Re: [PHP-DB] calling a column into a popup
That's almost it. I want to populate the menu automatically from the
database without having to manually modify the HTML. I believe the below
will pull the data but it doesn't modify the menu in HTML. I tried a few
things like using the variable name as the same as the co_name but the menu
always comes up blank unless I put an option name in the list. it doesn't
pass the variable for $co_name in the HTML.
thanks,
boo
From: Johannes Janson [EMAIL PROTECTED]
Date: Sun, 22 Apr 2001 21:49:36 +0200
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] calling a column into a popup
Hi,
let me get this right. What I understood is the following:
1. Your admin page where you enter the information of a company
2. A Page with a drop-down list where all the company names are listed.
(These names should then be linked to an information page with more
info?)
3. A page where new companies can be added.
well the drop-down list you can do with a simple while and a select(html)
like
this:
$result = mysql_query(SELECT companyName FROM companyTable ORDER BY
companyName);
echo select name=company;
while (list($c_name) = mysql_fetch_array($result)) {
echo option name=$c_name]$c_name/option;
}
If this is what you want (what I'm not sure about) I'm glad I understood
you. If not supply more detailed information.
Cheers
Johannes
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RE: [PHP-DB] calling a column into a popup
becky the code below is spot on (nearly)
stick it on a form that calls itself change the option name= to option
value= and when you refresh the form $company will contain the comany name.
you can also use selected in the folowing manner
$regs = pg_Exec($conn, select region_id, region_name from region
where .$depdisplay. is not null order by .$depdisplay. );
$numregs = pg_numrows($regs);
echo 'nbsp;Regionnbsp;nbsp;select name=reg
onChange=document.thisForm.submit()';
echo 'option value=0Choose a Region/option';
for ($s=0; $s$numregs; $s++)
{
$reglist = pg_fetch_array($regs,$s);
echo'option value='.$reglist[region_name].'';
print ($region == $reglist[region_name]) ? SELECTED:
;
echo ''.$reglist[region_name].'/option';
}
echo'/select';
-Original Message-
From: Beckie Pack [mailto:[EMAIL PROTECTED]]
Sent: 23 April 2001 15:29
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] calling a column into a popup
That's almost it. I want to populate the menu automatically from the
database without having to manually modify the HTML. I
believe the below
will pull the data but it doesn't modify the menu in HTML. I
tried a few
things like using the variable name as the same as the
co_name but the menu
always comes up blank unless I put an option name in the
list. it doesn't
pass the variable for $co_name in the HTML.
thanks,
boo
From: Johannes Janson [EMAIL PROTECTED]
Date: Sun, 22 Apr 2001 21:49:36 +0200
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] calling a column into a popup
Hi,
let me get this right. What I understood is the following:
1. Your admin page where you enter the information of a company
2. A Page with a drop-down list where all the company names
are listed.
(These names should then be linked to an information page with more
info?)
3. A page where new companies can be added.
well the drop-down list you can do with a simple while and
a select(html)
like
this:
$result = mysql_query(SELECT companyName FROM companyTable ORDER BY
companyName);
echo select name=company;
while (list($c_name) = mysql_fetch_array($result)) {
echo option name=$c_name]$c_name/option;
}
If this is what you want (what I'm not sure about) I'm glad
I understood
you. If not supply more detailed information.
Cheers
Johannes
--
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]
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To unsubscribe, e-mail: [EMAIL PROTECTED]
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Re: [PHP-DB] calling a column into a popup
So a certain company is previously selected? And then you want to populate the form with the info of the company? to do this link the menu with a onChange action to a display page. But I'm not sure if I get the layout. Could you describe it again? Cheers Johannes Beckie Pack [EMAIL PROTECTED] schrieb im Newsbeitrag [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... That's almost it. I want to populate the menu automatically from the database without having to manually modify the HTML. I believe the below will pull the data but it doesn't modify the menu in HTML. I tried a few things like using the variable name as the same as the co_name but the menu always comes up blank unless I put an option name in the list. it doesn't pass the variable for $co_name in the HTML. thanks, boo -- 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] calling a column into a popup
yup, now i realize it is a form calling itself with a php doc. i feel like a
dufus(sp?). i didn't recognize the echo as opposed to print so I didn't
see the html coming. learning php, mysql, and database design all together.
