Re: [PHP] php forms - select menu selected behavior

2009-04-30 Thread Marcus Gnaß
Troy Oltmanns wrote:
 I have the code below being used to rifle through a list of available
 categories and create select options for them. The code is being used to
 query the database and compare the product category to the current
 iteration, if there's a match, then add selected code so the category is
 prechosen. More code (not included) does the saving and all that, I've check
 phpmyadmin. But when the page submits, the old category appears in the drop
 down as selected. If I leave the page and come back it's fine, it's just
 right after it is saved. The form script is being used on itself, in that
 there is only one file for the form, the submission, etc. All of the other
 input elements will load the data after being saved, is it something
 specific to dropdowns, or it is the way the code is being instatiated?
 
 All help is much appreciated. Please let me know if anymore info is needed.
 

//MAKE CATEGORIES DROPDOWN

$catlist1 = ;

// read product
$catmatch = SELECT prod_cat0 FROM product WHERE dbi='$dbi';;
$catresult = mysql_query($catmatch);
$catquery = mysql_fetch_array($catresult);

// read categories
$sql = SELECT category FROM categories ORDER BY category;;
$result = mysql_query($sql);
while ($col2 = mysql_fetch_array($result)) {

$id = $col2[category];

if ($id == $catquery['prod_cat0']){

$catlist1 .= option value=\$id\ 
selected=\selected\$id/option;

}   else {

$catlist1 .= option value=\$id\$id/option;

}

}

 
 to instantiate ?=$catlist1?
 

The only data you need from table product is the column prod_cat0, from
table categories it's category, so you should read only the needed data
instead of using * for better performance.

Take the SQL and verify if it returns what you want it to return then.

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Re: [PHP] php forms - select menu selected behavior

2009-04-30 Thread Ashley Sheridan
On Thu, 2009-04-30 at 11:06 +0200, Marcus Gnaß wrote:
 Troy Oltmanns wrote:
  I have the code below being used to rifle through a list of available
  categories and create select options for them. The code is being used to
  query the database and compare the product category to the current
  iteration, if there's a match, then add selected code so the category is
  prechosen. More code (not included) does the saving and all that, I've check
  phpmyadmin. But when the page submits, the old category appears in the drop
  down as selected. If I leave the page and come back it's fine, it's just
  right after it is saved. The form script is being used on itself, in that
  there is only one file for the form, the submission, etc. All of the other
  input elements will load the data after being saved, is it something
  specific to dropdowns, or it is the way the code is being instatiated?
  
  All help is much appreciated. Please let me know if anymore info is needed.
  
 
 //MAKE CATEGORIES DROPDOWN
 
 $catlist1 = ;
 
 // read product
 $catmatch = SELECT prod_cat0 FROM product WHERE dbi='$dbi';;
 $catresult = mysql_query($catmatch);
 $catquery = mysql_fetch_array($catresult);
 
 // read categories
 $sql = SELECT category FROM categories ORDER BY category;;
 $result = mysql_query($sql);
 while ($col2 = mysql_fetch_array($result)) {
 
   $id = $col2[category];
 
   if ($id == $catquery['prod_cat0']){
 
   $catlist1 .= option value=\$id\ 
 selected=\selected\$id/option;
 
   }   else {
 
   $catlist1 .= option value=\$id\$id/option;
 
   }
 
 }
 
  
  to instantiate ?=$catlist1?
  
 
 The only data you need from table product is the column prod_cat0, from
 table categories it's category, so you should read only the needed data
 instead of using * for better performance.
 
 Take the SQL and verify if it returns what you want it to return then.
 
I tend to do my loops like this:

while ($col2 = mysql_fetch_array($result))
{
$id = $col2[category];
$selected =($id == $catquery['prod_cat0'])?'selected=selected':'';

$catlist1 .= option value=\$id\ $selected$id/option;
}

Just looks a little neater. 'Course, you could remove the $id line and
chuck the value straight into the short if statement there.


