RE: [PHP-DB] Forms question..
When I have an input form I must click on the first input field. Is there anyway to have it active when the form loads? Javascript, not PHP. Something like document.form.element.setfocus(), I think. #2 When I tab down to the next input field, is there anyway to have it highlight what is in the field (ie the default data) so that it will overwrite it? Client side...not PHP. #3 How can I get the form to submit when I hit the return button? Right now I have to click on the button with my mouse. Client side, depends on the browser, not PHP #4 How can I get a column of numbers to left justify? HTML, not PHP. Align=left ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Forms question..
1. document.formname.elementname.focus(); Example: if my form has a name of testform and the first element is fname... then in the body tag i would do this... body onload=document.testform.fname.focus(); 2. Yeah, you need to use the select() method... so document.testform.element.select(); Example: input name=example value=N/A onfocus=document.testform.example.select(); That should work...or it might have to be in a function... 3. It should do that for you with the submit button...what browser are you using? 4. use the align attributealign=left On Sun, 2002-10-20 at 12:27, James Hatridge wrote: Hi all... When I have an input form I must click on the first input field. Is there anyway to have it active when the form loads? #2 When I tab down to the next input field, is there anyway to have it highlight what is in the field (ie the default data) so that it will overwrite it? #3 How can I get the form to submit when I hit the return button? Right now I have to click on the button with my mouse. #4 How can I get a column of numbers to left justify? Thanks JIM -- Jim Hatridge Linux User #88484 -- BayerWulf Linux System # 129656 The Recycled Beowulf Project Looking for throw-away or obsolete computers and parts to recycle into a Linux super computer -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- .: B i g D o g :. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Forms Question
Hi there,
Take a look at the following snippet below. It is similar to what you're
looking for without having to bugger about duplicating loads of lines for
each store. Just add a new one to the array if required.
Best Regards,
- Paul -
select name=MonthSelected size=1
?php
$MonthArr = array (
January,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December,
);
reset($MonthArr);
foreach ($MonthArr as $MonthName) {
echo option VALUE=\$MonthName\;
if ($MonthName == $MonthSelected) echo selected;
echo ;
echo $MonthName;
}
?
/select
- Original Message -
From: Steve Cayford [EMAIL PROTECTED]
To: Jeff Grossman [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Wednesday, September 05, 2001 11:36 PM
Subject: Re: [PHP-DB] Forms Question
On Wednesday, September 5, 2001, at 04:50 PM, Jeff Grossman wrote:
Hello,
Here is the code I have:
while ($row=mysql_fetch_array($result)) {
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:
Select NAME=\$store\
option VALUE=\Signal Hill\Signal Hill
option VALUE=\Reseda\Reseda
option VALUE=\Orange\Orange
option VALUE=\West Covina\West Covina
option VALUE=\Riverside\Riverside
option VALUE=\Norwalk\Norwalk
option VALUE=\Fountain Valley\Fountain Valley
option VALUE=\Pasadena\Pasadena
option VALUE=\Redondo Beach\Redondo Beach
option VALUE=\San Bernardino\San Bernardino
option VALUE=\Kearny Mesa\Kearny Mesa
option VALUE=\San Marcos\San Marcos
option VALUE=\Chino\Chino
option VALUE=\Coporate Office\Corporate Office
/select/P;
echo PINPUT TYPE=text SIZE=35 NAME=\Jobdesc\
VALUE=\$jobdesc\/P;
echo pINPUT TYPE=submit VALUE=\submit\ LABEL=\Save
Changes\/P;
}
Is want I am trying to do possible? I want the value which is stored in
$store to automatically fill in on the drop down list. But, for some
reason it is defaulting to the first option, and not using the value
that is in the database.
Can I use a drop down menu, or should I just go to radio buttons?
Thanks,
Jeff
If I understand your question...
In order to have your value preset in the drop down list you need
indicate that with
option value=\blahblah\ selectedblahblah
What I've been using for this is a hash like this:
while ($row=mysql_fetch_array($result)) {
$selected = array();
$selected[$row[store]] = selected;
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:;
echo Select NAME=NameOfVariableToBePosted
echo option VALUE=\Signal Hill\ . $selected[Signal
Hill] . Signal Hill;
echo option VALUE=\Reseda\ . $selected[Reseda] .
Reseda;
...etc, etc., etc.
something along those lines, anyway. So, if $row[store] == Signal
Hill, then $selected[Signal Hill] will be set to selected, while
$selected[Reseda] and all the others will be blank.
This is a very keen thing about php.
-Steve
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Re: [PHP-DB] Forms Question
Hello,
Here is the code I have:
sure, your query is something like:
select ..., store, jobdesc .. FROM ..
why using:
$store=$row[store];
$jobdesc=$row[jobdesc];
echodhdhdh.$row[store].blabla would also work
i should use:
while (LIST($store,$jobdesc)=mysql_fetch_row($result)) {
.
