I want to create a session variable but the session variable name I want to be dynamic. So what I need is the proper syntax for a variable name that is a $_SESSION variable.
I have tried the code below which creates a variable variable name and stored a value to it and it works just fine.
$xvalue = $_SESSION['delivery_id'];
$delivery_string = 'this_delivery_id_' . $xvalue ;
$$delivery_string = $_SESSION['delivery_id'];
echo "Variable name: " . $delivery_string . "<br/>";
echo "Session Variable Value: " . $this_delivery_id_29 . "<br/>";
The above code echos 29 for line 5; the desired result.
So working upon what worked with a variable variable I then just tried to make the variable name a $_SESSION variable name instead.
$value = $_SESSION['delivery_id'];
$xsession = "$_SESSION[\"" . $value . "\"]"; // this gives compilation error so dead stop. I also tried without the escapes and also a dead stop. I also tried escaping the [ and ] and got rid of the compilation error so the code runs but it does not give the desired result.
$$xsession = $_SESSION['delivery_id'];
echo "Variable name: " . $xsession . "<br/>";
echo "Session Variable Value: " . $_SESSION["delivery_id_29"] . "<br/>";
So line 2 of the code is where the problem is.
The issue comes from trying to interpolate the variable a little "too hard," if you escape $_SESSION[$value] you'll be left with the string "$_SESSION[$value]", which is not a valid name for a variable - you're attempting to access the variable as if it were defined like so: $$_SESSION[$value] = 'foo';. What you want to be doing is taking the value of that array element and using that in the variable variable, which needs to be done by referencing it.
Either of the following seem to give the result you are going for; a straight variable variable:
$value = 'foo';
$_SESSION['foo'] = 'bar';
$bar = 'baz';
echo ${$_SESSION[$value]}; //prints baz
One with another step, aiding in making it more clear:
$identifier = $_SESSION[$value];
echo $$identifier //prints baz
I don't understand what you would be storing in this manner, but you may also investigate alternatives to achieve a cleaner, more straight-forward approach. If you clarify your use behind this, maybe someone will be able to suggest an alternative methodology.
Related
So variable variables are existing. Meaning that this is working
$a = 'test';
$$a = 'Hello';
echo ${'test'}; //outputs 'Hello'
But now I've come across some rather strange code using a variable without a name:
function test(&$numRows) {
$numRows = 5;
echo ' -- done test';
}
$value = 0;
test($value);
echo ' -- result is '.$value;
test(${''}); //variable without name
http://ideone.com/gTvayV Code fiddle
Output of this is:
-- done test -- result is 5 -- done test
That means, the code is not crashing.
Now my question is: what exactly happens if $numRows value is changed when the parameter is a variable without name? Will the value be written into nirvana? Is that the PHP variable equivalent to /dev/null?
I wasn't able to find anything specific about this.
Thanks in advance
${''} is a valid variable which name happens to be an empty string. If you have never set it before, it is undefined.
var_dump(isset(${''})); // if you have never set it before, it is undefined.
You don't see any error because you disabled the NOTICE error message.
error_reporting(E_ALL);
ini_set('display_errors', 1);
echo ${''}; // Notice: Undefined variable:
You can set it like this:
${''} = 10;
echo ${''}; // shows 10
Now my question is: what exactly happens if $numRows value is changed
when the parameter is a variable without name?
There's no such thing as a variable without name, an empty string in PHP is a totally valid name.
Maybe I'm wrong, but in PHP, all varibles can be accessed by their names (or more precisely, the string representation of their name), and since an empty string is still a string, it counts as a valid name.
Think about variables like an array key-value pair. You can create an array key with an empty string:
$arr = [];
$arr[''] = 'appul';
var_dump($arr['']); // prints: string(5) "appul"
$arr[''] = 'ponka';
var_dump($arr['']); // prints: string(5) "ponka"
Whenever you access $arr[''], you address the same value.
You can access all variables as a string using the $GLOBAL variable too, so you can examine what happens to your "nameless" variable:
${''} = 'ponka';
var_dump($GLOBALS['']); // prints: string(5) "ponka"
${''} = 'appul';
var_dump($GLOBALS['']); // prints: string(5) "appul"
Will the value be written into nirvana? Is that the PHP variable equivalent to /dev/null? I wasn't able to find anything specific about this.
No, it doesn't go to nirvana, it sits quietly in the global space, and it's a little bit trickier to access it, but otherways, it's a normal variable like any others.
I'm not even sure if what I am trying to do is possible, I have a simple php echo line as below..
<?php echo $T1R[0]['Site']; ?>
This works well but I want to make the "1" in the $T1R to be fluid, is it possible to do something like ..
<?php echo $T + '$row_ColNumC['ColNaumNo']' + R[0]['Site']; ?>
Where the 1 is replaced with the content of ColNaumNo i.e. the returned result might be..
<?php echo $T32R[0]['Site']; ?>
It is possible in PHP. The concept is called "variable variables".
The idea is simple: you generate the variable name you want to use and store it in another variable:
$name = 'T'.$row_ColNumC['ColNaumNo'].'R';
Pay attention to the string concatenation operator. PHP uses a dot (.) for this, not the plus sign (+).
