Re: [PHP] $$var

2011-03-06 Thread NetEmp
As per my experience so far, there is no such depth limit existing. The only
limit is imposed by the system resources (like script execution time etc.)
but not by PHP.

Cheers
NetEmp

On Sun, Mar 6, 2011 at 8:42 PM, shiplu  wrote:

> Just being curious, I have a question.
> How many times PHP interpreter will replace this variables? I mean how deep
> it will be?
>
> If I use variable variables like
>
> $$a
> how long it will be evaluated?
>
> --
> Shiplu Mokadd.im
> My talks, http://talk.cmyweb.net
> Follow me, http://twitter.com/shiplu
> Innovation distinguishes between follower and leader
>


Re: [PHP] $$var

2011-03-06 Thread Mujtaba Arshad
If $a = 'foo'
and $$a = nothing (i.e. no value assigned to $foo) you will get an error if
you tried to use this to do something else.

On Sun, Mar 6, 2011 at 3:21 PM, tedd  wrote:

> At 6:42 PM +0530 3/6/11, Ashim Kapoor wrote:
>
>> Dear All,
>>
>> I was reading the php manual for session_register, and I found the
>> following
>> line there : -
>>
>>
>> $_SESSION[$var] = $$var;
>>
>> Why do I need $$ there ? Can someone explain?
>>
>> Thank you,
>> Ashim
>>
>
> Ashim:
>
> You don't need to user session_register().
>
> Cheers,
>
> tedd
>
>
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Re: [PHP] $$var

2011-03-06 Thread tedd

At 6:42 PM +0530 3/6/11, Ashim Kapoor wrote:

Dear All,

I was reading the php manual for session_register, and I found the following
line there : -


$_SESSION[$var] = $$var;

Why do I need $$ there ? Can someone explain?

Thank you,
Ashim


Ashim:

You don't need to user session_register().

Cheers,

tedd


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Re: [PHP] $$var

2011-03-06 Thread sexyprout
∞

2011/3/6 shiplu 

> Just being curious, I have a question.
> How many times PHP interpreter will replace this variables? I mean how deep
> it will be?
>
> If I use variable variables like
>
> $$a
> how long it will be evaluated?
>
> --
> Shiplu Mokadd.im
> My talks, http://talk.cmyweb.net
> Follow me, http://twitter.com/shiplu
> Innovation distinguishes between follower and leader
>


Re: [PHP] $$var

2011-03-06 Thread shiplu
Just being curious, I have a question.
How many times PHP interpreter will replace this variables? I mean how deep
it will be?

If I use variable variables like
$$a
how long it will be evaluated?

-- 
Shiplu Mokadd.im
My talks, http://talk.cmyweb.net
Follow me, http://twitter.com/shiplu
Innovation distinguishes between follower and leader


Re: [PHP] $$var

2011-03-06 Thread Ashim Kapoor
> Hi Ashim,
>
> These are called Variable Variables. Ideally they should be avoided,
> as they introduce unnecessary legibility issues.
>
> This is what it does in a nutshell, it's actually quite simple:
>
> $foo = 'bar';
> $bar = 'foobar';
> echo $$foo;//This prints foobar
>
> What it does is, take the value of $foo (which is 'bar') and if a
> variable exists by that name, it will go forth and print the value of
> $bar; In this case foobar.
>

Alright Russel, Thank you,
Ashim.


Re: [PHP] $$var

2011-03-06 Thread Russell Dias
Hi Ashim,

These are called Variable Variables. Ideally they should be avoided,
as they introduce unnecessary legibility issues.