Ash
www.ashleysheridan.co.uk


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[PHP] php forms - select menu selected behavior

2009-04-28 Thread Troy Oltmanns
I have the code below being used to rifle through a list of available
categories and create select options for them. The code is being used to
query the database and compare the product category to the current
iteration, if there's a match, then add selected code so the category is
prechosen. More code (not included) does the saving and all that, I've check
phpmyadmin. But when the page submits, the old category appears in the drop
down as selected. If I leave the page and come back it's fine, it's just
right after it is saved. The form script is being used on itself, in that
there is only one file for the form, the submission, etc. All of the other
input elements will load the data after being saved, is it something
specific to dropdowns, or it is the way the code is being instatiated?

All help is much appreciated. Please let me know if anymore info is needed.

//MAKE CATEGORIES DROPDOWN
$sql=SELECT * FROM categories ORDER BY category;

$catmatch=SELECT * FROM product WHERE dbi='$dbi';
$catresult=mysql_query($catmatch);
$catquery=mysql_fetch_array($catresult);
//for($a=0;$amysql_fetch_array($catresult)){
//echo $catquery['prod_cat0'];
//}
$result=mysql_query($sql);

$catlist1=;
while ($col2=mysql_fetch_array($result)) {
$id=$col2[category];
if($id==$catquery['prod_cat0']){
$catlist1.=option value=\$id\
selected=\selected\.$id./option;
  }
else {
$catlist1.=option value=\$id\.$id./option;
  }

}

to instantiate ?=$catlist1?


Re: [PHP] PHP Forms

2006-05-06 Thread Richard Lynch
On Thu, May 4, 2006 9:57 pm, R. Van Tassel wrote:
 I am having an issue with a form, basically an order form, with 10
 rows.
 Each row is the same, the rows are being generated by a loop and I am
 appending the counter of the loop to the name of the form elements
 (i.e.
 quantity1, type1, next row = quantity2, type2, etc)

Save yourself a lot of headaches and use:

NAME=quantity[1]

It will be an array when you process it with $_POST

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[PHP] PHP Forms

2006-05-04 Thread R. Van Tassel
I am having an issue with a form, basically an order form, with 10 rows.
Each row is the same, the rows are being generated by a loop and I am
appending the counter of the loop to the name of the form elements (i.e.
quantity1, type1, next row = quantity2, type2, etc)

 

I can't seem to receive the variables without using sessions, which I'm
trying not to do. If only the first 2 rows are being filled out and
submitted I need to be able to loop through, grab all the variables and
print them out. Can anyone give any suggestion? 

 

 

 



Re: [PHP] PHP Forms

2006-05-04 Thread Chris

R. Van Tassel wrote:

I am having an issue with a form, basically an order form, with 10 rows.
Each row is the same, the rows are being generated by a loop and I am
appending the counter of the loop to the name of the form elements (i.e.
quantity1, type1, next row = quantity2, type2, etc)

 


I can't seem to receive the variables without using sessions, which I'm
trying not to do. If only the first 2 rows are being filled out and
submitted I need to be able to loop through, grab all the variables and
print them out. Can anyone give any suggestion? 


Make them an array:

input type=text name=quantity[1] value=X
input type=text name=quantity[2] value=X

etc etc.

Then when you post you get an array of quantities which you can loop 
through:


foreach($_POST['quantity'] as $p = $value) {
  if (empty($value)) continue; // they didn't fill this field in.
  
}

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Re: [PHP] PHP Forms

2006-05-04 Thread Chris

R. Van Tassel wrote:

Thanks for the quick response, that seems simple enough, but each row has
about 12 fields, so I just create 12 foreach statements?


Always CC the list..

Number the fields in the same way and you can do something like this:

$number_of_fields = sizeof($_POST['quantity']);
for($i = 1; $i = $number_of_fields; $i++)
$name = $_POST['name'][$i];
$description = $_POST['description'][$i];
$qty = $_POST['quantity'][$i];
.
}

and so on.

You'll need to add your own checks to make sure the fields are set 
correctly..




