank you, Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: php-db foreign key
On 9 August 2011 22:25, Chris Stinemetz chrisstinem...@gmail.com wrote: Your php code had more than one query running (one inside the other). It's the outer query that runs twice, not the inner one returning double the results Thanks Peter, Do you have any suggestions on how to fix this? Thank you, Chris Yes, debug your code and figure out why it's looping twice instead. For instance, try the other query in the mysql console. -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: php-db foreign key
Yes, debug your code and figure out why it's looping twice instead. For instance, try the other query in the mysql console. Thank you! It was the first query. I put a LIMIT 1 on it and now it is working correctly. I appreciate your help! Thank you, Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: php-db foreign key
On 9 August 2011 22:38, Chris Stinemetz chrisstinem...@gmail.com wrote: Yes, debug your code and figure out why it's looping twice instead. For instance, try the other query in the mysql console. Thank you! It was the first query. I put a LIMIT 1 on it and now it is working correctly. I appreciate your help! You're fixing the symptom, not the problem. Your query was returning multiple values because something is wrong with the query or your data. If you don't correct it, the problem will likely just grow bigger. -- hype WWW: plphp.dk / plind.dk LinkedIn: plind BeWelcome/Couchsurfing: Fake51 Twitter: kafe15 /hype -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Foreign Key Versus Table Index
Hi,
I'm slightly confused about foriegn keys and indexes on mysql innodb tables.
Foreign key constraints create a reference between two tables and
indexes make queries on a particular table faster if the index is on a
field in the where or order by clause.
My question was whether say for the following two tables:
Person Car
Id Id
Name PersonId
Address Make
Phone Number Colour
If I create a foriegn key linking the id field in person and the
personid field in car, do I need to create another index in car table
specifically for the personid field if I was running a query such as:
SELECT Id FROM car WHERE personid={personkeynumber}?
Thanks for your help.
Jemma
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Re: [PHP-DB] Foreign Key Versus Table Index
I am not up to scratch with innodb, but in oracle
you would have a Primary key on both the id fields in the Car and person
table and a Foreign key on PersonId linking it to Id in the Person table.
IN your select an index on PersonId would be beneficial if the tables get
large.
Jack
2008/10/2 J Hussein [EMAIL PROTECTED]
Hi,
I'm slightly confused about foriegn keys and indexes on mysql innodb
tables.
Foreign key constraints create a reference between two tables and indexes
make queries on a particular table faster if the index is on a field in the
where or order by clause.
My question was whether say for the following two tables:
Person Car
Id Id
Name PersonId
Address Make
Phone Number Colour
If I create a foriegn key linking the id field in person and the personid
field in car, do I need to create another index in car table specifically
for the personid field if I was running a query such as:
SELECT Id FROM car WHERE personid={personkeynumber}?
Thanks for your help.
Jemma
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--
J.A. van Zanen
Re: [PHP-DB] Foreign Key Versus Table Index
On Thu, Oct 2, 2008 at 6:50 AM, Jack van Zanen [EMAIL PROTECTED] wrote:
I am not up to scratch with innodb, but in oracle
you would have a Primary key on both the id fields in the Car and person
table and a Foreign key on PersonId linking it to Id in the Person table.
IN your select an index on PersonId would be beneficial if the tables get
large.
Jack
2008/10/2 J Hussein [EMAIL PROTECTED]
Hi,
I'm slightly confused about foriegn keys and indexes on mysql innodb
tables.
Foreign key constraints create a reference between two tables and indexes
make queries on a particular table faster if the index is on a field in
the
where or order by clause.
My question was whether say for the following two tables:
Person Car
Id Id
Name PersonId
Address Make
Phone Number Colour
If I create a foriegn key linking the id field in person and the personid
field in car, do I need to create another index in car table
specifically
for the personid field if I was running a query such as:
SELECT Id FROM car WHERE personid={personkeynumber}?
Thanks for your help.
Jemma
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--
J.A. van Zanen
Jenna,
FKs are not to create relationships but act as constraints inside the data
base. What that means is that for a record to be created that has a FK in
another table, that FK'ed record MUST exist in the table before you can add
that record.
Ex. AN Order table and a Customer table
If the Order table references the Customer table as a FK on the CustomerID
field, the CustomerId record MUST exist in the table before the order table
can be filled.
It does not alleviate the need to have a primary key on the other table.
http://dev.mysql.com/doc/refman/5.0/en/innodb-foreign-key-constraints.htmlfor
more details
--
Bastien
Cat, the other other white meat
RE: [PHP-DB] foreign key problem
A foreign key value can be null, if it suits the data application - or that is what I've always been taught. Here is a text book definition of Referential Integrity which was spoon fed to me by the Open University when studying their RDBMS course: "Referential Integrity - If a relation (table), R2 has a foreign key, F, that references the primary key, P, in another relation (table), R1, then every R2.F entry must either be a value equal to an R1.P primary key value or be null" Regards, Chris -Original Message- From: Bob Hall [mailto:[EMAIL PROTECTED]] Sent: 31 January 2001 01:53 To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] foreign key problem hello all! i have a little problem ,hope someone can help me out the problem is : let i have two tables T1 and T2 ,now T1 has following fields cntryid cntryname cntrycode etc. now cntryid is the primary key now i want to make the cntryid a foreign key in table T2 .so how can i do it plz help me . msjamal A column is a foreign key because it contains only values found in the referenced key, and no NULLs. Design your database so that the foreign key column contains only values from cntryid. If you want to know how to declare a referenced key/foreign key relationship, you will have to specify what RDBMS you are using. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] foreign key problem
Yes sir, you are correct. Thank you for correcting me. In practice, a NULL in a foreign key almost always means an orphan record. In most applications the designer will want to use the NOT NULL constraint. Bob Hall A foreign key value can be null, if it suits the data application - or that is what I've always been taught. Here is a text book definition of Referential Integrity which was spoon fed to me by the Open University when studying their RDBMS course: "Referential Integrity - If a relation (table), R2 has a foreign key, F, that references the primary key, P, in another relation (table), R1, then every R2.F entry must either be a value equal to an R1.P primary key value or be null" Regards, Chris -Original Message- From: Bob Hall [mailto:[EMAIL PROTECTED]] Sent: 31 January 2001 01:53 To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] foreign key problem hello all! i have a little problem ,hope someone can help me out the problem is : let i have two tables T1 and T2 ,now T1 has following fields cntryid cntryname cntrycode etc. now cntryid is the primary key now i want to make the cntryid a foreign key in table T2 .so how can i do it plz help me . msjamal A column is a foreign key because it contains only values found in the referenced key, and no NULLs. Design your database so that the foreign key column contains only values from cntryid. If you want to know how to declare a referenced key/foreign key relationship, you will have to specify what RDBMS you are using. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