This is what it does in a nutshell, it's actually quite simple:

$foo = 'bar';
$bar = 'foobar';
echo $$foo;//This prints foobar

What it does is, take the value of $foo (which is 'bar') and if a
variable exists by that name, it will go forth and print the value of
$bar; In this case foobar.
On Sun, Mar 6, 2011 at 11:12 PM, Ashim Kapoor  wrote:
> Dear All,
>
> I was reading the php manual for session_register, and I found the following
> line there : -
>
>
> $_SESSION[$var] = $$var;
>
> Why do I need $$ there ? Can someone explain?
>
> Thank you,
> Ashim
>

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Re: [PHP] Var within a var

2006-04-09 Thread Kevin Waterson
This one time, at band camp, bob pilly <[EMAIL PROTECTED]> wrote:

> Hi All
>  
>  Im trying to store a document template in mysql that has php var names 
> within it and then find it in the datebase and print it out with the var 
> names replaced with the var values.

eval()

Kevin

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Re: [PHP] var references question

2004-09-19 Thread Curt Zirzow
* Thus wrote Brandon Goodin:
> Greetings,
>  
> 
> In the following example when I run the myscript.php I pass the $myarr into
> the addVal function of the MyClass class. The var_dump in the addVal
> function appropriately displays all of the elements of the array a,b,c and
> d. However, the var_dump that is done in the script immediately following
> the addVal method call displays only elements a,b and c with NO d. From what
> I can see, php does not retain the same reference to the array object when
> it passes it into the function. It appears that it clones the array and
> retains only the changes within the scope of the function. In java I am used
> to creating an array and passing it into a method which accomplishes
> operations against the array. The passed in array is also modified because
> it is a reference to the same array object.
>

PHP pass by value by default, you want to pass by reference, see:

  http://php.net/functions.arguments.php



Curt
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Re: [PHP] var references

2004-02-06 Thread Arthur Pelkey
Thanks!, my syntax was a bit off, havn't had much sleep, but it kept 
slipping by me ;)

Chris W. Parker wrote:

Arthur Pelkey 
on Friday, February 06, 2004 9:46 AM said:

$blat = $tue_5a;


[snip]


while($row = mysql_fetch_array($result)) {
switch($blat) {
case tue_5a:
$min_1 = $row[29];
break;
}


[snip]


Why can I not refence the row[29] from min_1? ive tried various ways
and nothing shows up unless i actually put the assoc array value. Any
ideas? 


i think you forgot your $ in front of the tue_5a in 'case tue_5a'?
otherwise the way you are using tue_5a the second time, appears to be a
constant and not a string or a regular variable.


hth,
chris.


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RE: [PHP] var references

2004-02-06 Thread Chris W. Parker
Arthur Pelkey 
on Friday, February 06, 2004 9:46 AM said:

> $blat = $tue_5a;

[snip]

> while($row = mysql_fetch_array($result)) {
>   switch($blat) {
>   case tue_5a:
>   $min_1 = $row[29];
>   break;
>   }

[snip]

> Why can I not refence the row[29] from min_1? ive tried various ways
> and nothing shows up unless i actually put the assoc array value. Any
> ideas? 

i think you forgot your $ in front of the tue_5a in 'case tue_5a'?
otherwise the way you are using tue_5a the second time, appears to be a
constant and not a string or a regular variable.



hth,
chris.

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RE: [PHP] /VAR is driving me crazy - SOLVED

2004-01-16 Thread Boaz Yahav
Actually i just found the problem.
My sysadmin recently added MOD Security and one of the lines that 
my conf file had was : 

SecFilter /var

Security people can drive me crazy :)

Sincerely
 
berber
 
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-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED] 
Sent: Friday, January 16, 2004 5:07 PM
To: Boaz Yahav
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] /VAR is driving me crazy


Boaz Yahav wrote:
> Hi
> 
> Does anyone have an idea what is so special about the string "/VAR" ? 
> If i create a form that submits to a php file and in the form i put 
> the string /VAR anywhere in the text or just /VAR I get an 404 error 
> from apache on an existing
> file. If i take the same form and remove the /VAR string all is ok.
> 
> I'm not sure if this is an apache issue or PHP. Did anyone encounter 
> such a behavior?

