Re: [PHP] Variable functions within an object

2004-07-30 Thread Julio Sergio Santana 
Curt Zirzow wrote:
  $this-{$this-fname}();
  or (what it actually is doing.. )
  $func = $this-fname;
  $this-$func();
Curt
The point here is that the named function is outside the object. That 
is, $this-foo() doesn't exist, so $this-{$this-fname}(), does not 
work either.
But if you look at your suggested construct, I wonder why
$this-{$this-fname}() is sintactically correct while
{$this-fname}() is not (since we just striped out the prefix '$this-' 
which means that the function is inside the object).

Look for instance at a similar case when dealing with variable variables 
  in arrays, this is what the online documentation says in this case:

In order to use variable variables with arrays, you have to resolve an 
ambiguity problem. That is, if you write $$a[1] then the parser needs to 
know if you meant to use $a[1] as a variable, or if you wanted $$a as 
the variable and then the [1] index from that variable. The syntax for 
resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] 
for the second.

So, I think the case is similar here
Clearly, $this-fname() means call the object's method 'fname', but
{$this-fname}(), would mean call the function whose name is 
'$this-fname'.
I think, for the sake of orthogonallity, this feature should be added to 
 the language.

Thank you.
Sergio.
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Re: [PHP] Variable functions within an object

2004-07-30 Thread Curt Zirzow 
* Thus wrote Julio Sergio Santana:
 
 Curt Zirzow wrote:
 
   $this-{$this-fname}();
 
   or (what it actually is doing.. )
 
   $func = $this-fname;
   $this-$func();
 
 Curt
 
 The point here is that the named function is outside the object. That 
 is, $this-foo() doesn't exist, so $this-{$this-fname}(), does not 
 work either.
 But if you look at your suggested construct, I wonder why
 $this-{$this-fname}() is sintactically correct while
 {$this-fname}() is not (since we just striped out the prefix '$this-' 
 which means that the function is inside the object).

Sorry, i did misread what actual function you were trying to
access, the problem is that {} isn't really the thing that expands
the variable its the special cases:

  -{} 
  ${}

{} by itself simply defines a  code block.


 ...
 
 Clearly, $this-fname() means call the object's method 'fname', but
 {$this-fname}(), would mean call the function whose name is 
 '$this-fname'.
 I think, for the sake of orthogonallity, this feature should be added to 
  the language.

One way to solve this without adding a feature like that would be:

  ${$this-fname} = $this-fname;
  ${$this-fname}();

or for the oneline purists :)

  ${ ${$this-fname} = $this-fname }();

wow.. ${} is more powerful than i had originally though.

Curt
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Re: [PHP] Variable functions within an object

2004-07-30 Thread Julio Sergio Santana 

Curt Zirzow wrote:
or for the oneline purists :)
  ${ ${$this-fname} = $this-fname }();
wow.. ${} is more powerful than i had originally though.
Thank you Curt,
With your suggestion, I finally re-wrote the example, and here it is:
?php
function foo() {
   echo In foo()br /\n;
}
class a {
  var $fname;
  function a() {
$this-fname = 'foo'; // the name of the function
  }
  function execute() { // method to execute the named function
${${$this-fname}=$this-fname}();
  }
}
$w = new a;
$w-execute();
?
And this outputs:
X-Powered-By: PHP/4.1.2
Content-type: text/html
In foo()br /
Thank you again,
Sergio.
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[PHP] Variable functions within an object

2004-07-29 Thread Julio Sergio Santana 
I need to record the names of functions, and then use them later.
Recently I found the following example within the on-line documentation:
?php
function foo() {
   echo In foo()br /\n;
}
$func = 'foo';
$func();// This calls foo()
?
then I supposed that it was easy to extend this concept to objects and 
wrote the following case:

?php
function foo() {
   echo In foo()br /\n;
}
class a {
  var $fname;
  function a() {
$this-fname = 'foo'; // the name of the function
  }
  function execute() { // method to execute the named function
$this-fname();
// I also tried here
// {$this-fname}();
// ${this-fname}();
// $this-fname();
// but none of these worked
  }
}
$w = new a;
$w-execute();
?
And this was the error I got:
X-Powered-By: PHP/4.1.2
Content-type: text/html
br
bFatal error/b:  Call to undefined function:  fname() in b-/b on 
line b14/bbr