-Original Message-
From: Chris [mailto:[EMAIL PROTECTED] 
Sent: Thursday, May 04, 2006 11:05 PM

To: R. Van Tassel
Cc: php-general@lists.php.net
Subject: Re: [PHP] PHP Forms

R. Van Tassel wrote:


I am having an issue with a form, basically an order form, with 10 rows.
Each row is the same, the rows are being generated by a loop and I am
appending the counter of the loop to the name of the form elements (i.e.
quantity1, type1, next row = quantity2, type2, etc)



I can't seem to receive the variables without using sessions, which I'm
trying not to do. If only the first 2 rows are being filled out and
submitted I need to be able to loop through, grab all the variables and
print them out. Can anyone give any suggestion? 



Make them an array:

input type=text name=quantity[1] value=X
input type=text name=quantity[2] value=X

etc etc.

Then when you post you get an array of quantities which you can loop 
through:


foreach($_POST['quantity'] as $p = $value) {
   if (empty($value)) continue; // they didn't fill this field in.
   
}




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[PHP] PHP forms that are valid XHTML

2003-12-30 Thread Tim Burgan
Hello,

I'm *very* new to PHP. I am working through the 'Professional PHP
Programming' book by Worx.

In their forms they use the name attribute (ie. name=example) instead of
XHTML's id attribute (ie. id=example).

If I use 'name' my results display on the next page after the submit
button is pressed, but if I change it 'id' the results do not display.

How can I fix this?

Thanks
Tim Burgan

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Re: [PHP] PHP forms that are valid XHTML

2003-12-30 Thread Tom Rogers
Hi,

Wednesday, December 31, 2003, 11:45:37 AM, you wrote:
TB Hello,

TB I'm *very* new to PHP. I am working through the 'Professional PHP
TB Programming' book by Worx.

TB In their forms they use the name attribute (ie. name=example) instead of
TB XHTML's id attribute (ie. id=example).

TB If I use 'name' my results display on the next page after the submit
TB button is pressed, but if I change it 'id' the results do not display.

TB How can I fix this?

TB Thanks
TB Tim Burgan


id is used locally on the client and not passed when you press submit,
there is nothing wrong with using both if id is needed like

input type=text id=test name=test value=/

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Re: [PHP] PHP forms that are valid XHTML

2003-12-30 Thread Leif K-Brooks
Tim Burgan wrote:

In their forms they use the name attribute (ie. name=example) instead of
XHTML's id attribute (ie. id=example).
How can I fix this?

(X)HTML still requires name to be used for forms. It's usually best to 
use both name and ID for forms.

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Re: [PHP] PHP forms that are valid XHTML

2003-12-30 Thread Tim Burgan
Thanks for your replies.

The name attribute is depreciated in XHTML for use with the input tag and
has been replaced with the id attribute.

The name element can still be used in the form tag though.

What I'm looking for is something similar to ASP's
GetElementByID(example);

Thanks
Tim Burgan

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[PHP] php, forms, mysql

2002-11-18 Thread Adrian Partenie
Hello,
I could use some help. 


I have two framed pages, upperframe.html and lowerframe.html.  In upper frame.html:

echo table border=1; 
echo trtd/tdtdID/tdtdSubject/tdtdOpen/tdtdClose/td/tr; 
while($row = MySQL_fetch_array($result)) { 
echo trtdforminput type=\checkbox\ method=\post\ 
name=\linia\/form/td; 
echo tda href=\\ target=\_parent\{$row['id']}/a/td; ??
echo td{$row['subject']}/td; 
echo td{$row['open']}/td;
echo td{$row['close']}/td/tr; 
} 
echo /table;

I display the content of the main table, which has an autoincrement index. For every 
index I have another table, something like  tableID. What I want is to press on  the 
id from a row in upperframe table and to display in lowerframe the tableID. How can I 
do that? 



Thanks a lot, 

Adrian



Re: [PHP] php, forms, mysql

2002-11-18 Thread Marek Kilimajer
You don't need to use forms (which you got wrong), but simple links with 
target=lowerframe and
href=lowerframe.php?tableID=$tableID, then you get in your 
lowerframe.php $_GET['tableID']

Adrian Partenie wrote:

Hello,
I could use some help. 


