[PHP-DB] Using multiple submits on a page and retaining $POST data
Hello, and I would like to thank everyone that attempts to help me out. I have looked long and hard online before I resorted to this. My problem is that I have a page that relies on multiple submits and data from what was submitted. example: // 1. some sort of drop down // 2. submit button // 3. display data retrieved from database // 4. user enters more information // 5. second submit button, run checks to vertify all data is entered // 6. enter data into database The problem is in step 5, when the user hits the second submit and data is checked before db submission, I am losing the data displayed by the first submit button This is what I have tried ive tried using POST and SESSION variables to display the data, but this doesnt work. this is because in my code i have an if isset statement to check if submit has been pressed, then the variables are retrieved from the db and used. when clicking the second submit button and checks are done, the top portion of the page of information retrieved by the first submit are nothing but blanks because the first submit is no longer set (using isset) my next idea was to save the variable selected in the first drop down (step 1) and requery with this selection at the second submit button to retrieve and redisplay the data. this did not work because this variable was set back to because the drop down resets itself on each submit. disabling the drop down if the first submit is set and then setting the variable did not work either because when the second submit is hit, the first submit is no longer set, so the drop down is displayed, the variable is set to and i get nothing in my variable. this also was the case if i made it a global variable. so basically i have resorted to just googling things and trying out different arrangements of code to no avail. I need a way to hold the data on the top half of the screen while the bottom half is being manipulated. any, and i mean any help at this point would be GREATLY appreciated! Thank you. Onochie
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help me out. I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits and data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit and data is
checked before db submission, I am losing the data displayed by the first
submit button
hmm i don't get it by reading at the first. i think in the 2nd page u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data, but this
doesnt work. this is because in my code i have an if isset statement to
check if submit has been pressed, then the variables are retrieved from the
db and used. when clicking the second submit button and checks are done, the
top portion of the page of information retrieved by the first submit are
nothing but blanks because the first submit is no longer set (using isset)
my next idea was to save the variable selected in the first drop down (step
1) and requery with this selection at the second submit button to retrieve
and redisplay the data. this did not work because this variable was set back
to because the drop down resets itself on each submit. disabling the drop
down if the first submit is set and then setting the variable did not work
either because when the second submit is hit, the first submit is no longer
set, so the drop down is displayed, the variable is set to and i get
nothing in my variable. this also was the case if i made it a global
variable.
i wish to see the offline from your pages.. if able send to me
so basically i have resorted to just googling things and trying out
different arrangements of code to no avail. I need a way to hold the data on
the top half of the screen while the bottom half is being manipulated. any,
and i mean any help at this point would be GREATLY appreciated! Thank you.
Onochie
--
akan ada dimana mulut terkunci dan suara tak ada lagi..
saat itu gunakanlah HP untuk melakukan SMS!!
- ini aliran bedul.. bukan aliran aneh.
tertawa sebelum tertawa didepan RSJ..
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
Hello Gunawan
On your first comment, im not sure what you were referring to when you said
it is a js problem
on your second comment, i am talking about the user entered data. data
submitted by the first submit button will always be predefined by means of
the user using a selection box, the data im concerned with checking is the
input boxes he will be seeing. and this wont necessarily be username/pass.
and im trying to do this in php if at all possible.
on your next comment, (6) i am talking about after i vertified the validity
of the user typed/entered data, i will enter that into my db
next comment - i knew this would be hard to describe, but i'll try again.
this is to be done all on one page. and ive tried using something like:
value=\$_POST['name']\
but i am having problems keeping the initial drop down at the selected
choice, so whenever either submit button is pressed and the post goes to php
self, the drop down is refreshed back to its starting value, which is NULL
I hope that helps define my problem a little more
Onochie
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help me out.
