RE: [PHP-DB] using a variable as values in an array
Instead of building a string in your loop do the assignment in the loop I have no idea what is being pulled from where, so I can't really give a good example -Original Message- From: Charles Kline [mailto:[EMAIL PROTECTED] Sent: Monday, March 31, 2003 1:25 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] using a variable as values in an array I am trying to build an array from the results of a query on a mySQL table. I am using a loop to create this string in a variable $mystring: '30'=true, '20'=true I then need to use it like this: $defaultValues['ib_article'] = array($mystring); It isn't working. I am sure there is a better method, can anyone point me in the right direction? thanks charles -- 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] using a variable as values in an array
Here is more details. Thanks for the reply.
I am using HTML_QuickForm (pear) to create my forms. In order to set
the default state of some of the elements (checkboxes in this case) I
need to pass an array as the value to QuickForm.
This array needs to be in the format of this:
array('30'=true, '29'=true) where the 30 and 29 are the values that
are coming back from my query. This could return up to 50 records.
This is what I was trying, but it isn't woring:
$mystring;
foreach ($sth_opt as $thekey = $thevalue){
$mystring .= '.$thevalue.'=true, ;
}
On Monday, March 31, 2003, at 04:32 PM, Jennifer Goodie wrote:
Instead of building a string in your loop do the assignment in the loop
I have no idea what is being pulled from where, so I can't really give
a
good example
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Monday, March 31, 2003 1:25 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] using a variable as values in an array
I am trying to build an array from the results of a query on a mySQL
table.
I am using a loop to create this string in a variable $mystring:
'30'=true, '20'=true
I then need to use it like this:
$defaultValues['ib_article'] = array($mystring);
It isn't working. I am sure there is a better method, can anyone point
me in the right direction?
thanks
charles
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RE: [PHP-DB] using a variable as values in an array
$mystring;
foreach ($sth_opt as $thekey = $thevalue){
$mystring .= '.$thevalue.'=true, ;
}
$defaultValues['ib_article'] = array($mystring);
Try this ...
foreach ($sth_opt as $thekey = $thevalue){
$defaultValues['ib_article'][$thevalue] = 'true';
}
I'm not sure if the indices line up exactly with what you had in mind, but
I'm confident you can play with it to get what you want.
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Monday, March 31, 2003 1:49 PM
To: Jennifer Goodie
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] using a variable as values in an array
Here is more details. Thanks for the reply.
I am using HTML_QuickForm (pear) to create my forms. In order to set
the default state of some of the elements (checkboxes in this case) I
need to pass an array as the value to QuickForm.
This array needs to be in the format of this:
array('30'=true, '29'=true) where the 30 and 29 are the values that
are coming back from my query. This could return up to 50 records.
This is what I was trying, but it isn't woring:
$mystring;
foreach ($sth_opt as $thekey = $thevalue){
$mystring .= '.$thevalue.'=true, ;
}
On Monday, March 31, 2003, at 04:32 PM, Jennifer Goodie wrote:
Instead of building a string in your loop do the assignment in the loop
I have no idea what is being pulled from where, so I can't really give
a
good example
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Monday, March 31, 2003 1:25 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] using a variable as values in an array
I am trying to build an array from the results of a query on a mySQL
table.
I am using a loop to create this string in a variable $mystring:
'30'=true, '20'=true
I then need to use it like this:
$defaultValues['ib_article'] = array($mystring);
It isn't working. I am sure there is a better method, can anyone point
me in the right direction?
thanks
charles
--
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To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP-DB] using a variable as values in an array
Hmm... that is not quite it. Here is a better example:
Here is how I would define default values for a checkbox in
HTML_QuickForm;
$defaultValues['ichkABC'] = array('A'=true,'B'=true);
$form-setDefaults($defaultValues);
The checkboxes are defined later like this:
$checkbox[] = HTML_QuickForm::createElement('checkbox', 'A', null,
'A');
$checkbox[] = HTML_QuickForm::createElement('checkbox', 'B', null,
'B');
$checkbox[] = HTML_QuickForm::createElement('checkbox', 'C', null,
'C');
$checkbox[] = HTML_QuickForm::createElement('checkbox', 'D', null,
'D');
$form-addGroup($checkbox, 'ichkABC', 'ABCD:', array('nbsp;', 'br
/'));
So what I am trying to do is built that array that gets passed to:
$defaultValues['ichkABC'] =
If I can create a variable that contains the array that is fine too.
the values 'A' or 'B' as in the above example will be the values in
quotes as I get them from my table query.
thanks
charles
On Monday, March 31, 2003, at 04:52 PM, Jennifer Goodie wrote:
$mystring;
foreach ($sth_opt as $thekey = $thevalue){
$mystring .= '.$thevalue.'=true, ;
}
$defaultValues['ib_article'] = array($mystring);
Try this ...
foreach ($sth_opt as $thekey = $thevalue){
$defaultValues['ib_article'][$thevalue] = 'true';
}
I'm not sure if the indices line up exactly with what you had in mind,
but
I'm confident you can play with it to get what you want.
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Monday, March 31, 2003 1:49 PM
To: Jennifer Goodie
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] using a variable as values in an array
Here is more details. Thanks for the reply.
I am using HTML_QuickForm (pear) to create my forms. In order to set
the default state of some of the elements (checkboxes in this case) I
need to pass an array as the value to QuickForm.
This array needs to be in the format of this:
array('30'=true, '29'=true) where the 30 and 29 are the values that
are coming back from my query. This could return up to 50 records.
This is what I was trying, but it isn't woring:
$mystring;
foreach ($sth_opt as $thekey = $thevalue){
$mystring .= '.$thevalue.'=true, ;
}
On Monday, March 31, 2003, at 04:32 PM, Jennifer Goodie wrote:
Instead of building a string in your loop do the assignment in the
loop
I have no idea what is being pulled from where, so I can't really give
a
good example
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Monday, March 31, 2003 1:25 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] using a variable as values in an array
I am trying to build an array from the results of a query on a mySQL
table.
I am using a loop to create this string in a variable $mystring:
'30'=true, '20'=true
I then need to use it like this:
$defaultValues['ib_article'] = array($mystring);
It isn't working. I am sure there is a better method, can anyone point
me in the right direction?
thanks
charles
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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