The only thing I can think of is that if you're using GET as your form 
method and end up with something like:

www.domain.com/file.php?v=/VAR

then maybe for some odd reason you're web server is actually looking for

a path of "file.php?v=" and "VAR" instead of realizing the "/VAR" is in 
a query string. That shouldn't be happening, though.

-- 
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Re: [PHP] /VAR is driving me crazy

2004-01-16 Thread John W. Holmes
Boaz Yahav wrote:
Hi

Does anyone have an idea what is so special about the string "/VAR" ?
If i create a form that submits to a php file and in the form i put the
string /VAR
anywhere in the text or just /VAR I get an 404 error from apache on an
existing
file. If i take the same form and remove the /VAR string all is ok.
I'm not sure if this is an apache issue or PHP. Did anyone encounter
such a behavior?
The only thing I can think of is that if you're using GET as your form 
method and end up with something like:

www.domain.com/file.php?v=/VAR

then maybe for some odd reason you're web server is actually looking for 
a path of "file.php?v=" and "VAR" instead of realizing the "/VAR" is in 
a query string. That shouldn't be happening, though.

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Re: [PHP] var static

2003-08-28 Thread Jordan S. Jones
If I were you, I would use the following:

if (!is_a ($miInstancia, 'db'))
$miInstancia=new db();
That way you can ensure that the variable has been instantiated and is an instance 
of the db class..
But it may not really matter..
Jordan S. Jones



Alvaro Martinez wrote:

I've found the solution myself.
The db class is the next:
class db{

function db (){
// funcion que se conecta con la BBDD
$result = @mysql_pconnect("inforalv", "discoteca", "password");
if (!$result)
return false;
if ([EMAIL PROTECTED]("discoteca"))
return false;
}
function &getInstancia(){
static $miInstancia;
if (!get_class($miInstancia) == 'db')
$miInstancia=new db();
return $miInstancia;
}
and the call to this class is the next:

$conexiondb1=&db::getInstancia();
$conexiondb2=&db::getInstancia();
In the second call I obtain the same object.

Muchas gracias por vuestra ayuda

Alvaro

"Dynamical.Biz" <[EMAIL PROTECTED]> escribió en el mensaje
news:[EMAIL PROTECTED]
Although PHP supports static variables in functions (see here), it has no
support for static variables in classes.
http://www.php.net/manual/en/language.variables.scope.php
(espero que te sirva)

saludos

aniceto lópez :: DYNAMICAL.BIZ
web development & host services
Barcelona - Spain


-Mensaje original-
Asunto: [PHP] var static
I want to obtain only one instance of one class. In other to do that, I call
to a static method, that creates the instance(if it doesnt exit) or returns
the reference of the instance.
The call to the static method is the next:
  $conexiondb=db::getInstancia();

Well, but if I call to db::getInstancia() another time, I obtain another new
object  :-(
The code of the db class is the next (it is a Singleton Pattern, but it
doesnt work)
class db{
  var $_miInstancia;
  function db (){
 // funcion que se conecta con la BBDD
static $miInstancia;
$this->_miInstancia=&$miInstancia;
$result = @mysql_pconnect("inforalv", "discoteca", "password");
if (!$result)
  return false;
if ([EMAIL PROTECTED]("discoteca"))
  return false;
  }
&function getInstancia(){
  if (!isset($this))
 $_miInstancia=new db();
  return $_miInstancia;
}
}
I think that the problem is that the var _miInstance is not static and I
dont know how to do it.
Could you please tell me if there is anything wrong?
Thanks.

Alvaro

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Re: [PHP] var static

2003-08-28 Thread Robert Cummings
Globals!? YUCK :) A better solution IMHO that maintains encapsulation,
is to use a static var in a function:

function &getClassVar( $name )
{
return getAndSetClassVar( $name, false );
}

function &setClassVar( $name, $value )
{
return getAndSetClassVar( $name, true, $value );
}

function &getAndSetClassVar( $name, $set, &$value )
{
static $classVars = array();

if( !$set )
{
return $classVars[$name];
}

$classVars[$name] = &$value;

return $value;
}

Admittedly this is slower than using a global, but it doesn't pollute
the global scope and ensures you'll never have naming conflicts.

