Re: [PHP] $$var

[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

[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

[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

[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

[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

[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

[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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[PHP] $$var

[Ashim Kapoor]

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

Re: [PHP] Var within a var

[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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[PHP] Var within a var

[bob pilly]

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.
 
 e.g i have this stored
 
 Dear $title $name
 we would like to..
 
 I declare the vars $title and $name
 
 $title = 'Mr';
 $name = 'Test';
 //retrieve the stored template
 $query = "select template from letters";
 $result = mysqli_query($dblink,$query);
 if($row = mysqli_fetch_array($result)){
 print $row['template'];
 }
 
 i would expect that to print:
 
 Dear Mr Test
 we would like to..
 
 but it doesnt it prints out
 
 Dear $title $name
  we would like to..
 
 can someone point out where i am going wrong?
 
 Thanks for any help in advance!
 
 Cheers
 
 Bob
 

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Re: [PHP] var references question

[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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[PHP] var references question

[Brandon Goodin]

Greetings,

 

This is a little difficult to explain. So, below is an example I will work
from. Forgive my ignorance here. I am a java developer using php4 and trying
to figure out a particular scenario.

 

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.

 

Is there a way to accomplish this behavior in php 4?

 

For example:

 

 myscript.php 

 

.

 

$myarr = array();

 

$myarr["valuea"]="a";

$myarr["valueb"]="b";

$myarr["valuec"]="c";

 

$myclass = new MyClass();

$myclass->addVal($myarr);

 

var_dump($myarr);

 

.

 

===end 

 

 

 

 myclass.php 

 

class MyClass {

 

function addVal($myarr){

  $myarr["valued"] = "d";

  var_dump($myarr);

}

 

}

===end 

 

 

Thanks,

Brandon

 

Re: [PHP] var references

[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

[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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[PHP] var references

[Arthur Pelkey]

I have the following:


$blat = $tue_5a;

$result = mysql_query("SELECT * FROM classes WHERE c_d_tue='1' AND 
c_s_tue_1_hr='5' AND c_s_tue_1_dn='am'");

while($row = mysql_fetch_array($result)) {
	switch($blat) {
 	case tue_5a:
 		$min_1 = $row[29];
 		break;
 	}
 	?>
 	class title:">
 	class minute:">
 	class description:
	
}
?>

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?

It should display "50".

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

[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
 
Visit http://www.weberdev.com/ & http://www.weberblog.com/ Today!!!
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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

[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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Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/

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[PHP] /VAR is driving me crazy

[Boaz Yahav]

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?

Sincerely
 
berber
 
Visit http://www.weberdev.com/ & http://www.weberblog.com/ Today!!!
To see where PHP might take you tomorrow.
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[PHP] var check not working...

[Jas]

For some reason this will not work... my $_POST variables check for no 
form variables and display a form... if they choose an item from a 
select box it queries the database for a field matching the $_post var 
and builds a new form creating 8 separate $_post variables to check 
for.. so far it works except when selecting an item from the select box 
it will only display the first form when there are $_post vars present. 
 I think I am missing something, could someone look over my code?
Thanks in advance,
Jas

function vlans_dhcp() {
$lvl = base64_decode($_SESSION['lvl']);
	if ($lvl != "admin") {
		$_SESSION['lvl'] = base64_encode($lvl);
		$_SESSION['msg'] = $dberrors[12];
		call_user_func("db_logs");
	} elseif ($lvl == "admin") {
		$_SESSION['lvl'] = base64_encode($lvl);
		if (($_POST == "") || (!isset($_POST['dhcp_vlans'])) || 
(!isset($_POST['sub'])) || (!isset($_POST['msk'])) || 
(!isset($_POST['dns01'])) || (!isset($_POST['dns02'])) || 
(!isset($_POST['rtrs'])) || (!isset($_POST['vlans']))) {
			create 1st form with dhcp_vlans as $_post var;
		} elseif (($_POST != "") || (isset($_POST['dhcp_vlans'])) || 
($_POST['dhcp_vlans'] != "--") || (!isset($_POST['sub'])) || 
(!isset($_POST['msk'])) || (!isset($_POST['dns01'])) || 
(!isset($_POST['dns02'])) || (!isset($_POST['rtrs'])) || 
(!isset($_POST['vlans']))) {
			create 2nd form based on dhcp_vlans $_post var and create 8 different 
$_post vars...;
		} elseif (($_POST != "") || (!isset($_POST['dhcp_vlans'])) || 
(isset($_POST['id'])) || (isset($_POST['sub'])) || 
(isset($_POST['msk'])) || (isset($_POST['dns01'])) || 
(isset($_POST['dns02'])) || (isset($_POST['rtrs'])) || 
(isset($_POST['rnge'])) || (isset($_POST['vlans']))) {
			now put 8 $_post vars in database...;
		} else {
			$_SESSION['lvl'] = base64_encode($lvl);
			header("Location: help.php"); }
	} else {
		$_SESSION['lvl'] = base64_encode($lvl);
		$_SESSION['msg'] = $dberrors[12];
		call_user_func("db_logs"); }
}

