Re: [PHP] function array problem
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
While reading this thread I've noticed that you all use venue in the
index of the parameter array. Is this intended or you actually mean value?
You may want to consider these functions:
|func_get_args
||func_get_arg|
|func_num_args|
to make the functions input more loose and have it accept multiple
formats as needed like one dimensional, two dimensional or simple
parameters the same time.
--
Thodoris
Re: [PHP] function array problem
Thodoris wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
While reading this thread I've noticed that you all use venue in the
index of the parameter array. Is this intended or you actually mean
value?
You may want to consider these functions:
|func_get_args
||func_get_arg|
|func_num_args|
to make the functions input more loose and have it accept multiple
formats as needed like one dimensional, two dimensional or simple
parameters the same time.
I'm not sure what the OP meant, but venue is a location, for example the
location of a Metallica concert or the location of a court proceeding.
Latin venire (to come)
--
Thanks!
-Shawn
http://www.spidean.com
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Re: [PHP] function array problem
On Wed, 2009-02-18 at 10:21 -0600, Shawn McKenzie wrote:
Thodoris wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
While reading this thread I've noticed that you all use venue in the
index of the parameter array. Is this intended or you actually mean
value?
You may want to consider these functions:
|func_get_args
||func_get_arg|
|func_num_args|
to make the functions input more loose and have it accept multiple
formats as needed like one dimensional, two dimensional or simple
parameters the same time.
I'm not sure what the OP meant, but venue is a location, for example the
location of a Metallica concert or the location of a court proceeding.
Latin venire (to come)
--
Thanks!
-Shawn
http://www.spidean.com
Yes it was intended, as the queries all relate to an event-based CMS I'm
putting together.
Ash
www.ashleysheridan.co.uk
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[PHP] function array problem
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
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Re: [PHP] function array problem
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
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To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
--
Bastien
Cat, the other other white meat
Re: [PHP] function array problem
On Tue, 2009-02-17 at 15:13 -0500, Bastien Koert wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
It's a shame really, because to me it just seems darn messy to have to
perform a check inside the function itself and initialise variables
there. Putting such initialisations inside the parentheses seems more
elegant. :(
Ash
www.ashleysheridan.co.uk
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] function array problem
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
It seems to work OK for me. I typically use your altered format or the
form that Bastien posted (except I usually default the parameters to
null rather than an empty string) though.
?php
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
print_r($values);
}
addEvent();
?
X-Powered-By: PHP/5.2.0
Content-type: text/html
Array
(
[name] =
[venue] =
[description] =
[errors] = Array
(
)
)
Andrew
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Re: [PHP] function array problem
On Tue, Feb 17, 2009 at 3:29 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Tue, 2009-02-17 at 15:13 -0500, Bastien Koert wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
It's a shame really, because to me it just seems darn messy to have to
perform a check inside the function itself and initialise variables
there. Putting such initialisations inside the parentheses seems more
elegant. :(
Ash
www.ashleysheridan.co.uk
To each their own. I think having an array (especially nested arrays)
embedded in the parameter list of a function declaration like that
looks kind of ugly, but that's just me.
Andrew
--
PHP General Mailing List (http://www.php.net/)
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Re: [PHP] function array problem
On Tue, 2009-02-17 at 15:19 -0500, Andrew Ballard wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
It seems to work OK for me. I typically use your altered format or the
form that Bastien posted (except I usually default the parameters to
null rather than an empty string) though.
?php
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
print_r($values);
}
addEvent();
?
X-Powered-By: PHP/5.2.0
Content-type: text/html
Array
(
[name] =
[venue] =
[description] =
[errors] = Array
(
)
)
Andrew
That's quite odd then! I might have to try it on another server, could
be my main one is having a hissy fit?!
Ash
www.ashleysheridan.co.uk
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Re: [PHP] function array problem
On Tue, 2009-02-17 at 15:21 -0500, Andrew Ballard wrote:
On Tue, Feb 17, 2009 at 3:29 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Tue, 2009-02-17 at 15:13 -0500, Bastien Koert wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
It's a shame really, because to me it just seems darn messy to have to
perform a check inside the function itself and initialise variables
there. Putting such initialisations inside the parentheses seems more
elegant. :(
Ash
www.ashleysheridan.co.uk
To each their own. I think having an array (especially nested arrays)
embedded in the parameter list of a function declaration like that
looks kind of ugly, but that's just me.
Andrew
You kidding? Nested arrays is what makes me get up in the morning!
Ash
www.ashleysheridan.co.uk
--
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Re: [PHP] function array problem
What about:
function addEvent($values='')
{
!is_array($values) $values = Array('name' = '', 'venue' = '',
'description' ='',
'errors' = Array());
//rest of the code
}
It's nice and sort.
