Re: [PHP-DB] Drop-down box in php
From: andy amol
I am using the following code, but it is not populating my script. If
you can help I would be grateful.
I am using mysql as database.
?
$sql = SELECT course_id FROM course;
$sql_result = mysql_query($sql)
or die(Couldn't execute query.);
while ($row = mysql_fetch_array($sql_result)) {
$type = $row[course_id];
$typedesc =$row[dept_id];
$option_block .= OPTION value=\$type\$typedesc/OPTION;
You're using [dept_id], but not selecting that column in your query.
---John Holmes...
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Re: [PHP-DB] Drop-down box in php
I use html forms with select field size 1 and the option values and/or what is supposed to appear in the box. This you populate by your DB query, maybe using an array and a loop (from 0 to ..). Torsten andy amol schrieb: hi, I would like to know how to create and populate drop down boxes in php. I want the value to be populated from database. What I am try to do is to provide the forign key value as combo box option, so that I do not have to check for referential integrity. Also if you can help me with a date validation program I will be grateful. Thanks in advance. - Do you Yahoo!? Yahoo! Photos: High-quality 4x6 digital prints for 25¢ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Drop-down box in php
You mean select/select html dropdown-s?
print 'select name=sg';
$r = mysql_query(SELECT id, field1, fieldn FROM table WHERE fieldn = 'sg');
while($RESULT = mysql_fetch_array($)) {
print 'option value='.$RESULT['id'].''.$RESULT['fieldn']'./option';
}
print '/select';
Validating: just check the $_POST['sg'] value, against is_numeric(), then you can do a SELECT, to
check whether the id exists.
You can find some more examples, on php net manual pages
andrej
andy amol wrote:
hi,
I would like to know how to create and populate drop down boxes in php.
I want the value to be populated from database.
What I am try to do is to provide the forign key value as combo box option, so that I do not have to check for referential integrity.
Also if you can help me with a date validation program I will be grateful.
Thanks in advance.
-
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Re: [PHP-DB] Drop-down box in php
From: andy amol [EMAIL PROTECTED]
I would like to know how to create and populate drop down boxes in php.
I want the value to be populated from database.
What I am try to do is to provide the forign key value as combo box
option, so that I do not have to check for referential integrity.
You still have to check. Just because you provide a discreet number of
options in a select box doesn't mean that's really all the user can choose
from. There are many ways to manipulate the data.
That being said, just create a loop as you draw items from your database.
?php
echo 'select name=something size=1';
$sql = SELECT name FROM products WHERE ...;
$result = query($sql);
while($row = fetch_assoc($result))
{ echo option value=\{$row['name']}\{$row['name']}/option\n; }
echo /select;
I don't know what database you're using, so query() and fetch_assoc() are
generic.
---John Holmes...
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RE: [PHP-DB] Drop-down box in php
If it is a MySWL database, the following code works great
?
$sql = select choice, description from views;
$result = mysql_query($sql) or die(mysql_error());
$viewing = ;
$viewing .= select name=\view_code\\n;
while ($view_list = mysql_fetch_array($result))
{
$view_code = $view_list[choice];
$view_desc = stripslashes($view_list[description]);
$viewing .= option value=\$view_code\$view_desc/option\n;
}
$viewing .= /select;
?
Of course you'll have to change it to fit your needs, but I just have a
while loop that gets the information and then dumps each line it gets
out to a temp variable. On the pages that include this one is where the
displaying comes out.
HTH,
Robert
-Original Message-
From: andy amol [mailto:[EMAIL PROTECTED]
Sent: Tuesday, April 20, 2004 11:53 AM
To: David Robley; [EMAIL PROTECTED]
Subject: [PHP-DB] Drop-down box in php
hi,
I would like to know how to create and populate drop down boxes in
php. I want the value to be populated from database. What I am try to do
is to provide the forign key value as combo box option, so that I do not
have to check for referential integrity.