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Re: [PHP] Trying to submit form to frame and pass php var

[Raditha Dissanayake]

Hi,
you cannot invoke php functions from javascript. You have to pass on any 
relevent variables (which can be done with hidden input fields) with 
your POST/GET data to the script for processing on the server.

[EMAIL PROTECTED] wrote:

Hello I was wondering if someone could help me out.

Trying to submit form to frame and pass php var at the same time.

<!--
function submitform(){
top.admin.document.adminform.submit();
}
//-->
and...


 
 

Link look like: Delete
What I need to do though, is run this exact script while passing
($delete=yes) as php at the same time?  Basically, run the form and if
delete=yes then use the following form tag :  
Any ideas?  Thanks Brandon



 



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[PHP] Trying to submit form to frame and pass php var

[info]

Hello I was wondering if someone could help me out.

Trying to submit form to frame and pass php var at the same time.

<!--
function submitform(){
top.admin.document.adminform.submit();
}
//-->

and...


  
  



Link look like: Delete

What I need to do though, is run this exact script while passing
($delete=yes) as php at the same time?  Basically, run the form and if
delete=yes then use the following form tag :  

Any ideas?  Thanks Brandon

Re: [PHP] var static

[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

[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

[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

[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

[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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[PHP] var static

[Alvaro Martinez]

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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[PHP] var static

[Alvaro Martinez]

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] Evaluate PHP var in stored sql statement?

[Noah]

Hm..

Guess I'll find out as I go deeper into PHP

--Noah


- Original Message -
From: "Maxim Maletsky" <[EMAIL PROTECTED]>
To: "Noah" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Saturday, January 25, 2003 9:47 AM
Subject: Re: [PHP] Evaluate PHP var in stored sql statement?