HTH,
Rob.

On Thu, 2003-08-28 at 13:48, Tom Rogers wrote:
> 
> static calls to a class will never have $this set so you have to set
> your reference globally something like this:
> 
>  $_miInstancia = array();
> class db{
> var $c;
> function db ($count=false){
> global $_miInstancia;
> $_miInstancia['db'] =& $this;
> if($count)$this->c = $count;
> // funcion que se conecta con la BBDD
> $result = @mysql_pconnect("inforalv", "discoteca", "password");
> if (!$result)
> return false;
> if ([EMAIL PROTECTED]("discoteca"))
> return false;
> }
> function &getInstancia($count=false){
> global $_miInstancia;
> if (!isset($_miInstancia['db'])){
> new db($count);
> }
> return $_miInstancia;
> }
> }
> $conexiondb =& db::getInstancia(1);
> print_r($conexiondb);
> $conexiondb2 =& db::getInstancia();
> print_r($conexiondb2);
> 
> ?>
> 
> (The $count was just to prove we only have the one instance)
> -- 
> regards,
> Tom
> 
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Re: [PHP] var static

2003-08-28 Thread Alvaro Martinez
I've found the solution myself.
The db class is the next:

class db{

function db (){
// funcion que se conecta con la BBDD
$result = @mysql_pconnect("inforalv", "discoteca", "password");
if (!$result)
return false;
if ([EMAIL PROTECTED]("discoteca"))
return false;
}

function &getInstancia(){
static $miInstancia;
if (!get_class($miInstancia) == 'db')
$miInstancia=new db();
return $miInstancia;
}


and the call to this class is the next:

$conexiondb1=&db::getInstancia();
$conexiondb2=&db::getInstancia();

In the second call I obtain the same object.

Muchas gracias por vuestra ayuda

Alvaro

"Dynamical.Biz" <[EMAIL PROTECTED]> escribió en el mensaje
news:[EMAIL PROTECTED]

Although PHP supports static variables in functions (see here), it has no
support for static variables in classes.
http://www.php.net/manual/en/language.variables.scope.php

(espero que te sirva)


saludos

aniceto lópez :: DYNAMICAL.BIZ
web development & host services
Barcelona - Spain




-Mensaje original-
Asunto: [PHP] var static


I want to obtain only one instance of one class. In other to do that, I call
to a static method, that creates the instance(if it doesnt exit) or returns
the reference of the instance.
The call to the static method is the next:

   $conexiondb=db::getInstancia();

Well, but if I call to db::getInstancia() another time, I obtain another new
object  :-(

The code of the db class is the next (it is a Singleton Pattern, but it
doesnt work)

class db{
   var $_miInstancia;

   function db (){
  // funcion que se conecta con la BBDD
 static $miInstancia;
 $this->_miInstancia=&$miInstancia;

 $result = @mysql_pconnect("inforalv", "discoteca", "password");
 if (!$result)
   return false;
 if ([EMAIL PROTECTED]("discoteca"))
   return false;
   }

&function getInstancia(){
   if (!isset($this))
  $_miInstancia=new db();
   return $_miInstancia;
}
}

I think that the problem is that the var _miInstance is not static and I
dont know how to do it.
Could you please tell me if there is anything wrong?

Thanks.