I know that this can be solved easily with an intermediate variable:
$temp = $this-fname;
$temp();
but I wonder if there is a more direct method.
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Re: [PHP] Variable functions within an object

2004-07-29 Thread Curt Zirzow 
* Thus wrote Julio Sergio Santana:
 
 class a {
   var $fname;
   function a() {
 $this-fname = 'foo'; // the name of the function
   }
 
   function execute() { // method to execute the named function
 $this-fname();
 // I also tried here
 // {$this-fname}();
 // ${this-fname}();
 // $this-fname();

  $this-{$this-fname}();

  or (what it actually is doing.. )

  $func = $this-fname;
  $this-$func();

Curt
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Re: [PHP] Variable functions within an object

2004-07-29 Thread Jason Davidson 
http://ca3.php.net/manual/en/function.call-user-func.php   ?

Jason

On Thu, 29 Jul 2004 17:11:50 -0500, Julio Sergio Santana
[EMAIL PROTECTED] wrote:
 I need to record the names of functions, and then use them later.
 Recently I found the following example within the on-line documentation:
 ?php
 function foo() {
 echo In foo()br /\n;
 }
 
 $func = 'foo';
 $func();// This calls foo()
 
 ?
 
 then I supposed that it was easy to extend this concept to objects and
 wrote the following case:
 
 ?php
 
 function foo() {
 echo In foo()br /\n;
 }
 
 class a {
var $fname;
function a() {
  $this-fname = 'foo'; // the name of the function
}
 
function execute() { // method to execute the named function
  $this-fname();
  // I also tried here
  // {$this-fname}();
  // ${this-fname}();
  // $this-fname();
  // but none of these worked
}
 }
 
 $w = new a;
 $w-execute();
 
 ?
 
 And this was the error I got:
 
 X-Powered-By: PHP/4.1.2
 Content-type: text/html
 
 br
 bFatal error/b:  Call to undefined function:  fname() in b-/b on
 line b14/bbr
 
 I know that this can be solved easily with an intermediate variable:
 
 $temp = $this-fname;
 $temp();
 
 but I wonder if there is a more direct method.
 
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Re: [PHP] Variable Functions...

2003-07-11 Thread Burhan Khalid 
On Friday, July 11, 2003, 3:11:51 AM, Michael wrote:

MS Smarty has a class method where it calls:

$this-$some_var(somevalue);

Are you sure about that syntax? I'm not too familiar with Smarty, only
used it once, but I think its $this-some_var(value);

MS and this throws errors on Windows versions of php that i've tried. why 
MS is that?

Does it error out on l/unix?



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Burhan Khalid
phplist[at]meidomus[dot]com
http://www.meidomus.com


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[PHP] Variable Functions...

2003-07-10 Thread Michael Smith 
Smarty has a class method where it calls:

$this-$some_var(somevalue);

and this throws errors on Windows versions of php that i've tried. why 
is that?

-Michael
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[PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Philip Olson 


This does not work as expected (as I expect it at least) and gives the
following error. This seems to result with use of empty(), isset(), and
unset(), perhaps others :

  Call to undefined function: empty()

When using :

  $foo = 'empty';
  if ($foo($var)) print 'worked.';

Of course the following works as expected.

  if (empty($var)) print 'worked.';

And with other functions :

  $foo = 'somefunction';
  if ($foo($var)) print 'worked.';

Why won't variable functions work with empty/isset/unset this way?  And
why the 'undefined function' fatal error?


regards,
philip



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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Altunergil, Oktay 

$foo is a string not a PHP function which makes 'empty' a string and not a
function/command.

Am I missing something?

oktay

-Original Message-
From: Philip Olson [mailto:[EMAIL PROTECTED]]
Sent: Monday, May 07, 2001 1:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP] variable functions: empty/isset/unset invalid?