- Original Message -
From: Jeff Grossman [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, September 05, 2001 11:50 PM
Subject: [PHP-DB] Forms Question
Hello,
Here is the code I have:
while ($row=mysql_fetch_array($result)) {
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:
Select NAME=\$store\
option VALUE=\Signal Hill\Signal Hill
option VALUE=\Reseda\Reseda
option VALUE=\Orange\Orange
option VALUE=\West Covina\West Covina
option VALUE=\Riverside\Riverside
option VALUE=\Norwalk\Norwalk
option VALUE=\Fountain Valley\Fountain Valley
option VALUE=\Pasadena\Pasadena
option VALUE=\Redondo Beach\Redondo Beach
option VALUE=\San Bernardino\San Bernardino
option VALUE=\Kearny Mesa\Kearny Mesa
option VALUE=\San Marcos\San Marcos
option VALUE=\Chino\Chino
option VALUE=\Coporate Office\Corporate Office
/select/P;
echo PINPUT TYPE=text SIZE=35 NAME=\Jobdesc\
VALUE=\$jobdesc\/P;
echo pINPUT TYPE=submit VALUE=\submit\ LABEL=\Save
Changes\/P;
}
Is want I am trying to do possible? I want the value which is stored in
$store to automatically fill in on the drop down list. But, for some
reason it is defaulting to the first option, and not using the value
that is in the database.
Can I use a drop down menu, or should I just go to radio buttons?
Thanks,
Jeff
--
Jeff Grossman ([EMAIL PROTECTED])
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
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To unsubscribe, e-mail: [EMAIL PROTECTED]
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Re: [PHP-DB] Forms Question
On Wednesday, September 5, 2001, at 04:50 PM, Jeff Grossman wrote:
Hello,
Here is the code I have:
while ($row=mysql_fetch_array($result)) {
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:
Select NAME=\$store\
option VALUE=\Signal Hill\Signal Hill
option VALUE=\Reseda\Reseda
option VALUE=\Orange\Orange
option VALUE=\West Covina\West Covina
option VALUE=\Riverside\Riverside
option VALUE=\Norwalk\Norwalk
option VALUE=\Fountain Valley\Fountain Valley
option VALUE=\Pasadena\Pasadena
option VALUE=\Redondo Beach\Redondo Beach
option VALUE=\San Bernardino\San Bernardino
option VALUE=\Kearny Mesa\Kearny Mesa
option VALUE=\San Marcos\San Marcos
option VALUE=\Chino\Chino
option VALUE=\Coporate Office\Corporate Office
/select/P;
echo PINPUT TYPE=text SIZE=35 NAME=\Jobdesc\
VALUE=\$jobdesc\/P;
echo pINPUT TYPE=submit VALUE=\submit\ LABEL=\Save
Changes\/P;
}
Is want I am trying to do possible? I want the value which is stored in
$store to automatically fill in on the drop down list. But, for some
reason it is defaulting to the first option, and not using the value
that is in the database.
Can I use a drop down menu, or should I just go to radio buttons?
Thanks,
Jeff
If I understand your question...
In order to have your value preset in the drop down list you need
indicate that with
option value=\blahblah\ selectedblahblah
What I've been using for this is a hash like this:
while ($row=mysql_fetch_array($result)) {
$selected = array();
$selected[$row[store]] = selected;
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:;
echo Select NAME=NameOfVariableToBePosted
echo option VALUE=\Signal Hill\ . $selected[Signal
Hill] . Signal Hill;
echo option VALUE=\Reseda\ . $selected[Reseda] .
Reseda;
...etc, etc., etc.
something along those lines, anyway. So, if $row[store] == Signal
Hill, then $selected[Signal Hill] will be set to selected, while
$selected[Reseda] and all the others will be blank.
This is a very keen thing about php.
-Steve
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Forms Question
The solution is easier than you think. You'll need to insert an if
statement in your option html tag that says
echo option VALUE=\Signal Hill\Signal Hill\n;
should become:
echo option VALUE=\Signal Hill\;
if ($store == 'Signal Hill') { print selected; }
echo Signal Hill/option\n;
Jeff Grossman wrote:
Hello,
Here is the code I have:
while ($row=mysql_fetch_array($result)) {
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:
Select NAME=\$store\
option VALUE=\Signal Hill\Signal Hill
option VALUE=\Reseda\Reseda
option VALUE=\Orange\Orange
option VALUE=\West Covina\West Covina
option VALUE=\Riverside\Riverside
option VALUE=\Norwalk\Norwalk
option VALUE=\Fountain Valley\Fountain Valley
option VALUE=\Pasadena\Pasadena
option VALUE=\Redondo Beach\Redondo Beach
option VALUE=\San Bernardino\San Bernardino
option VALUE=\Kearny Mesa\Kearny Mesa
option VALUE=\San Marcos\San Marcos
option VALUE=\Chino\Chino
option VALUE=\Coporate Office\Corporate Office
/select/P;
echo PINPUT TYPE=text SIZE=35 NAME=\Jobdesc\
VALUE=\$jobdesc\/P;
echo pINPUT TYPE=submit VALUE=\submit\ LABEL=\Save
Changes\/P;
}
Is want I am trying to do possible? I want the value which is stored in
$store to automatically fill in on the drop down list. But, for some
reason it is defaulting to the first option, and not using the value
that is in the database.
Can I use a drop down menu, or should I just go to radio buttons?
Thanks,
Jeff
--
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]