Alvaro

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Re: [PHP] var static

2003-08-28 Thread Tom Rogers
Hi,

Friday, August 29, 2003, 3:01:45 AM, you wrote:
AM> I want to obtain only one instance of one class. In other to do that, I call
AM> to a static method, that creates the instance(if it doesnt exit) or returns
AM> the reference of the instance.
AM> The call to the static method is the next:

AM>$conexiondb=db::getInstancia();

AM> Well, but if I call to db::getInstancia() another time, I obtain another new
AM> object  :-(

AM> The code of the db class is the next (it is a Singleton Pattern, but it
AM> doesnt work)

AM> class db{
AM>var $_miInstancia;

AM>function db (){
AM>   // funcion que se conecta con la BBDD
AM>  static $miInstancia;
AM>  $this->_miInstancia=&$miInstancia;

AM>  $result = @mysql_pconnect("inforalv", "discoteca", "password");
AM>  if (!$result)
AM>return false;
AM>  if ([EMAIL PROTECTED]("discoteca"))
AM>return false;
AM>}

AM> &function getInstancia(){
AM>if (!isset($this))
AM>   $_miInstancia=new db();
AM>return $_miInstancia;
AM> }
AM> }

AM> I think that the problem is that the var _miInstance is not static and I
AM> dont know how to do it.
AM> Could you please tell me if there is anything wrong?

AM> Thanks.

AM> Alvaro

static calls to a class will never have $this set so you have to set
your reference globally something like this:

c = $count;
// funcion que se conecta con la BBDD
$result = @mysql_pconnect("inforalv", "discoteca", "password");
if (!$result)
return false;
if ([EMAIL PROTECTED]("discoteca"))
return false;
}
function &getInstancia($count=false){
global $_miInstancia;
if (!isset($_miInstancia['db'])){
new db($count);
}
return $_miInstancia;
}
}
$conexiondb =& db::getInstancia(1);
print_r($conexiondb);
$conexiondb2 =& db::getInstancia();
print_r($conexiondb2);

?>

(The $count was just to prove we only have the one instance)
-- 
regards,
Tom

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RE: [PHP] var static

2003-08-28 Thread Dynamical.biz
Although PHP supports static variables in functions (see here), it has no
support for static variables in classes.
http://www.php.net/manual/en/language.variables.scope.php

(espero que te sirva)


saludos

aniceto lópez :: DYNAMICAL.BIZ
web development & host services
Barcelona - Spain




-Mensaje original-
Asunto: [PHP] var static


I want to obtain only one instance of one class. In other to do that, I call
to a static method, that creates the instance(if it doesnt exit) or returns
the reference of the instance.
The call to the static method is the next:

   $conexiondb=db::getInstancia();

Well, but if I call to db::getInstancia() another time, I obtain another new
object  :-(

The code of the db class is the next (it is a Singleton Pattern, but it
doesnt work)

class db{
   var $_miInstancia;

   function db (){
  // funcion que se conecta con la BBDD
 static $miInstancia;
 $this->_miInstancia=&$miInstancia;

 $result = @mysql_pconnect("inforalv", "discoteca", "password");
 if (!$result)
   return false;
 if ([EMAIL PROTECTED]("discoteca"))
   return false;
   }

&function getInstancia(){
   if (!isset($this))
  $_miInstancia=new db();
   return $_miInstancia;
}
}

I think that the problem is that the var _miInstance is not static and I
dont know how to do it.
Could you please tell me if there is anything wrong?

Thanks.

Alvaro

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RE: [PHP] Var question

2002-10-24 Thread Clint Tredway
Thanks guys..

I am moving away from ColdFusion to PHP and so I still forget about the
isset() function.

-Original Message-
From: 1LT John W. Holmes [mailto:holmes072000@;charter.net] 
Sent: Thursday, October 24, 2002 8:53 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP] Var question


You can just check for 

if(isset($_POST['go']))

You don't really care what the value is since it's just a button.

---John Holmes...

- Original Message - 
From: "Clint Tredway" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, October 24, 2002 9:33 AM
Subject: [PHP] Var question


> I am building a form that posts to itself.
> 
> I have the following to detect the submit button being clicked: 
> If($_POST["go"] == "add link")
> 
> I am getting a warning that says 'go' is undefined. How do I define 
> this?
> 
> Thanks,
> Clint
> 
> 
> 
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Re: [PHP] Var question

2002-10-24 Thread Maxim Maletsky

is go a variable that is passed via post? I guess yes but you get this
error without submitting yet.