2009/2/17 Ashley Sheridan a...@ashleysheridan.co.uk:
On Tue, 2009-02-17 at 15:21 -0500, Andrew Ballard wrote:
On Tue, Feb 17, 2009 at 3:29 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Tue, 2009-02-17 at 15:13 -0500, Bastien Koert wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
It's a shame really, because to me it just seems darn messy to have to
perform a check inside the function itself and initialise variables
there. Putting such initialisations inside the parentheses seems more
elegant. :(
Ash
www.ashleysheridan.co.uk
To each their own. I think having an array (especially nested arrays)
embedded in the parameter list of a function declaration like that
looks kind of ugly, but that's just me.
Andrew
You kidding? Nested arrays is what makes me get up in the morning!
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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Re: [PHP] function array problem
On Tue, 2009-02-17 at 20:35 +, Lewis Wright wrote:
What about:
function addEvent($values='')
{
!is_array($values) $values = Array('name' = '', 'venue' = '',
'description' ='',
'errors' = Array());
//rest of the code
}
It's nice and sort.
2009/2/17 Ashley Sheridan a...@ashleysheridan.co.uk:
On Tue, 2009-02-17 at 15:21 -0500, Andrew Ballard wrote:
On Tue, Feb 17, 2009 at 3:29 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Tue, 2009-02-17 at 15:13 -0500, Bastien Koert wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.ukwrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is
there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
i tend to do
function addEvent($values='')
{
if (!is_array($values))
{
$values = Array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
}
//rest of the code
}
It's a shame really, because to me it just seems darn messy to have to
perform a check inside the function itself and initialise variables
there. Putting such initialisations inside the parentheses seems more
elegant. :(
Ash
www.ashleysheridan.co.uk
To each their own. I think having an array (especially nested arrays)
embedded in the parameter list of a function declaration like that
looks kind of ugly, but that's just me.
Andrew
You kidding? Nested arrays is what makes me get up in the morning!
Ash
www.ashleysheridan.co.uk
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That would make it look a little bit neater than what I already have,
might go with that.
Ash
www.ashleysheridan.co.uk
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Re: [PHP] function array problem
Andrew Ballard wrote:
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
It seems to work OK for me. I typically use your altered format or the
form that Bastien posted (except I usually default the parameters to
null rather than an empty string) though.
?php
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
print_r($values);
}
addEvent();
?
X-Powered-By: PHP/5.2.0
Content-type: text/html
Array
(
[name] =
[venue] =
[description] =
[errors] = Array
(
)
)
Andrew
Works great for me also: PHP 5.2.4-2ubuntu5.5 with Suhosin-Patch 0.9.6.2
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Thanks!
-Shawn
http://www.spidean.com
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Re: [PHP] function array problem
On Tue, Feb 17, 2009 at 3:10 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
I've had a bit of a problem with a function I'm using for a form.
Essentially, the function looks like this:
function addEvent($values = Array('name' = '', 'venue' = '',
'description' = '', 'errors' = Array()))
{
// code here displays the form
}
The function is used to both display an empty form, and the form
populated with values again should there be any validation errors.
Now this works fine when the form has been filled out and there are
errors present, as I can call the function with the correct array
values. However, when I call the function with no arguments (intending
the function to populate the $values array itself) all it does is
present me with an empty array. A print_r($values) just returns
Array( ), no key values defined.
I altered the function to this:
function addEvent($values = Array())
{
if(count($values) == 0)
{
$values = Array('name' = '', 'venue' = '', 'description' =
'', 'errors' = Array());
}
// code here displays the form
}
then all works as intended. Question is, am I being dense, or is there a
reason why this shouldn't work?
Ash
www.ashleysheridan.co.uk
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Here's how I handle making sure I have the correct structure.
function something(array $array=array()) {
$defs = array('name' = '', 'venue' = '', 'description' ='',
'errors' = Array());
$array = array_merge($defs, $array);
}
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[PHP] Function, array - problem (probably easy to solve)
I am constructing a function for retrieving records from a mysql database
and putting them into an array called $print_field[name][number].
My problem is that I only seem to get the first two records and I think I
know where the problem is, but I can´t seem to be able to solve it.
Any ideas anyone?
?
function print_query_array($query, $fields) {
global $print_field;
global $linkid;
// Determine what fields are selected
$field = explode("|",$fields);
$mysql_res = mysql_query($query,$linkid);
while ($strow = mysql_fetch_array($mysql_res))
$print_field[$field[0]][] = $strow[$field[0]]; // Problem
$print_field[$field[1]][] = $strow[$field[1]]; // Problem
}
}
print_query_array("SELECT field1, field2 FROM table LIMIT 10",
"field1|field2");
for ($i=0; $isizeof($print_field); $i++)
echo $print_field["field1"][$i]."br";
echo $print_field["field2"][$i]."br";
}
?
Thanks,
// Tobias
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