Also if you can help me with a date validation program I will be
grateful.
Thanks in advance.
-
Do you Yahoo!?
Yahoo! Photos: High-quality 4x6 digital prints for 25¢
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Re: [PHP-DB] Drop-down box in php
hi,
I am using the following code, but it is not populating my script. If you can help
I would be grateful.
I am using mysql as database.
?
$sql = SELECT course_id FROM course;
$sql_result = mysql_query($sql)
or die(Couldn't execute query.);
while ($row = mysql_fetch_array($sql_result)) {
$type = $row[course_id];
$typedesc =$row[dept_id];
$option_block .= OPTION value=\$type\$typedesc/OPTION;
}
?
SELECT name=selecttype id=selecttype
? echo $option_block; ?
/SELECT
thanks.
John W. Holmes [EMAIL PROTECTED] wrote:
From: andy amol
I would like to know how to create and populate drop down boxes in php.
I want the value to be populated from database.
What I am try to do is to provide the forign key value as combo box
option, so that I do not have to check for referential integrity.
You still have to check. Just because you provide a discreet number of
options in a box doesn't mean that's really all the user can choosefrom. There are
many ways to manipulate the data.That being said, just create a loop as you draw items
from your database.echo '';
$sql = SELECT name FROM products WHERE ...;
$result = query($sql);
while($row = fetch_assoc($result))
{ echo {$row['name']}\n; }
echo ;
I don't know what database you're using, so query() and fetch_assoc() are
generic.
---John Holmes...
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-
Do you Yahoo!?
Yahoo! Photos: High-quality 4x6 digital prints for 25¢
Re: [PHP-DB] Drop down box NOT populated
Jsut a guess...
Your row has a capital 'A' in the SQL statement, and a lower case 'a' in
teh $row[] call..
does that matter?
Larry Sandwick [EMAIL PROTECTED]
22/01/2004 15:46
Please respond to
[EMAIL PROTECTED]
To
[EMAIL PROTECTED]
cc
Subject
[PHP-DB] Drop down box NOT populated
Can anyone tell me what I am missing here,
The Select statement does not populate my drop down window?
My drop down box is empty !
[snip]
?
include '../db.php';
$how = mysql_query(SELECT count(distinct(account)) from Backlog) or die
(Something bad happened: . mysql_error());
$how_many = mysql_result($how, 0);
echo h3$how_many Accounts to pick from !!!/h3;
$sql = mysql_query(SELECT distinct(Account) FROM Backlog)or die
(Something bad happened here: . mysql_error()) ;
echo select name=\account\\n;
echo option\n;
while ($row = mysql_fetch_array($sql))
{
echo ' option
value='.$row[account].''.$row[account]./option\n;
}
echo /select\n;
?
[end snip]
Larry Sandwick
Sarreid, Ltd.
Network/System Administrator
phone: (252) 291-1414 x223
fax : (252) 237-1592
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Re: [PHP-DB] Drop down box NOT populated
From: [EMAIL PROTECTED]
Jsut a guess...
Your row has a capital 'A' in the SQL statement, and a lower case 'a' in
teh $row[] call..
does that matter?
Yep, that would matter, but not the exact problem. I don't know if this
thread has already been answered or not, so...
The real problem is with this:
$sql = mysql_query(SELECT distinct(Account) FROM Backlog)or die
(Something bad happened here: . mysql_error()) ;
echo select name=\account\\n;
echo option\n;
while ($row = mysql_fetch_array($sql))
{
echo ' option
value='.$row[account].''.$row[account]./option\n;
because there is no account index in $row. You're not selecting account,
you're selecting distinct(Account).
So, you could do it the hard way and use $row['distinct(Account)'] as your
value or change your SQL to:
$sql = mysql_query(SELECT distinct(Account) AS acc FROM Backlog)or die
and use $row['acc']. This is called making an alias. You alias the
distinct(account) column to be called acc. You can name the alias what
ever you want.