There are three solutions:

1. Change your error reporting level to 55:  error_reporting(55). This
will stop warning your undefined variables.

2. Prefix the variable with an at-mark:  @$_POST['go']. This is a
quickie.

3. Most elegant way: add one more check: if(isset($_POST['go']) and $_POST['go']
== 'add link').


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"Clint Tredway" <[EMAIL PROTECTED]> wrote... :

> I am building a form that posts to itself. 
> 
> I have the following to detect the submit button being clicked:
> If($_POST["go"] == "add link")
> 
> I am getting a warning that says 'go' is undefined. How do I define
> this?
> 
> Thanks,
> Clint
> 
> 
> 
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Re: [PHP] Var question

2002-10-24 Thread 1LT John W. Holmes
You can just check for 

if(isset($_POST['go']))

You don't really care what the value is since it's just a button.

---John Holmes...

- Original Message - 
From: "Clint Tredway" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, October 24, 2002 9:33 AM
Subject: [PHP] Var question


> I am building a form that posts to itself. 
> 
> I have the following to detect the submit button being clicked:
> If($_POST["go"] == "add link")
> 
> I am getting a warning that says 'go' is undefined. How do I define
> this?
> 
> Thanks,
> Clint
> 
> 
> 
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Re: [PHP] $var , '$var'

2001-08-18 Thread ReDucTor

nothing, execpt, think there would be about 0.1 time difference... :D I
use $var = $blah."bleh'.$foo because syntax highlighting looks better :D
  - James "ReDucTor" Mitchelll
- Original Message -
From: "nafiseh saberi" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, August 19, 2001 4:33 PM
Subject: [PHP] $var , '$var'


>
> hi.
> what is the difference between $var and '$var' ?/
> thanks.
>
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Re: [PHP] $var , '$var'

2001-08-18 Thread Chris Lambert

In what context?

print $var; // prints the value of $var
print "$var"; // prints the value of $var
print '$var'; // prints $var

/* Chris Lambert, CTO - [EMAIL PROTECTED]
WhiteCrown Networks - More Than White Hats
Web Application Security - www.whitecrown.net
*/

- Original Message - 
From: nafiseh saberi <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, August 19, 2001 2:33 AM
Subject: [PHP] $var , '$var'


| 
| hi.
| what is the difference between $var and '$var' ?/
| thanks.
| 
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| 
| 


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Re: [PHP] var question

2001-04-16 Thread Steve Edberg

You can also look at environment variable $QUERY_STRING. For example,

 if ($HTTP_SERVER_VARS['QUERY_STRING'] == 'login') {
 ...do log in procedure

If register_globals is on, all you need is

 if ($QUERY_STRING == 'login')

If you might be passing other parameters in the URL, and want to ignore 
case, you could use something like:

 if (eregi('login', $QUERY_STRING)) {

This would match

 www.blah.com?login
 www.blah.com?LogIn
 www.blah.com?login&user=herman_hollerith

but it would also match

 www.blah.com?sloginthemud
 www.blah.com?login0

So, a little regular expression twiddling might be in order.

For more info, see:

http://www.php.net/manual/en/language.variables.predefined.php
http://www.php.net/manual/en/ref.regex.php
http://www.php.net/manual/en/ref.pcre.php