I have two framed pages, upperframe.html and lowerframe.html.  In upper frame.html:

echo table border=1; 
echo trtd/tdtdID/tdtdSubject/tdtdOpen/tdtdClose/td/tr; while($row = MySQL_fetch_array($result)) { 
echo trtdforminput type=\checkbox\ method=\post\ name=\linia\/form/td; 
echo tda href=\\ target=\_parent\{$row['id']}/a/td; ??
echo td{$row['subject']}/td; 
echo td{$row['open']}/td;
echo td{$row['close']}/td/tr; 
} 
echo /table;

I display the content of the main table, which has an autoincrement index. For every index I have another table, something like  tableID. What I want is to press on  the id from a row in upperframe table and to display in lowerframe the tableID. How can I do that? 



Thanks a lot, 

Adrian

 



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RE: [PHP] PHP Forms

2002-02-26 Thread John Day

Hi

I think your mysql_query is wrong. You need to include the $db variable
in the query otherwise it doesn't know where to put the stuff.

$result = mysql_query($sql, $db);

Cheers
JD

-Original Message-
From: Chiew, Richard [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 25, 2002 6:31 PM
To: [EMAIL PROTECTED]
Subject: [PHP] PHP Forms


I have the following php code to asks for a new student to fill out a
registration form.  Using php, take what the user enters and put it into
the students table. But when i select * from students in mysql, it
didn't show, why?
 
HTML
BODY bgColor=#ff
?php
if ($submit) {
  // process form
  $db = mysql_connect (localhost, root, mollier);
  $select=mysql_select_db (mydb); 
  $sql = INSERT INTO students 
(student_id, firstname,lastname, address) VALUES 
('$student_id', '$firstname', '$lastname', '$address');
  $result = mysql_query($sql);
  echo Thank you! Information entered.\n;
} else{
 
  // display form
 
  ?
FORM onsubmit=return formvalidation(this) 
action=?php echo $PHP_SELF?
method=postINPUT type=hidden value=New Student Registration - BBS 
name=FORM_NAME 
P style=LINE-HEIGHT: 100%; BACKGROUND-COLOR: #ff
align=centerBFONT 
color=#ff size=5BRREGISTRATION FORMBRBR/FONT/B/P
  PREB
FONT color=#ffREQUIRED QUESTIONS:/FONT/B
   BR
   1. What is your student id number?nbsp;INPUT tabIndex=2
maxLength=46 size=46 name=student_idBR
   2. What is your name?BR
   First name:input type=Text name=firstnamebr
 Last name:input type=Text name=lastnamebr
 Address: input type=Text name=address size=60BR
/PRE
UL
  LI
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Please
click 
the Submit Button only once/font /P
LI 
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Then
wait for 
your confirmation screen/font /P
LI 
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Save
the confirmation 
screen because it has information you will need/font /P
/LI/UL
P style=BACKGROUND-COLOR: #ffINPUT type=submit value=Submit
Registrationnbsp;/P
P style=BACKGROUND-COLOR: #ff align=centerEnd of Form-Thank 
You!/P/FORM
HR
H5 style=BACKGROUND-COLOR: #ff /H5
?php
} // end if
?
/BODY/HTML




[PHP] PHP Forms

2002-02-25 Thread Chiew, Richard

I have the following php code to asks for a new student to fill out a
registration form.  Using php, take what the user enters and put it into
the students table. But when i select * from students in mysql, it
didn't show, why?
 
HTML
BODY bgColor=#ff
?php
if ($submit) {
  // process form
  $db = mysql_connect (localhost, root, mollier);
  $select=mysql_select_db (mydb); 
  $sql = INSERT INTO students 
(student_id, firstname,lastname, address) VALUES 
('$student_id', '$firstname', '$lastname', '$address');
  $result = mysql_query($sql);
  echo Thank you! Information entered.\n;
} else{
 