> no, much less instead
>
> ---
> Maxim Maletsky
> [EMAIL PROTECTED]
>
>
> On Sat, 25 Jan 2003 09:18:47 -0800 "Noah" <[EMAIL PROTECTED]> wrote:
>
> > Thanks for your help Maxim.
> >
> > I've got quite a bit to learn.  PHP is far more challenging than Cold
> > Fusion.
> >
> > --Noah
> >
> > - Original Message -
> > From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> > To: "Noah" <[EMAIL PROTECTED]>
> > Cc: <[EMAIL PROTECTED]>
> > Sent: Friday, January 24, 2003 11:02 AM
> > Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> >
> >
> > >
> > > eval takes the string and executes PHP in that string. So,
> > >
> > > eval("\$myrow[2] = \"$myrow[2]\";");
> > >
> > > is equivalent to:
> > >
> > > $myrow[2] = "$myrow[2]";
> > >
> > > I'd reccomend you using single quotes and then escape only backslashes
> > > and single quotes within a string
> > >
> > > --
> > > Maxim Maletsky
> > > [EMAIL PROTECTED]
> > >
> > >
> > >
> > > "Noah" <[EMAIL PROTECTED]> wrote... :
> > >
> > > > Don't quite understand why this works, but it does:
> > > >
> > > > /* Escape quotes in sql_query so our header_id gets evaluated */
> > > > eval("\$myrow[2] = \"$myrow[2]\";");
> > > >
> > > > What's the deal with the leading backslash in ' eval("\$myrow[2] '?
> > > >
> > > > Does eval() tell php to evaluate the string as php code; i.e. text
> > without
> > > > quotes?
> > > >
> > > > Thanks for the help...
> > > >
> > > > --Noah
> > > >
> > > >
> > > > - Original Message -
> > > > From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> > > > To: "CF High" <[EMAIL PROTECTED]>
> > > > Cc: <[EMAIL PROTECTED]>
> > > > Sent: Friday, January 24, 2003 9:59 AM
> > > > Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> > > >
> > > >
> > > > >
> > > > > use eval()
> > > > >
> > > > > www.php.net/eval
> > > > >
> > > > >
> > > > > --
> > > > > Maxim Maletsky
> > > > > [EMAIL PROTECTED]
> > > > >
> > > > >
> > > > >
> > > > > "CF High" <[EMAIL PROTECTED]> wrote... :
> > > > >
> > > > > > Hey all.
> > > > > >
> > > > > > I need to find out how to get PHP to evaluate a PHP variable
stored
> > in
> > > > our
> > > > > > MySql database.
> > > > > >
> > > > > > Currently the variable is being read as a string; i.e. select *
from
> > > > table
> > > > > > where column = $myrow[0] -- instead of evaluating $myrow[0] as a
> > > > variable.
> > > > > >
> > > > > >
> > > > > > The SQL statement is retrieved in the first call to dbConnect
(see
> > code
> > > > > > below)
> > > > > >
> > > > > >
> > > > > >
> > > > > > dbConnect("SELECT header_id, header, sql_query FROM cat_headers
> > WHERE
> > > > > > section_id = $_REQUEST[section_id] order by header");
> > > > > >
> > > > > >   $set = $result;   /* Give initial result set a unique name
*/
> > > > > >
> > > > > >   while ($myrow = mysql_fetch_row($set)) {
> > > > > >
> > > > > > echo(Display the header title here);
> > > > > >
> > > > > >
> > > > > >   dbConnect($myrow[2]);  /*  $myrow[2] is the
sql_query
> > > > gotten
> > > > > > from the first dbConnect() call  */
> > > > > >
> > > > > >  $set1 = $result;/* Give next result set a
unique
> > name
> > > > */
> > > > > >
> > > > > > while ($myrow = mysql_fetch_row($set1)) {
> > > > > >
> > > > > > echo(Display category rows here);
> > > > > >
> > > > > > }
> > > > > >
> > > > > >   }
> > > > > >
> > > > > > --
> > > > > >  Any leads/suggestions much appreciated..
> > > > > >
> > > > > > --Noah
> > > > > >
> > > > > >
> > > > > >
> > > > > > --
> > > > > > PHP General Mailing List (http://www.php.net/)
> > > > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > > > >
> > > > >
> > > >
> > > >
> > > > --
> > > > PHP General Mailing List (http://www.php.net/)
> > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > >
> > >
> >
>


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Re: [PHP] Evaluate PHP var in stored sql statement?

[Maxim Maletsky]

no, much less instead

---
Maxim Maletsky
[EMAIL PROTECTED]


On Sat, 25 Jan 2003 09:18:47 -0800 "Noah" <[EMAIL PROTECTED]> wrote:

