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 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 ' Region onChange="document.thisForm.submit()">';
> echo 'Choose a Region';
>
> for ($s=0; $s<$numregs; $s++)
> {
> $reglist = pg_fetch_array($regs,$s);
> echo' print ($region == $reglist["region_name"]) ? " SELECTED":"
> ";
> echo '>'.$reglist["region_name"].'';
> }
>
> echo'';
>
>> -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:
>>> $result = mysql_query("SELECT companyName FROM companyTable ORDER BY
>>> companyName");
>>> echo "";
>>> while (list($c_name) = mysql_fetch_array($result)) {
>>> echo "$c_name";
>>> }
>>>
>>> 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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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 $co_name in the HTML. > > thanks, > boo > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] calling a column into a popup
becky the code below is spot on (nearly)
stick it on a form that calls itself change the ';
echo 'Choose a Region';
for ($s=0; $s<$numregs; $s++)
{
$reglist = pg_fetch_array($regs,$s);
echo''.$reglist["region_name"].'';
}
echo'';
> -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:
> > $result = mysql_query("SELECT companyName FROM companyTable ORDER BY
> > companyName");
> > echo "";
> > while (list($c_name) = mysql_fetch_array($result)) {
> > echo "$c_name";
> > }
> >
> > 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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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:
> $result = mysql_query("SELECT companyName FROM companyTable ORDER BY
> companyName");
> echo "";
> while (list($c_name) = mysql_fetch_array($result)) {
> echo "$c_name";
> }
>
> 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
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:
$result = mysql_query("SELECT companyName FROM companyTable ORDER BY
companyName");
echo "";
while (list($c_name) = mysql_fetch_array($result)) {
echo "$c_name";
}
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
>
>
> --
> 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]
[PHP-DB] calling a column into a popup
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 -- 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]