 -steve



At 02:43 PM 4/16/01 , Plutarck wrote:
>Add an "=" on the end of your url. Viola.
>
>
>--
>Plutarck
>Should be working on something...
>...but forgot what it was.
>
>
>""Jeroen Geusebroek"" <[EMAIL PROTECTED]> wrote in message
>[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > Hi Guys,
> >
> > I have a question about the way PHP handles var/strings.
> >
> > Let's say i have this URL: http://foo.bar.com/?login.
> > And in my script i have this code:
> >
> > if($login) { echo "blab"; } or
> > if(isset($login)) { echo "blab"; }
> >
> > It always returns FALSE. I think that is because the string
> > is empty. Shouldn't PHP, even if a var is empty, put it in
> > his var-list?
> >
> > Is there another way to do what i want?
> >
> > Thanks,
> >
> > Jeroen Geusebroek
> >


++
| Steve Edberg   University of California, Davis |
| [EMAIL PROTECTED] (530)754-9127 |
| http://aesric.ucdavis.edu/  http://pgfsun.ucdavis.edu/ |
+-- Gort, Klaatu barada nikto! --+


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Re: [PHP] var question

2001-04-16 Thread Plutarck

Note: Yes, I meant string instrument. Any references to YoYo Ma will be
dealt with swiftly and severly.


--
Plutarck
Should be working on something...
...but forgot what it was.


""Plutarck"" <[EMAIL PROTECTED]> wrote in message
9bfp5e$169$[EMAIL PROTECTED]">news:9bfp5e$169$[EMAIL PROTECTED]...
> Add an "=" on the end of your url. Viola.
>
>
> --
> Plutarck
> Should be working on something...
> ...but forgot what it was.
>
>
> ""Jeroen Geusebroek"" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > Hi Guys,
> >
> > I have a question about the way PHP handles var/strings.
> >
> > Let's say i have this URL: http://foo.bar.com/?login.
> > And in my script i have this code:
> >
> > if($login) { echo "blab"; } or
> > if(isset($login)) { echo "blab"; }
> >
> > It always returns FALSE. I think that is because the string
> > is empty. Shouldn't PHP, even if a var is empty, put it in
> > his var-list?
> >
> > Is there another way to do what i want?
> >
> > Thanks,
> >
> > Jeroen Geusebroek
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED]
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> > To contact the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
>
>
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Re: [PHP] var question

2001-04-16 Thread Aemuli

Actually, I meant the instrument LOL ;)

So as I would pronounce it IRL, "vee-O-luh". I use to spell it wrong, and
was notified of it, but I've decided to just use viola on purpose. I
actually litterally pronounce it "viola".

Some would argue that the majority of people think I just do it to annoy
them. I just like the way "viola" sounds :)

Plutarck
Should be working on something...
...but forgot what it was.


- Original Message -
From: "Brian Clark" <[EMAIL PROTECTED]>
To: "Plutarck" <[EMAIL PROTECTED]>
Sent: Monday, 16. April 2001 16:54
Subject: Re: [PHP] var question


> Hi Plutarck,
>
> @ 5:43:35 PM on 4/16/2001, Plutarck wrote:
>
> > Add an "=" on the end of your url. Viola.
>
> Heheh, I think you meant Voila which is like "there you are" in
> French?
>
> Viola is either a) Any of the numerous plants of the genus Viola, or
> b) Slightly larger than a violin, tuned a fifth lower.
>
> ;-)
>
> -Brian
>


_
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Get your free @yahoo.com address at http://mail.yahoo.com


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Re: [PHP] var question

2001-04-16 Thread Plutarck

Add an "=" on the end of your url. Viola.


--
Plutarck
Should be working on something...
...but forgot what it was.


""Jeroen Geusebroek"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi Guys,
>
> I have a question about the way PHP handles var/strings.
>
> Let's say i have this URL: http://foo.bar.com/?login.
> And in my script i have this code:
>
> if($login) { echo "blab"; } or
> if(isset($login)) { echo "blab"; }
>
> It always returns FALSE. I think that is because the string
> is empty. Shouldn't PHP, even if a var is empty, put it in
> his var-list?
>
> Is there another way to do what i want?
>
> Thanks,
>
> Jeroen Geusebroek
>
>
>
> --
> 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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RE: [PHP] VAR and variables