> Thanks for your help Maxim.
> 
> I've got quite a bit to learn.  PHP is far more challenging than Cold
> Fusion.
> 
> --Noah
> 
> - Original Message -
> From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> To: "Noah" <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Friday, January 24, 2003 11:02 AM
> Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> 
> 
> >
> > eval takes the string and executes PHP in that string. So,
> >
> > eval("\$myrow[2] = \"$myrow[2]\";");
> >
> > is equivalent to:
> >
> > $myrow[2] = "$myrow[2]";
> >
> > I'd reccomend you using single quotes and then escape only backslashes
> > and single quotes within a string
> >
> > --
> > Maxim Maletsky
> > [EMAIL PROTECTED]
> >
> >
> >
> > "Noah" <[EMAIL PROTECTED]> wrote... :
> >
> > > Don't quite understand why this works, but it does:
> > >
> > > /* Escape quotes in sql_query so our header_id gets evaluated */
> > > eval("\$myrow[2] = \"$myrow[2]\";");
> > >
> > > What's the deal with the leading backslash in ' eval("\$myrow[2] '?
> > >
> > > Does eval() tell php to evaluate the string as php code; i.e. text
> without
> > > quotes?
> > >
> > > Thanks for the help...
> > >
> > > --Noah
> > >
> > >
> > > - Original Message -
> > > From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> > > To: "CF High" <[EMAIL PROTECTED]>
> > > Cc: <[EMAIL PROTECTED]>
> > > Sent: Friday, January 24, 2003 9:59 AM
> > > Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> > >
> > >
> > > >
> > > > use eval()
> > > >
> > > > www.php.net/eval
> > > >
> > > >
> > > > --
> > > > Maxim Maletsky
> > > > [EMAIL PROTECTED]
> > > >
> > > >
> > > >
> > > > "CF High" <[EMAIL PROTECTED]> wrote... :
> > > >
> > > > > Hey all.
> > > > >
> > > > > I need to find out how to get PHP to evaluate a PHP variable stored
> in
> > > our
> > > > > MySql database.
> > > > >
> > > > > Currently the variable is being read as a string; i.e. select * from
> > > table
> > > > > where column = $myrow[0] -- instead of evaluating $myrow[0] as a
> > > variable.
> > > > >
> > > > >
> > > > > The SQL statement is retrieved in the first call to dbConnect (see
> code
> > > > > below)
> > > > >
> > > > >
> > > > >
> > > > > dbConnect("SELECT header_id, header, sql_query FROM cat_headers
> WHERE
> > > > > section_id = $_REQUEST[section_id] order by header");
> > > > >
> > > > >   $set = $result;   /* Give initial result set a unique name */
> > > > >
> > > > >   while ($myrow = mysql_fetch_row($set)) {
> > > > >
> > > > > echo(Display the header title here);
> > > > >
> > > > >
> > > > >   dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query
> > > gotten
> > > > > from the first dbConnect() call  */
> > > > >
> > > > >  $set1 = $result;/* Give next result set a unique
> name
> > > */
> > > > >
> > > > > while ($myrow = mysql_fetch_row($set1)) {
> > > > >
> > > > > echo(Display category rows here);
> > > > >
> > > > > }
> > > > >
> > > > >   }
> > > > >
> > > > > --
> > > > >  Any leads/suggestions much appreciated..
> > > > >
> > > > > --Noah
> > > > >
> > > > >
> > > > >
> > > > > --
> > > > > PHP General Mailing List (http://www.php.net/)
> > > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > > >
> > > >
> > >
> > >
> > > --
> > > PHP General Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> 


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Re: [PHP] Evaluate PHP var in stored sql statement?

[Noah]

Thanks for your help Maxim.

I've got quite a bit to learn.  PHP is far more challenging than Cold
Fusion.

--Noah

- Original Message -
From: "Maxim Maletsky" <[EMAIL PROTECTED]>
To: "Noah" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Friday, January 24, 2003 11:02 AM
Subject: Re: [PHP] Evaluate PHP var in stored sql statement?


>
> eval takes the string and executes PHP in that string. So,
>
> eval("\$myrow[2] = \"$myrow[2]\";");
>
> is equivalent to:
>
> $myrow[2] = "$myrow[2]";
>
> I'd reccomend you using single quotes and then escape only backslashes
> and single quotes within a string
>
> --
> Maxim Maletsky
> [EMAIL PROTECTED]
>
>
>
> "Noah" <[EMAIL PROTECTED]> wrote... :
>
> > Don't quite understand why this works, but it does:
> >
> > /* Escape quotes in sql_query so our header_id gets evaluated */
> > eval("\$myrow[2] = \"$myrow[2]\";");
> >
> > What's the deal with the leading backslash in ' eval("\$myrow[2] '?
> >
> > Does eval() tell php to evaluate the string as php code; i.e. text
without
> > quotes?
> >
> > Thanks for the help...
> >
> > --Noah
> >
> >
> > - Original Message -
> > From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> > To: "CF High" <[EMAIL PROTECTED]>
> > Cc: <[EMAIL PROTECTED]>
> > Sent: Friday, January 24, 2003 9:59 AM
> > Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> >
> >
> > >
> > > use eval()
> > >
> > > www.php.net/eval
> > >
> > >
> > > --
> > > Maxim Maletsky
> > > [EMAIL PROTECTED]
> > >
> > >
> > >
> > > "CF High" <[EMAIL PROTECTED]> wrote... :
> > >
> > > > Hey all.
> > > >
> > > > I need to find out how to get PHP to evaluate a PHP variable stored
in
> > our
> > > > MySql database.
> > > >
> > > > Currently the variable is being read as a string; i.e. select * from
> > table
> > > > where column = $myrow[0] -- instead of evaluating $myrow[0] as a
> > variable.
> > > >
> > > >
> > > > The SQL statement is retrieved in the first call to dbConnect (see
code
> > > > below)
> > > >
> > > >
> > > >
> > > > dbConnect("SELECT header_id, header, sql_query FROM cat_headers
WHERE
> > > > section_id = $_REQUEST[section_id] order by header");
> > > >
> > > >   $set = $result;   /* Give initial result set a unique name */
> > > >
> > > >   while ($myrow = mysql_fetch_row($set)) {
> > > >
> > > > echo(Display the header title here);
> > > >
> > > >
> > > >   dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query
> > gotten
> > > > from the first dbConnect() call  */
> > > >
> > > >  $set1 = $result;/* Give next result set a unique
name
> > */
> > > >
> > > > while ($myrow = mysql_fetch_row($set1)) {
> > > >
> > > > echo(Display category rows here);
> > > >
> > > > }
> > > >
> > > >   }
> > > >
> > > > --
> > > >  Any leads/suggestions much appreciated..
> > > >
> > > > --Noah
> > > >
> > > >
> > > >
> > > > --
> > > > PHP General Mailing List (http://www.php.net/)
> > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > >
> > >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>