This does not work as expected (as I expect it at least) and gives the
following error. This seems to result with use of empty(), isset(), and
unset(), perhaps others :

  Call to undefined function: empty()

When using :

  $foo = 'empty';
  if ($foo($var)) print 'worked.';

Of course the following works as expected.

  if (empty($var)) print 'worked.';

And with other functions :

  $foo = 'somefunction';
  if ($foo($var)) print 'worked.';

Why won't variable functions work with empty/isset/unset this way?  And
why the 'undefined function' fatal error?


regards,
philip



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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Johnson, Kirk 

Change the parens around $var to curly braces:

if ($foo{$var}) print 'worked.';

Kirk

 -Original Message-
 From: Philip Olson [mailto:[EMAIL PROTECTED]]
 Subject: [PHP] variable functions: empty/isset/unset invalid?
 
 This does not work as expected (as I expect it at least) and gives the
 following error. This seems to result with use of empty(), 
 isset(), and
 unset(), perhaps others :
 
   Call to undefined function: empty()
 
 When using :
 
   $foo = 'empty';
   if ($foo($var)) print 'worked.';
 

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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Philip Olson 


I wish it were that easy.  Also, I'm looking for words on WHY this
behavior exists.

  http://www.php.net/manual/en/functions.variable-functions.php

?php

  // works
  if (empty($var)) print '$var is emptybr';
  
  // works
  $foo = 'is_string';
  $var = 'abcdef';   
  if ($foo($var)) print '$var is a stringbr';
  
  // works
  $foo = 'strlen';
  $var = 'abcdef';
  if ($foo($var)  5) print '$var is over 5 charsbr';
  
  // doesn't work : Fatal Error : Call to undefined function: empty()
  // same with isset() and unset()
  $foo = 'empty';
  if ($foo($var)) print '$var is empty';
  
?

In otherwords, only these few functions aren't working as variable
functions but result in a Fatal Error instead.  Why?


Regards,
Philip

On Mon, 7 May 2001, Johnson, Kirk wrote:

 Change the parens around $var to curly braces:
 
 if ($foo{$var}) print 'worked.';
 
 Kirk
 
  -Original Message-
  From: Philip Olson [mailto:[EMAIL PROTECTED]]
  Subject: [PHP] variable functions: empty/isset/unset invalid?
  
  This does not work as expected (as I expect it at least) and gives the
  following error. This seems to result with use of empty(), 
  isset(), and
  unset(), perhaps others :
  
Call to undefined function: empty()
  
  When using :
  
$foo = 'empty';
if ($foo($var)) print 'worked.';
  
 
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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Johnson, Kirk 

Thanks for the additional code, now I see what you are after. Sorry, I don't
know the answer, other than using curly braces will fix the problem for
empty(). Also, a User Contributed Note at
http://www.php.net/manual/en/functions.php#functions.user-defined has this
to say:

quote
there are tons of good uses for this sort of functionality. But it should be
noted that this will not work with 
include() 
include_once() 
require() 
require_once() 

it's safe to assume that this is for safty.
/quote

Kirk

 -Original Message-
 From: Philip Olson [mailto:[EMAIL PROTECTED]]
 Sent: Monday, May 07, 2001 11:45 AM
 To: [EMAIL PROTECTED]
 Subject: RE: [PHP] variable functions: empty/isset/unset invalid?
 
 
 
 I wish it were that easy.  Also, I'm looking for words on WHY this
 behavior exists.
 
   http://www.php.net/manual/en/functions.variable-functions.php
 
 ?php
 
   // works
   if (empty($var)) print '$var is emptybr';
   
   // works
   $foo = 'is_string';
   $var = 'abcdef';   
   if ($foo($var)) print '$var is a stringbr';
   
   // works
   $foo = 'strlen';
   $var = 'abcdef';
   if ($foo($var)  5) print '$var is over 5 charsbr';
   
   // doesn't work : Fatal Error : Call to undefined function: empty()
   // same with isset() and unset()
   $foo = 'empty';
   if ($foo($var)) print '$var is empty';
   
 ?
 
 In otherwords, only these few functions aren't working as variable
 functions but result in a Fatal Error instead.  Why?
 