I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits and
data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit and data
is
checked before db submission, I am losing the data displayed by the
first
submit button
hmm i don't get it by reading at the first. i think in the 2nd page u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data, but this
doesnt work. this is because in my code i have an if isset statement to
check if submit has been pressed, then the variables are retrieved from
the
db and used. when clicking the second submit button and checks are done,
the
top portion of the page of information retrieved by the first submit are
nothing but blanks because the first submit is no longer set (using
isset)
my next idea was to save the variable selected in the first drop down
(step
1) and requery with this selection at the second submit button to
retrieve
and redisplay the data. this did not work because this variable was set
back
to because the drop down resets itself on each submit. disabling the
drop
down if the first submit is set and then setting the variable did not
work
either because when the second submit is hit, the first submit is no
longer
set, so the drop down is displayed, the variable is set to and i get
nothing in my variable. this also was the case if i made it a global
variable.
i wish to see the offline from your pages.. if able send to me
so basically i have resorted to just googling things and trying out
different arrangements of code to no avail. I need a way to hold the
data on
the top half of the screen while the bottom half is being manipulated.
any,
and i mean any help at this point would be GREATLY appreciated! Thank
you.
Onochie
--
akan ada dimana mulut terkunci dan suara tak ada lagi..
saat itu gunakanlah HP untuk melakukan SMS!!
- ini aliran bedul.. bukan aliran aneh.
tertawa sebelum tertawa didepan RSJ..
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
solved
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello Gunawan
On your first comment, im not sure what you were referring to when you said
it is a js problem
on your second comment, i am talking about the user entered data. data
submitted by the first submit button will always be predefined by means of
the user using a selection box, the data im concerned with checking is the
input boxes he will be seeing. and this wont necessarily be username/pass.
and im trying to do this in php if at all possible.
on your next comment, (6) i am talking about after i vertified the validity
of the user typed/entered data, i will enter that into my db
next comment - i knew this would be hard to describe, but i'll try again.
this is to be done all on one page. and ive tried using something like:
value=\$_POST['name']\
but i am having problems keeping the initial drop down at the selected
choice, so whenever either submit button is pressed and the post goes to php
self, the drop down is refreshed back to its starting value, which is NULL
hahaha..
that's simple.. drop down menu always return to the null because u
don't give selected on one of option.
i have a function (build in) that's answer your problem.. if you on
hurry.. i'm sory, i cant give it now.. my skrip in other comp..
select
option value=base selectedsale options/option
option value=base2 Buy options/option
/select
try this on offline
i believe.. the first you enter the page.. the sale option will be selected
now add 2 option then try one of the option you add have selected..
what happen next you will see the default will be change, remember
erase selected
forgive me.. i think it matter of js but it matter of html.
fyi.. try using ajax .. i believe it make your life for comfort..
(since it don't need to refresh the page) and make your skill up.
I hope that helps define my problem a little more
Onochie
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help me out.
I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits and
data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit and data
is
checked before db submission, I am losing the data displayed by the
first
submit button
hmm i don't get it by reading at the first. i think in the 2nd page u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data, but this
doesnt work. this is because in my code i have an if isset statement to
check if submit has been pressed, then the variables are retrieved from
the
db and used. when clicking the second submit button and checks are done,
the
top portion of the page of information retrieved by the first submit are
nothing but blanks because the first submit is no longer set (using
isset)
my next idea was to save the variable selected in the first drop down
(step
1) and requery with this selection at the second submit button to
retrieve
and redisplay the data. this did not work because this variable was set
back
to because the drop down resets itself on each submit. disabling the
drop
down if the first submit is set and then setting the variable did not
work
either because when the second submit is hit, the first submit is no
longer
set, so the drop down is displayed, the variable is set to and i get
nothing in my variable. this also was the case if i made it a global
variable.
i wish to see the offline from your pages.. if able send to me
so basically i have resorted to just googling things and trying out
different arrangements of code to no avail. I need a way to hold the
data on
the top half of the screen while the bottom half is being manipulated.
any,
and i mean any help at this point would be GREATLY appreciated! Thank
you.