2001-01-29 Thread Steve Edberg

At 10:50 AM + 1/29/01, Tim Ward wrote:
>  > -Original Message-
>>  From: Steve Edberg [mailto:[EMAIL PROTECTED]]
>>  Sent: 25 January 2001 22:02
>>  To: Matt; [EMAIL PROTECTED]
>>  Subject: Re: [PHP] VAR and variables
>>
>>
>>  At 3:00 PM -0600 1/25/01, Matt wrote:
>>  >I have a question that may seem kind of silly, but I'm curious...
>>  >
>>  >When using PHP why would one use "var" to define a variable as
>>  >opposed to just regularly creating it?
>>
>>
>>  Because that's the way it is ;).
>>
>>  The var is part of the syntax of a class definition; it isn't used
>>  anywhere else. I don't know the actual deep reason for having it, as
>>  far as the parser is concerned, but it does make it clear - at least
>>  to me - what the class variables are.
>
>class definitions are the fundamental building blocks of object orientated
>programming. They define an object type which your object is an example of.
>The defined variables should be regarded as properties of the class rather
>than variables in the usual sense. If you're going to be strict about it you
>should not even refer to them directly outside the class definition but use
>methods to access and change them.


Yes --- I was referring to the use of the keyword 'var' here (as 
opposed to nothing, or some _other_ construct), which I think was the 
original question.



>  >
>>  You can also initialize the variable here, too:
>>
>>  var $a = 5;
>>
>
>not any more you can't ... use the constructor


According to the docs, you can still use a _constant_ initializer in 
PHP4 (I use them in 4.0.4), just not a variable one. From 
http://www.php.net/manual/en/language.oop.php:

Note: In PHP 4, only constant initializers for var variables 
are allowed. Use constructors for non-constant initializers.


Of course, this might not be the official OOP usage, but this is 
PHP's way (I'd call it Pseudo-OOP, but the acronym for that isn't all 
that pleasant ;)

- steve


-- 
+--- "They've got a cherry pie there, that'll kill ya" --+
| Steve Edberg   University of California, Davis |
| [EMAIL PROTECTED]   Computer Consultant |
| http://aesric.ucdavis.edu/  http://pgfsun.ucdavis.edu/ |
+-- FBI Special Agent Dale Cooper ---+

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Re: [PHP] VAR and variables

2001-01-25 Thread Steve Edberg

At 3:00 PM -0600 1/25/01, Matt wrote:
>I have a question that may seem kind of silly, but I'm curious...
>
>When using PHP why would one use "var" to define a variable as 
>opposed to just regularly creating it?


Because that's the way it is ;).

The var is part of the syntax of a class definition; it isn't used 
anywhere else. I don't know the actual deep reason for having it, as 
far as the parser is concerned, but it does make it clear - at least 
to me - what the class variables are.

You can also initialize the variable here, too:

var $a = 5;


>For example, if I have a class defined as below:
>
>class Simple {
>   var $a;
>
>   function Simple() {
> $this->a = 5;
>   }
>
>   function first() {
> return $this->a;
>   }
>
>}
>
>
>This class, when created, sets the variable $a to 5, and the 
>function first returns the value in $a my question is could I 
>omit the var and still have it function in this fashion? What is the 
>purpose of it?
>
>Also, can I access this variable through the same -> convention, 
>i.e. would the following be allowed?
>
>$test = new Simple();
>echo $test->a;
>
>

Yep.

You can also _set_ $a this way:

$test = new simple;
$test->a = 77;
echo $test->first();

will display 77.

See
http://www.php.net/manual/en/language.oop.php

for more info.

- steve

-- 
+--- "They've got a cherry pie there, that'll kill ya" --+
| Steve Edberg   University of California, Davis |
| [EMAIL PROTECTED]   Computer Consultant |
| http://aesric.ucdavis.edu/  http://pgfsun.ucdavis.edu/ |
+-- FBI Special Agent Dale Cooper ---+

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