it's kinda fun, given time for a project you know.
thanks to both of you.
beckie
From: Steve Brett [EMAIL PROTECTED]
Date: Mon, 23 Apr 2001 16:02:02 +0100
To: Beckie Pack [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [PHP-DB] calling a column into a popup
becky the code below is spot on (nearly)
stick it on a form that calls itself change the option name= to option
value= and when you refresh the form $company will contain the comany name.
you can also use selected in the folowing manner
$regs = pg_Exec($conn, select region_id, region_name from region
where .$depdisplay. is not null order by .$depdisplay. );
$numregs = pg_numrows($regs);
echo 'nbsp;Regionnbsp;nbsp;select name=reg
onChange=document.thisForm.submit()';
echo 'option value=0Choose a Region/option';
for ($s=0; $s$numregs; $s++)
{
$reglist = pg_fetch_array($regs,$s);
echo'option value='.$reglist[region_name].'';
print ($region == $reglist[region_name]) ? SELECTED:
;
echo ''.$reglist[region_name].'/option';
}
echo'/select';
-Original Message-
From: Beckie Pack [mailto:[EMAIL PROTECTED]]
Sent: 23 April 2001 15:29
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] calling a column into a popup
That's almost it. I want to populate the menu automatically from the
database without having to manually modify the HTML. I
believe the below
will pull the data but it doesn't modify the menu in HTML. I
tried a few
things like using the variable name as the same as the
co_name but the menu
always comes up blank unless I put an option name in the
list. it doesn't
pass the variable for $co_name in the HTML.
thanks,
boo
From: Johannes Janson [EMAIL PROTECTED]
Date: Sun, 22 Apr 2001 21:49:36 +0200
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] calling a column into a popup
Hi,
let me get this right. What I understood is the following:
1. Your admin page where you enter the information of a company
2. A Page with a drop-down list where all the company names
are listed.
(These names should then be linked to an information page with more
info?)
3. A page where new companies can be added.
well the drop-down list you can do with a simple while and
a select(html)
like
this:
$result = mysql_query(SELECT companyName FROM companyTable ORDER BY
companyName);
echo select name=company;
while (list($c_name) = mysql_fetch_array($result)) {
echo option name=$c_name]$c_name/option;
}
If this is what you want (what I'm not sure about) I'm glad
I understood
you. If not supply more detailed information.
Cheers
Johannes
--
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] calling a column into a popup
Hi,
let me get this right. What I understood is the following:
1. Your admin page where you enter the information of a company
2. A Page with a drop-down list where all the company names are listed.
(These names should then be linked to an information page with more
info?)
3. A page where new companies can be added.
well the drop-down list you can do with a simple while and a select(html)
like
this:
$result = mysql_query("SELECT companyName FROM companyTable ORDER BY
companyName");
echo "select name=company";
while (list($c_name) = mysql_fetch_array($result)) {
echo "option name=$c_name]$c_name/option";
}
If this is what you want (what I'm not sure about) I'm glad I understood
you. If not supply more detailed information.
Cheers
Johannes
"Beckie Pack" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
I'm creating a database where I'd like to relate a company name field to
all
the field in the database (mysql). what i'm trying to do is have a field
pop up with all the current company names in a pop up list on a browser.
so,
for example, using a form i can enter all the data for a company including
name, address, etc. what i'm trying to do is have a popup for the company
so
if the information is already entered it is there for selection or the
person can enter a new company if it's not there.
any ideas?
thanks,
beckie
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For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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