Onochie
--
akan ada dimana mulut terkunci dan suara tak ada lagi..
saat itu gunakanlah HP untuk melakukan SMS!!
- ini aliran bedul.. bukan aliran aneh.
tertawa sebelum tertawa didepan RSJ..
--
akan ada dimana mulut terkunci dan suara tak ada lagi..
saat itu gunakanlah HP untuk melakukan SMS!!
- ini aliran bedul.. bukan aliran aneh.
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
Almost what I wanted to do.
I am looking for a way to set the default option value to the selection the
user just chose. for example, something like
select name=cars value=$_POST['cars']
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
solved
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello Gunawan
On your first comment, im not sure what you were referring to when you
said
it is a js problem
on your second comment, i am talking about the user entered data. data
submitted by the first submit button will always be predefined by means
of
the user using a selection box, the data im concerned with checking is
the
input boxes he will be seeing. and this wont necessarily be
username/pass.
and im trying to do this in php if at all possible.
on your next comment, (6) i am talking about after i vertified the
validity
of the user typed/entered data, i will enter that into my db
next comment - i knew this would be hard to describe, but i'll try
again.
this is to be done all on one page. and ive tried using something like:
value=\$_POST['name']\
but i am having problems keeping the initial drop down at the selected
choice, so whenever either submit button is pressed and the post goes to
php
self, the drop down is refreshed back to its starting value, which is
NULL
hahaha..
that's simple.. drop down menu always return to the null because u
don't give selected on one of option.
i have a function (build in) that's answer your problem.. if you on
hurry.. i'm sory, i cant give it now.. my skrip in other comp..
select
option value=base selectedsale options/option
option value=base2 Buy options/option
/select
try this on offline
i believe.. the first you enter the page.. the sale option will be
selected
now add 2 option then try one of the option you add have selected..
what happen next you will see the default will be change, remember
erase selected
forgive me.. i think it matter of js but it matter of html.
fyi.. try using ajax .. i believe it make your life for comfort..
(since it don't need to refresh the page) and make your skill up.
I hope that helps define my problem a little more
Onochie
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help me
out.
I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits and
data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is
entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit and
data
is
checked before db submission, I am losing the data displayed by the
first
submit button
hmm i don't get it by reading at the first. i think in the 2nd page u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data, but
this
doesnt work. this is because in my code i have an if isset statement
to
check if submit has been pressed, then the variables are retrieved
from
the
db and used. when clicking the second submit button and checks are
done,
the
top portion of the page of information retrieved by the first submit
are
nothing but blanks because the first submit is no longer set (using
isset)
my next idea was to save the variable selected in the first drop
down
(step
1) and requery with this selection at the second submit button to
retrieve
and redisplay the data. this did not work because this variable was
set
back
to because the drop down resets itself on each submit. disabling
the
drop
down if the first submit is set and then setting the variable did
not
work
either because when the second submit is hit, the first submit is no
longer
set, so the drop down is displayed, the variable is set to and i
get
nothing in my variable. this also was the case if i made it a global
variable.
i wish to see the offline from your pages.. if able send to me
so basically i have resorted to just googling things and trying out
different arrangements of code to no avail. I need a way to hold the
data on
the top half of the screen while the bottom half is being
manipulated.
any,
and i mean any help at this point would be GREATLY appreciated!
Thank
you.
Onochie
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
Try something like this:
Your query here to get your dropdown values
**
*// This first option value will set your default dropdown to display as
blank*
echo OPTION value=\\--SELECT--/OPTION\n;
*// This will create a loop to return each option of your dropdown*
foreach ($query_result as $q)
{
*// This if statement says that IF the value in your database matches your
$_POST(selected) value then mark it as SELECTED*
if ($q['my_column_value'] == $_POST['my_selected_value'])
echo OPTION value=\{$q['my_column_value']}\
SELECTED{$q['my_column_value']}/OPTION\n;
*// This else statement will return your default(unselected) dropdown list
if it can't match the selected value with a value in your database*
else
echo OPTION
value=\{$q['my_column_value']}\{$q['my_column_value']}/OPTION\n;
}
Hope that helps.