If you developed with your error_reporting() set to E_ALL, you'd have gotten
a notice about Undefined index 'account' in $row that may have tipped you
off to all of this.
Hope this helps.
---John Holmes...
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RE: [PHP-DB] Drop down box NOT populated
Instead of:
$row = mysql_fetch_array($sql)
Use:
$row = mysql_fetch_assoc($sql)
Humberto Silva
World Editing
Portugal
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: sexta-feira, 23 de Janeiro de 2004 15:42
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Drop down box NOT populated
From: [EMAIL PROTECTED]
Jsut a guess...
Your row has a capital 'A' in the SQL statement, and a lower case 'a'
in teh $row[] call..
does that matter?
Yep, that would matter, but not the exact problem. I don't know if this
thread has already been answered or not, so...
The real problem is with this:
$sql = mysql_query(SELECT distinct(Account) FROM Backlog)or die
(Something bad happened here: . mysql_error()) ;
echo select name=\account\\n;
echo option\n;
while ($row = mysql_fetch_array($sql))
{
echo ' option
value='.$row[account].''.$row[account]./option\n;
because there is no account index in $row. You're not selecting
account, you're selecting distinct(Account).
So, you could do it the hard way and use $row['distinct(Account)'] as
your value or change your SQL to:
$sql = mysql_query(SELECT distinct(Account) AS acc FROM Backlog)or die
and use $row['acc']. This is called making an alias. You alias the
distinct(account) column to be called acc. You can name the alias what
ever you want.
If you developed with your error_reporting() set to E_ALL, you'd have
gotten a notice about Undefined index 'account' in $row that may have
tipped you off to all of this.
Hope this helps.
---John Holmes...
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Re: [PHP-DB] Drop down box NOT populated
From: Humberto Silva [EMAIL PROTECTED] Instead of: $row = mysql_fetch_array($sql) Use: $row = mysql_fetch_assoc($sql) While I'm in the habit of doing that, using fetch_array() isn't going to cause any problems with regard to the original post. Which one you use is really a matter of personal opinion and rarely affects the code. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Drop down box NOT populated
I confess i didn't read the original post ... You're absolutely right ... Humberto Silva World Editing Portugal -Original Message- From: John W. Holmes [mailto:[EMAIL PROTECTED] Sent: sexta-feira, 23 de Janeiro de 2004 16:03 To: Humberto Silva Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Drop down box NOT populated From: Humberto Silva [EMAIL PROTECTED] Instead of: $row = mysql_fetch_array($sql) Use: $row = mysql_fetch_assoc($sql) While I'm in the habit of doing that, using fetch_array() isn't going to cause any problems with regard to the original post. Which one you use is really a matter of personal opinion and rarely affects the code. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Drop down box NOT populated
Larry Sandwick wrote:
Can anyone tell me what I am missing here,
The Select statement does not populate my drop down window?
My drop down box is empty !
?
include '../db.php';
$how = mysql_query(SELECT count(distinct(account)) from Backlog) or
die (Something bad happened: . mysql_error());
$how_many = mysql_result($how, 0);
echo h3$how_many Accounts to pick from !!!/h3;
$sql = mysql_query(SELECT distinct(Account) FROM Backlog)or die
(Something bad happened here: . mysql_error()) ;
echo select name=\account\\n;
echo option\n;
^^
Try removing this line.
while ($row = mysql_fetch_array($sql))
{
echo ' option
value='.$row[account].''.$row[account]./option\n;
}
echo /select\n;
?
--
Stuart
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Re: [PHP-DB] Drop down box NOT populated
- Original Message -
From: Larry Sandwick
To: [EMAIL PROTECTED]
Sent: Thursday, January 22, 2004 3:46 PM
Subject: [PHP-DB] Drop down box NOT populated
Can anyone tell me what I am missing here,
The Select statement does not populate my drop down window?