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Re: [PHP] Evaluate PHP var in stored sql statement?

[Matt]

> From: "Noah" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Friday, January 24, 2003 4:39 PM
> Subject: Re: [PHP] Evaluate PHP var in stored sql statement?


> eval("\$myrow[2] = \"$myrow[2]\";");
>
> What's the deal with the leading backslash in ' eval("\$myrow[2] '?

It's escaping the $ so you keep the literal on the left hand side of the
expression.



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Re: [PHP] Evaluate PHP var in stored sql statement?

[Maxim Maletsky]

eval takes the string and executes PHP in that string. So,

eval("\$myrow[2] = \"$myrow[2]\";");

is equivalent to:

$myrow[2] = "$myrow[2]";

I'd reccomend you using single quotes and then escape only backslashes
and single quotes within a string

--
Maxim Maletsky
[EMAIL PROTECTED]



"Noah" <[EMAIL PROTECTED]> wrote... :

> Don't quite understand why this works, but it does:
> 
> /* Escape quotes in sql_query so our header_id gets evaluated */
> eval("\$myrow[2] = \"$myrow[2]\";");
> 
> What's the deal with the leading backslash in ' eval("\$myrow[2] '?
> 
> Does eval() tell php to evaluate the string as php code; i.e. text without
> quotes?
> 
> Thanks for the help...
> 
> --Noah
> 
> 
> - Original Message -
> From: "Maxim Maletsky" <[EMAIL PROTECTED]>
> To: "CF High" <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Friday, January 24, 2003 9:59 AM
> Subject: Re: [PHP] Evaluate PHP var in stored sql statement?
> 
> 
> >
> > use eval()
> >
> > www.php.net/eval
> >
> >
> > --
> > Maxim Maletsky
> > [EMAIL PROTECTED]
> >
> >
> >
> > "CF High" <[EMAIL PROTECTED]> wrote... :
> >
> > > Hey all.
> > >
> > > I need to find out how to get PHP to evaluate a PHP variable stored in
> our
> > > MySql database.
> > >
> > > Currently the variable is being read as a string; i.e. select * from
> table
> > > where column = $myrow[0] -- instead of evaluating $myrow[0] as a
> variable.
> > >
> > >
> > > The SQL statement is retrieved in the first call to dbConnect (see code
> > > below)
> > >
> > >
> > >
> > > dbConnect("SELECT header_id, header, sql_query FROM cat_headers WHERE
> > > section_id = $_REQUEST[section_id] order by header");
> > >
> > >   $set = $result;   /* Give initial result set a unique name */
> > >
> > >   while ($myrow = mysql_fetch_row($set)) {
> > >
> > > echo(Display the header title here);
> > >
> > >
> > >   dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query
> gotten
> > > from the first dbConnect() call  */
> > >
> > >  $set1 = $result;/* Give next result set a unique name
> */
> > >
> > > while ($myrow = mysql_fetch_row($set1)) {
> > >
> > > echo(Display category rows here);
> > >
> > > }
> > >
> > >   }
> > >
> > > --
> > >  Any leads/suggestions much appreciated..
> > >
> > > --Noah
> > >
> > >
> > >
> > > --
> > > PHP General Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> 
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 


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Re: [PHP] Evaluate PHP var in stored sql statement?