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Almost what I wanted to do.
I am looking for a way to set the default option value to the selection
the
user just chose. for example, something like
select name=cars value=$_POST['cars']
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
solved
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello Gunawan
On your first comment, im not sure what you were referring to when you
said
it is a js problem
on your second comment, i am talking about the user entered data. data
submitted by the first submit button will always be predefined by
means
of
the user using a selection box, the data im concerned with checking is
the
input boxes he will be seeing. and this wont necessarily be
username/pass.
and im trying to do this in php if at all possible.
on your next comment, (6) i am talking about after i vertified the
validity
of the user typed/entered data, i will enter that into my db
next comment - i knew this would be hard to describe, but i'll try
again.
this is to be done all on one page. and ive tried using something
like:
value=\$_POST['name']\
but i am having problems keeping the initial drop down at the selected
choice, so whenever either submit button is pressed and the post goes
to
php
self, the drop down is refreshed back to its starting value, which is
NULL
hahaha..
that's simple.. drop down menu always return to the null because u
don't give selected on one of option.
i have a function (build in) that's answer your problem.. if you on
hurry.. i'm sory, i cant give it now.. my skrip in other comp..
select
option value=base selectedsale options/option
option value=base2 Buy options/option
/select
try this on offline
i believe.. the first you enter the page.. the sale option will be
selected
now add 2 option then try one of the option you add have selected..
what happen next you will see the default will be change, remember
erase selected
forgive me.. i think it matter of js but it matter of html.
fyi.. try using ajax .. i believe it make your life for comfort..
(since it don't need to refresh the page) and make your skill up.
I hope that helps define my problem a little more
Onochie
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help me
out.
I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits
and
data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is
entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit and
data
is
checked before db submission, I am losing the data displayed by
the
first
submit button
hmm i don't get it by reading at the first. i think in the 2nd page
u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data,
but
this
doesnt work. this is because in my code i have an if isset
statement
to
check if submit has been pressed, then the variables are
retrieved
from
the
db and used. when clicking the second submit button and checks are
done,
the
top portion of the page of information retrieved by the first
submit
are
nothing but blanks because the first submit is no longer set
(using
isset)
my next idea was to save
Re: [PHP-DB] Re: do not display dublicated entries
On 3/14/07, Niel Archer [EMAIL PROTECTED] wrote: Hi I have tried everything except of the one that forms the date in one field and use the DISTINCT for ip and date. It works for me. I use it for essentially the same job, listing most recent visit to a location (not IP, but physical venue). Have you tried adapting it to your needs, such as: $query = SELECT DISTINCT ip, day, month, year FROM tracker WHERE page = 'index' ORDER BY `tracker`.`year` DESC , `tracker`.`month` DESC , `tracker`.`day` DESC LIMIT 0, 20; Niel -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php try grouping by IP and selecting the max date value SELECT IP,max(date) AS latestvisit FROM table GROUP BY IP -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Using multiple submits on a page and retaining $POST data
solved.
Thanks Dan, this was pretty much the solution I came up with. Thank you
Gunawan for the idea that got the ball rolling as well
Onochie
On 3/16/07, Dan Shirah [EMAIL PROTECTED] wrote:
Try something like this:
Your query here to get your dropdown values
**
*// This first option value will set your default dropdown to display as
blank*
echo OPTION value=\\--SELECT--/OPTION\n;
*// This will create a loop to return each option of your dropdown*
foreach ($query_result as $q)
{
*// This if statement says that IF the value in your database matches your
$_POST(selected) value then mark it as SELECTED*
if ($q['my_column_value'] == $_POST['my_selected_value'])
echo OPTION value=\{$q['my_column_value']}\
SELECTED{$q['my_column_value']}/OPTION\n;
*// This else statement will return your default(unselected) dropdown list
if it can't match the selected value with a value in your database*
else
echo OPTION
value=\{$q['my_column_value']}\{$q['my_column_value']}/OPTION\n;
}
Hope that helps.