My drop down box is empty !
[snip]
?
include '../db.php';
$how = mysql_query(SELECT count(distinct(account)) from Backlog) or die
(Something bad happened: . mysql_error());
$how_many = mysql_result($how, 0);
echo h3$how_many Accounts to pick from !!!/h3;
$sql = mysql_query(SELECT distinct(Account) FROM Backlog)or die (Something bad
happened here: . mysql_error()) ;
echo select name=\account\\n;
echo option\n;-- WHAT IS THIS DOING HERE !
while ($row = mysql_fetch_array($sql))
{
echo ' option value='.$row[account].''.$row[account]./option\n;
}
echo /select\n;
?
[end snip]
Larry Sandwick
Sarreid, Ltd.
Network/System Administrator
phone: (252) 291-1414 x223
fax : (252) 237-1592
RE: [PHP-DB] Drop down box
From: Ben Cairns [EMAIL PROTECTED] To: php-db [EMAIL PROTECTED] Subject: RE: [PHP-DB] Drop down box Date: Thu, 1 Mar 2001 14:06:21 + Hope this helps -- Ben Cairns - Head Of Technical Operations intasept.COM Tel: 01332 365333 Fax: 01332 346010 E-Mail: [EMAIL PROTECTED] Web: http://www.intasept.com "MAKING sense of the INFORMATION TECHNOLOGY age @ WORK.." Hope what helps? *lol* _ Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com. -- 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] Drop down box
The action you want is only available using JavaScript. You need to submit the form without a submit button. The "select" tag has the "OnChange" event which is scriptable. You can use the JavaScript statement "this.form.submit();" associated with the "OnChange" event. form... select name="x" OnChange="this.form.submit();"option.. /select.../form It's a good idea to include the submit button as well, for browers that don't support javascript, or for those who turn it off. That way your site will always work properly. -- 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] Drop down box
At 06:32 PM 3/1/2001 -0300, you wrote: The action you want is only available using JavaScript. True but the action *you* want is most definitely not what many of your *users* are going to want. If you use Javascript this way your site will be un-navigable if Javascript is turned off or the browser does not support it. What's the big deal about a single button click? Also, using Javascript this way means that if a user is just curious what the choices are, you force a refresh without them necessarily wanting one. IMHO of course. Cheers - Island Net AMT Solutions Group Inc. Telephone: 250 383-0096 1412 Quadra Toll Free:1 800 331-3055 Victoria, B.C. Fax:250 383-6698 V8W 2L1 E-Mail:[EMAIL PROTECTED] Canada WWW: http://www.islandnet.com/ - -- 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] Drop down box
The event "OnChange" only happens when the user open the drop down list and change ( select ) a new value, this is not "just looking" . I don't like this kind of solution. If I have to put a button on the page I think the selection list must disappear. Why not give the user a interface with less "clicks"? May be we can reach another interface approach which turn it possible. Jayme. -Mensagem Original- De: Ron Brogden [EMAIL PROTECTED] Para: [EMAIL PROTECTED] Enviada em: quinta-feira, 1 de maro de 2001 18:48 Assunto: Re: [PHP-DB] Drop down box At 06:32 PM 3/1/2001 -0300, you wrote: The action you want is only available using JavaScript. True but the action *you* want is most definitely not what many of your *users* are going to want. If you use Javascript this way your site will be un-navigable if Javascript is turned off or the browser does not support it. What's the big deal about a single button click? Also, using Javascript this way means that if a user is just curious what the choices are, you force a refresh without them necessarily wanting one. IMHO of course. Cheers -- --- Island Net AMT Solutions Group Inc. Telephone: 250 383-0096 1412 Quadra Toll Free:1 800 331-3055 Victoria, B.C. Fax:250 383-6698 V8W 2L1 E-Mail: [EMAIL PROTECTED] Canada WWW: http://www.islandnet.com/ -- --- -- 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]