 
 Regards,
 Philip
 
 On Mon, 7 May 2001, Johnson, Kirk wrote:
 
  Change the parens around $var to curly braces:
  
  if ($foo{$var}) print 'worked.';
  
  Kirk

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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Philip Olson 


Solved.

empty, isset and unset are not functions, they are language contructs,
which results in the error.  This makes sense, a workaround is creating
functions like isEmpty (or something similar) and using them.  I'll be
submitting a future request soon ;)

Thanks everyone, especially OpenSrc in #php

regards,
Philip

snip

  ?php
  
// works
if (empty($var)) print '$var is emptybr';

// works
$foo = 'is_string';
$var = 'abcdef';   
if ($foo($var)) print '$var is a stringbr';

// works
$foo = 'strlen';
$var = 'abcdef';
if ($foo($var)  5) print '$var is over 5 charsbr';

// doesn't work : Fatal Error : Call to undefined function: empty()
// same with isset() and unset()
$foo = 'empty';
if ($foo($var)) print '$var is empty';

  ?
  
  In otherwords, only these few functions aren't working as variable
  functions but result in a Fatal Error instead.  Why?



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RE: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Altunergil, Oktay 

Though I didn't know about 'variable functions' when answered, this is
exactly what I said.. Interesting :)

$foo is a string not a PHP function which makes 'empty' a string and not a
function/command.

Am I missing something?

oktay

-Original Message-
From: Philip Olson [mailto:[EMAIL PROTECTED]]
Sent: Monday, May 07, 2001 2:43 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP] variable functions: empty/isset/unset invalid?



Solved.

empty, isset and unset are not functions, they are language contructs,
which results in the error.  This makes sense, a workaround is creating
functions like isEmpty (or something similar) and using them.  I'll be
submitting a future request soon ;)

Thanks everyone, especially OpenSrc in #php

regards,
Philip

snip

  ?php
  
// works
if (empty($var)) print '$var is emptybr';

// works
$foo = 'is_string';
$var = 'abcdef';   
if ($foo($var)) print '$var is a stringbr';

// works
$foo = 'strlen';
$var = 'abcdef';
if ($foo($var)  5) print '$var is over 5 charsbr';

// doesn't work : Fatal Error : Call to undefined function: empty()
// same with isset() and unset()
$foo = 'empty';
if ($foo($var)) print '$var is empty';

  ?
  
  In otherwords, only these few functions aren't working as variable
  functions but result in a Fatal Error instead.  Why?



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Re: [PHP] variable functions: empty/isset/unset invalid?

2001-05-07 Thread Gyozo Papp 


 Thanks for the additional code, now I see what you are after. Sorry, I don't
 know the answer, other than using curly braces will fix the problem for
 empty(). Also, a User Contributed Note at
 http://www.php.net/manual/en/functions.php#functions.user-defined has this
 to say:
 
 quote
 there are tons of good uses for this sort of functionality. But it should be
 noted that this will not work with 
 include() 
 include_once() 
 require() 
 require_once() 
 
 it's safe to assume that this is for safty.
 /quote
These funtions are directives rather than normal php functions as the manual says. 
I'm  just guessing, maybe empty(), isset() and other variable-testing functions are 
also directives with a function-syntax wrapper and this is the reason why PHP doesn't 
find them as functions (not in the target-list :).

 
 Kirk
 
  -Original Message-
  From: Philip Olson [mailto:[EMAIL PROTECTED]]
  Sent: Monday, May 07, 2001 11:45 AM
  To: [EMAIL PROTECTED]
  Subject: RE: [PHP] variable functions: empty/isset/unset invalid?
  
  
  
  I wish it were that easy.  Also, I'm looking for words on WHY this
  behavior exists.
  
http://www.php.net/manual/en/functions.variable-functions.php
  
  ?php
  
// works
if (empty($var)) print '$var is emptybr';

// works
$foo = 'is_string';
$var = 'abcdef';   
if ($foo($var)) print '$var is a stringbr';

// works
$foo = 'strlen';
$var = 'abcdef';
if ($foo($var)  5) print '$var is over 5 charsbr';

// doesn't work : Fatal Error : Call to undefined function: empty()
// same with isset() and unset()
$foo = 'empty';
if ($foo($var)) print '$var is empty';

  ?
  
  In otherwords, only these few functions aren't working as variable
  functions but result in a Fatal Error instead.  Why?
  
  
  Regards,
  Philip
  
  On Mon, 7 May 2001, Johnson, Kirk wrote:
  
   Change the parens around $var to curly braces:
   
   if ($foo{$var}) print 'worked.';
   
   Kirk
 
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