  // display form
 
  ?
FORM onsubmit=return formvalidation(this) 
action=?php echo $PHP_SELF?
method=postINPUT type=hidden value=New Student Registration - BBS 
name=FORM_NAME 
P style=LINE-HEIGHT: 100%; BACKGROUND-COLOR: #ff
align=centerBFONT 
color=#ff size=5BRREGISTRATION FORMBRBR/FONT/B/P
  PREB
FONT color=#ffREQUIRED QUESTIONS:/FONT/B
   BR
   1. What is your student id number?nbsp;INPUT tabIndex=2
maxLength=46 size=46 name=student_idBR
   2. What is your name?BR
   First name:input type=Text name=firstnamebr
 Last name:input type=Text name=lastnamebr
 Address: input type=Text name=address size=60BR
/PRE
UL
  LI
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Please
click 
the Submit Button only once/font /P
LI 
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Then
wait for 
your confirmation screen/font /P
LI 
  P style=BACKGROUND-COLOR: #fffont color=#FF3300Save
the confirmation 
screen because it has information you will need/font /P
/LI/UL
P style=BACKGROUND-COLOR: #ffINPUT type=submit value=Submit
Registrationnbsp;/P
P style=BACKGROUND-COLOR: #ff align=centerEnd of Form-Thank 
You!/P/FORM
HR
H5 style=BACKGROUND-COLOR: #ff /H5
?php
} // end if
?
/BODY/HTML




Re: [PHP] PHP Forms

2002-02-25 Thread Jason Wong

On Tuesday 26 February 2002 02:31, Chiew, Richard wrote:
 I have the following php code to asks for a new student to fill out a
 registration form.  Using php, take what the user enters and put it into
 the students table. But when i select * from students in mysql, it
 didn't show, why?
  
 HTML
 BODY bgColor=#ff
 ?php
 if ($submit) {
   // process form
   $db = mysql_connect (localhost, root, mollier);
   $select=mysql_select_db (mydb); 
   $sql = INSERT INTO students 
 (student_id, firstname,lastname, address) VALUES 
 ('$student_id', '$firstname', '$lastname', '$address');
   $result = mysql_query($sql);
   echo Thank you! Information entered.\n;

Try adding some debug code. Look at the examples in the manual chapter 
MySQL functions.

echo $sql to see what it contains.


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Re: [PHP] PHP Forms

2002-02-25 Thread Stewart G.

Your submit button does not have name=submit set. So $submit is never 
being set.

Also your html is full of errors, I would clean that up and put it thru a 
validator, try html tidy (www.w3c.org).

And, make sure that formvalidation() is passing true, is that fails the 
form will never be submitted.

-- Stewart

On Mon, 25 Feb 2002, Chiew, Richard wrote:

 I have the following php code to asks for a new student to fill out a
 registration form.  Using php, take what the user enters and put it into
 the students table. But when i select * from students in mysql, it
 didn't show, why?
  
 HTML
 BODY bgColor=#ff
 ?php
 if ($submit) {
   // process form
   $db = mysql_connect (localhost, root, mollier);
   $select=mysql_select_db (mydb); 
   $sql = INSERT INTO students 
 (student_id, firstname,lastname, address) VALUES 
 ('$student_id', '$firstname', '$lastname', '$address');
   $result = mysql_query($sql);
   echo Thank you! Information entered.\n;
 } else{
  
   // display form
  
   ?
 FORM onsubmit=return formvalidation(this) 
 action=?php echo $PHP_SELF?
 method=postINPUT type=hidden value=New Student Registration - BBS 
 name=FORM_NAME 
 P style=LINE-HEIGHT: 100%; BACKGROUND-COLOR: #ff
 align=centerBFONT 
 color=#ff size=5BRREGISTRATION FORMBRBR/FONT/B/P
   PREB
 FONT color=#ffREQUIRED QUESTIONS:/FONT/B
BR
1. What is your student id number?nbsp;INPUT tabIndex=2
 maxLength=46 size=46 name=student_idBR
2. What is your name?BR
First name:input type=Text name=firstnamebr
  Last name:input type=Text name=lastnamebr
  Address: input type=Text name=address size=60BR
 /PRE
 UL
   LI
   P style=BACKGROUND-COLOR: #fffont color=#FF3300Please
 click 
 the Submit Button only once/font /P
 LI 
   P style=BACKGROUND-COLOR: #fffont color=#FF3300Then
 wait for 
 your confirmation screen/font /P
 LI 
   P style=BACKGROUND-COLOR: #fffont color=#FF3300Save
 the confirmation 
 screen because it has information you will need/font /P
 /LI/UL
 P style=BACKGROUND-COLOR: #ffINPUT type=submit value=Submit
 Registrationnbsp;/P
 P style=BACKGROUND-COLOR: #ff align=centerEnd of Form-Thank 
 You!/P/FORM
 HR
 H5 style=BACKGROUND-COLOR: #ff /H5
 ?php
 } // end if
 ?
 /BODY/HTML
 