[Noah]

Don't quite understand why this works, but it does:

/* Escape quotes in sql_query so our header_id gets evaluated */
eval("\$myrow[2] = \"$myrow[2]\";");

What's the deal with the leading backslash in ' eval("\$myrow[2] '?

Does eval() tell php to evaluate the string as php code; i.e. text without
quotes?

Thanks for the help...

--Noah


- Original Message -
From: "Maxim Maletsky" <[EMAIL PROTECTED]>
To: "CF High" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Friday, January 24, 2003 9:59 AM
Subject: Re: [PHP] Evaluate PHP var in stored sql statement?


>
> use eval()
>
> www.php.net/eval
>
>
> --
> Maxim Maletsky
> [EMAIL PROTECTED]
>
>
>
> "CF High" <[EMAIL PROTECTED]> wrote... :
>
> > Hey all.
> >
> > I need to find out how to get PHP to evaluate a PHP variable stored in
our
> > MySql database.
> >
> > Currently the variable is being read as a string; i.e. select * from
table
> > where column = $myrow[0] -- instead of evaluating $myrow[0] as a
variable.
> >
> >
> > The SQL statement is retrieved in the first call to dbConnect (see code
> > below)
> >
> >
> >
> > dbConnect("SELECT header_id, header, sql_query FROM cat_headers WHERE
> > section_id = $_REQUEST[section_id] order by header");
> >
> >   $set = $result;   /* Give initial result set a unique name */
> >
> >   while ($myrow = mysql_fetch_row($set)) {
> >
> > echo(Display the header title here);
> >
> >
> >   dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query
gotten
> > from the first dbConnect() call  */
> >
> >  $set1 = $result;/* Give next result set a unique name
*/
> >
> > while ($myrow = mysql_fetch_row($set1)) {
> >
> > echo(Display category rows here);
> >
> > }
> >
> >   }
> >
> > --
> >  Any leads/suggestions much appreciated..
> >
> > --Noah
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>


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Re: [PHP] Evaluate PHP var in stored sql statement?

[Maxim Maletsky]

use eval()

www.php.net/eval


--
Maxim Maletsky
[EMAIL PROTECTED]



"CF High" <[EMAIL PROTECTED]> wrote... :

> Hey all.
> 
> I need to find out how to get PHP to evaluate a PHP variable stored in our
> MySql database.
> 
> Currently the variable is being read as a string; i.e. select * from table
> where column = $myrow[0] -- instead of evaluating $myrow[0] as a variable.
> 
> 
> The SQL statement is retrieved in the first call to dbConnect (see code
> below)
> 
> 
> 
> dbConnect("SELECT header_id, header, sql_query FROM cat_headers WHERE
> section_id = $_REQUEST[section_id] order by header");
> 
>   $set = $result;   /* Give initial result set a unique name */
> 
>   while ($myrow = mysql_fetch_row($set)) {
> 
> echo(Display the header title here);
> 
> 
>   dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query gotten
> from the first dbConnect() call  */
> 
>  $set1 = $result;/* Give next result set a unique name */
> 
> while ($myrow = mysql_fetch_row($set1)) {
> 
> echo(Display category rows here);
> 
> }
> 
>   }
> 
> --
>  Any leads/suggestions much appreciated..
> 
> --Noah
> 
> 
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 


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[PHP] Evaluate PHP var in stored sql statement?

[CF High]

Hey all.

I need to find out how to get PHP to evaluate a PHP variable stored in our
MySql database.

Currently the variable is being read as a string; i.e. select * from table
where column = $myrow[0] -- instead of evaluating $myrow[0] as a variable.


The SQL statement is retrieved in the first call to dbConnect (see code
below)



dbConnect("SELECT header_id, header, sql_query FROM cat_headers WHERE
section_id = $_REQUEST[section_id] order by header");

  $set = $result;   /* Give initial result set a unique name */

  while ($myrow = mysql_fetch_row($set)) {

echo(Display the header title here);


  dbConnect($myrow[2]);  /*  $myrow[2] is the sql_query gotten
from the first dbConnect() call  */

 $set1 = $result;/* Give next result set a unique name */

while ($myrow = mysql_fetch_row($set1)) {

echo(Display category rows here);

}

  }

--
 Any leads/suggestions much appreciated..