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Almost what I wanted to do.
I am looking for a way to set the default option value to the selection
the
user just chose. for example, something like
select name=cars value=$_POST['cars']
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
solved
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello Gunawan
On your first comment, im not sure what you were referring to when
you
said
it is a js problem
on your second comment, i am talking about the user entered data.
data
submitted by the first submit button will always be predefined by
means
of
the user using a selection box, the data im concerned with checking
is
the
input boxes he will be seeing. and this wont necessarily be
username/pass.
and im trying to do this in php if at all possible.
on your next comment, (6) i am talking about after i vertified the
validity
of the user typed/entered data, i will enter that into my db
next comment - i knew this would be hard to describe, but i'll try
again.
this is to be done all on one page. and ive tried using something
like:
value=\$_POST['name']\
but i am having problems keeping the initial drop down at the
selected
choice, so whenever either submit button is pressed and the post
goes to
php
self, the drop down is refreshed back to its starting value, which
is
NULL
hahaha..
that's simple.. drop down menu always return to the null because u
don't give selected on one of option.
i have a function (build in) that's answer your problem.. if you on
hurry.. i'm sory, i cant give it now.. my skrip in other comp..
select
option value=base selectedsale options/option
option value=base2 Buy options/option
/select
try this on offline
i believe.. the first you enter the page.. the sale option will be
selected
now add 2 option then try one of the option you add have selected..
what happen next you will see the default will be change, remember
erase selected
forgive me.. i think it matter of js but it matter of html.
fyi.. try using ajax .. i believe it make your life for comfort..
(since it don't need to refresh the page) and make your skill up.
I hope that helps define my problem a little more
Onochie
On 3/16/07, Gunawan Wibisono [EMAIL PROTECTED] wrote:
that's not php problem that's was javascript problem
On 3/16/07, Onochie Anyanetu [EMAIL PROTECTED] wrote:
Hello, and I would like to thank everyone that attempts to help
me
out.
I
have looked long and hard online before I resorted to this.
My problem is that I have a page that relies on multiple submits
and
data
from what was submitted. example:
// 1. some sort of drop down
// 2. submit button
// 3. display data retrieved from database
// 4. user enters more information
// 5. second submit button, run checks to vertify all data is
entered
u mean when clicked.. the info like username are check whenever is
valid or not?? perhaps u should try ajax
// 6. enter data into database
i don't get it what u mean in here??
The problem is in step 5, when the user hits the second submit
and
data
is
checked before db submission, I am losing the data displayed by
the
first
submit button
hmm i don't get it by reading at the first. i think in the 2nd
page u
should try to make input type=hidden name=user value=landa
this input contain what previous data enter...
foreach($_POST as $nm=$val){
$txt.=input type=hidden name=$nm value=$val;
}
echo $txt;
This is what I have tried
ive tried using POST and SESSION variables to display the data,
but
this
doesnt work. this is because in my code i have an if isset
statement
to
check if submit has been
[PHP-DB] dst
Hi quick question (off topic a bit) MySQL seems to be not running on DST while the machine it is on is. any ideas ? if I use a mysql_query(select now(), connection); it looks like it is an hour earlier then the system time. Ron -- = It's is not, it isn't ain't, and it's it's, not its, if you mean it is. If you don't, it's its. Then too, it's hers. It isn't her's. It isn't our's either. It's ours, and likewise yours and theirs. -- Oxford Uni Press = Ron Croonenberg | | Phone: 1 765 658 4761 Lab Instructor | Fax: 1 765 658 4732 Technology Coordinator| | Department of Computer Science| e-mail: [EMAIL PROTECTED] DePauw University | 275 Julian Science Math Center | 602 South College Ave.| Greencastle, IN 46135| = http://www.csc.depauw.edu/RonCroonenberg.html = -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