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[PHP] PHP Forms

2002-01-02 Thread Matt Obstgarten

 Hello,
I am new to PHP. I have PHP (cgi version) configured on my
machine running XP Pro and IIS. It works fine. The first script I wanted
to write was the ever-useful feedback form script where the contents of
a form are e-mailed to the webmaster. I am sure that I have a working
script but when I test on my machine I get a server error. I realize
that this is most likely because I don't have any mail settings
configured on IIS (new to this as well). I was wondering if someone
could give me some advice on how to properly set this up so I can
locally test mail scripts like this. Thanks,
Matt


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Re: [PHP] PHP Forms

2002-01-02 Thread Valentin V. Petruchek

Mail() function works fine on win32, you need enter smtp server in your
php.ini and sendfrom address.
example:

SMTP = 10.1.25.1 ; for Win32 only
sendmail_from = [EMAIL PROTECTED]

Valentin Petruchek (aki Zliy Pes)
*** ??? ?? ***
http://zliypes.com.ua
mailto:[EMAIL PROTECTED]
- Original Message -
From: Matt Obstgarten [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, January 02, 2002 3:48 PM
Subject: [PHP] PHP Forms


 Hello,
 I am new to PHP. I have PHP (cgi version) configured on my
 machine running XP Pro and IIS. It works fine. The first script I wanted
 to write was the ever-useful feedback form script where the contents of
 a form are e-mailed to the webmaster. I am sure that I have a working
 script but when I test on my machine I get a server error. I realize
 that this is most likely because I don't have any mail settings
 configured on IIS (new to this as well). I was wondering if someone
 could give me some advice on how to properly set this up so I can
 locally test mail scripts like this. Thanks,
 Matt


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 To contact the list administrators, e-mail: [EMAIL PROTECTED]





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[PHP] PHP Forms and String Limitations

2001-12-06 Thread Jeremy Reed

I've set up a news database with a PHP front-end and am using MS SQL Server.
The news table is set up correctly with the pertinent field being of type
'text' which, according to documentation, should support well over 10 megs
of text data.

This is the problem: The user tries to submit a news article of 2+ pages
(approx 4400 characters) but the article gets truncated to about 4050
characters.  Is there some sort of limitation on PHP variables that is
causing this?

Thanks!

Best regards,

Jeremy Reed



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[PHP] PHP forms

2001-01-31 Thread Victor Hamutenya

Hi, my name is Victor from Namibia, I do web development with PHP, I 
hapenned to find your email address on one of the PHP sites on the Net, 
as one of the contributors on the PHP notes. 

Can you please, if it is possible, tell me how to write in a form field 
with PHP script. Like in Javascript, if one has a form named members and 
a text field called memberid, one can write in the memberid field as 
follows:
script language =javascript
member.memberid.value='M100';
/script

I will appreciate your help very much.

Thanking you in advance.

Vict



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Re: [PHP] PHP forms

2001-01-31 Thread Johannes Janson

e.g.
input type=password name=memberid !!NOW!! value=?php echo "whatever"; ? 

Johannes

"Victor Hamutenya" [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Hi, my name is Victor from Namibia, I do web development with PHP, I
 hapenned to find your email address on one of the PHP sites on the Net,
 as one of the contributors on the PHP notes.

 Can you please, if it is possible, tell me how to write in a form field
 with PHP script. Like in Javascript, if one has a form named members and
 a text field called memberid, one can write in the memberid field as
 follows:
 script language =javascript
 member.memberid.value='M100';
 /script

 I will appreciate your help very much.

 Thanking you in advance.

 Vict



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 PHP General 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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