--Noah



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RE: [PHP] Var question

[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
> 
> 
> 
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 

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Re: [PHP] Var question

[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').


--
Maxim Maletsky
[EMAIL PROTECTED]


www.PHPBeginner.com  // PHP for Beginners
www.maxim.cx // my Home

// my Wish List: ( Get me something! )
http://www.amazon.com/exec/obidos/registry/2IXE7SMI5EDI3



"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
> 
> 
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 


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Re: [PHP] Var question

[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
> 
> 
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 

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[PHP] Var question

[Clint Tredway]

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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[PHP] var-vars question

[heinisch]

Hi folks,
Linux, PHP3

I have the following problem:
I have an object, which creates a form, wherein there are selects, which
names accord to values in the database.

But I didn´t get it to have a variable var, I´m sure it´s just a small 
thing, but I cannot see it
Any suggestions ?
TIA Oliver

echo '';
// Already tried post, no difference
echo '';
while($da=$MKA -> getDbAns()) // This gives back an array named $da 
while there are answers
{

   $xwert=$da[1];

//  $da[1] contains "abc" the var  $xwert will be "abc" 

// now I´d like to see the value of $abc but there´s nothing in

   echo ' '.$$xwert.'';

// set the new select named "abc"
   echo '';

// forget the rest, just for understanding what while does
   for($i=0; $i < 11 ; $i++)
   {
  if(isset($$da[1])) // just for testing
  {
 echo "".$i."";
  }
  else
  {
 echo "".$i."";
  }
   } // end for
   echo '';
   echo '';
   echo ''.$da[1].'';
   echo ''.$da[2].'';
   echo ''.$da[3].' EUR';
   echo '';
} // end while
echo '';
echo '';
echo "";


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Re: [PHP] $var , '$var'

[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.
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
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>


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Re: [PHP] $var , '$var'

[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.
| 
| -- 
| 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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[PHP] $var , '$var'

[nafiseh saberi]

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

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[PHP] Re: javascript var on a php var...

[George Whiffen]

Romeo Manzur wrote:
> 
> hi, I want to know how could I save a javascript variable on a php
> variable???
> Thanks...

It depends how the user will get to the php page: 

1. Form
If the user is about to submit a form and you want some
Javascript variable from
your page to end up as a php variable after the form is
submitted then:

Create a hidden form variable e.g. 
Set this formvariable to your Javascript variable in the
Javascript
e.g.   document.form.myvariable.value =
myjavascriptvariable; 

After submission $myvariable will be a php variable in the
target page with the value you gave it.

2. Link
If you want to set a php variable in a page which the user
will get to by a link,
then you need to add a GET query to the link e.g. your
Javascript will have something like this:
document.myhref.location =
document.myhref.location+"?myvariable="+myjavascriptvariable;
i.e. the link becomes   myoriginallink?myvariable=...
php will automatically become up the value you specify and
set $myvariable to that value.

 
George

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RE: [PHP] javascript var on a php var...

[Adrian Ciutureanu]

window.location = 'http://url?yourVar=' + yourVar;

> -Original Message-
> From: Romeo Manzur [mailto:[EMAIL PROTECTED]]
> Sent: 5 iulie 2001 07:56
> To: [EMAIL PROTECTED]
> Subject: [PHP] javascript var on a php var...
> 
> 
> hi, I want to know how could I save a javascript variable on a php
> variable???
> Thanks...
> 
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
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> To contact the list administrators, e-mail: 
> [EMAIL PROTECTED]
> 
> 

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[PHP] javascript var on a php var...

[Romeo Manzur]

hi, I want to know how could I save a javascript variable on a php
variable???
Thanks...


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Re: [PHP] var question

[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

[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]
> > For additional commands, e-mail: [EMAIL PROTECTED]
> > To contact the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
>
>
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Re: [PHP] var question

[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

[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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[PHP] var question

[Jeroen Geusebroek]

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



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RE: [PHP] VAR and variables

[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

[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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[PHP] VAR and variables

[Matt]

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?

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;


Thanks. 



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