If the value of $row_ColNumc['ColNaumNo'] is 32 then the value stored in variable $name is 'T32R';
You can then prepend the variable $name with an extra $ to use it as the name of another variable (indirection). The code echo($$name); prints the content of variable $T32R (if any).
If the variable $T32R stores an array then the syntax $$name[0] is ambiguous and the parser needs a hint to interpret it. It is well explained in the documentation page (of the variable variables):
In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.
You can do like this
$T1R[0]['Site'] = "test";
$c = 1;
$a = "T".$c."R";
$b = $$a;
echo "<pre>";
print_r($b[0]['Site']);
Or more simpler like this
$T1R[0]['Site'] = "test";
$c = 1;
$a = "T".$c."R";
echo "<pre>";
print_r(${$a}[0]['Site']);
I'm working through a course on PHP and I'm completely stuck on one of the tasks.
The bit i'm stuck on is:
"PalprimeChecker [a function created just for this task] objects have a property called number. This task has two parts. First, assign that property a value of 17...."
The code I've written is returning an error and won't let me progress...
include('class.palprimechecker.php');
$checker = new PalprimeChecker();
$checker->number = '17';
echo "The number " . "$checker";
echo "(is|is not)";
echo " a palprime.";
I'm not sure at all where I'm going wrong with this. Anyone know the correct way to assign this value?
Hope you can help as I'm tearing my hair out!
Thanks!
Modify this :
echo "The number " . "$checker";
By This :
echo "The number " . $checker->number;
You were trying to print the entire object (which isn't possible without creating a method for it, check orangePill's answer for this), what you wanted to do is simply print the number inside the object.
Also note that you don't need to use quotes when assigning numbers. It might cause issues later on. You should simply assign it like this :
$checker->number = 17;
You can also add a __toString method on the PalprimeChecker class.
public function __toString(){
return (string)$this->number;
}
This will allow echo "The number " . $checker; to produce a string.
Be careful with the concatenation.
<?php
include("class.palprimechecker.php");
$checker = new PalprimeChecker;
$checker->number = 17;
echo "The number" . $checker->number . "";
echo "(is|is not)";
echo " a palprime.";
?>
Well, you are assigning a string, and not a number.
$checker->number = 17; //Note that there are no quotes around '17'.
The title may be a little confusing. This is my problem:
I know you can hold a variable name in another variable and then read the content of the first variable. This is what I mean:
$variable = "hello"
$variableholder = 'variable'
echo $$variableholder;
That would print: "hello". Now, I've got a problem with this:
$somearray = array("name"=>"hello");
$variableholder = "somearray['name']"; //or $variableholder = 'somearray[\'name\']';
echo $$variableholder;
That gives me a PHP error (it says $somearray['name'] is an undefined variable). Can you tell me if this is possible and I'm doing something wrong; or this if this is plain impossible, can you give me another solution to do something similar?
Thanks in advance.
For the moment, I could only think of something like this:
<?php
// literal are simple
$literal = "Hello";
$vv = "literal";
echo $$vv . "\n";
// prints "Hello"
// for containers it's not so simple anymore
$container = array("Hello" => "World");
$vv = "container";
$reniatnoc = $$vv;
echo $reniatnoc["Hello"] . "\n";
// prints "World"
?>
The problem here is that (quoting from php: access array value on the fly):
the Grammar of the PHP language only allows subscript notation on the end of variable expressions and not expressions in general, which is how it works in most other languages.
Would PHP allow the subscript notation anywhere, one could write this more dense as
echo $$vv["Hello"]
Side note: I guess using variable variables isn't that sane to use in production.
How about this? (NOTE: variable variables are as bad as goto)
$variablename = 'array';
$key = 'index';
echo $$variablename[$key];
I know this is not exactly reflection, but kind of.
I want to make a debug function that gets a variable and prints a var_dump and the variable name.
Of course, when the programmer writes a call to the function, they already know the variable's name, so they could write something like:
debug( $myvar, 'myvar' );
But I want it to be quick and easy to write, just the function name, the variable, and voilà !
debug( $myvar ); // quicker and easier :)
You can do it by converting the variable to a key/value set before passing it to the function.
function varName($theVar) {
$variableName = key($theVar);
$variableValue = $theVar[$variableName];
echo ('The name of the variable used in the function call was '.$variableName.'<br />');
echo ('The value of the variable used in the function call was '.$variableValue.'<br />');
}
$myVar = 'abc';
varName(compact('myVar'));
Though I don't recommend creating a reference to a nameless variable, function varName(&$theVar) works too.
Since compact() takes the variable name as a string rather than the actual variable, iterating over a list of variable names should be easy.
As to why you would want to do this -- don't ask me but it seems like a lot of people ask the question so here's my solution.
I know I'm answering a 4 year old question but what the hell...
compact() might help you is your friend here!
I made a similar function to quickly dump out info on a few chosen variables into a log for debugging errors and it goes something like this:
function vlog() {
$args = func_get_args();
foreach ($args as $arg) {
global ${$arg};
}
return json_encode(compact($args));
}
I found JSON to be the cleanest and most readable form for these dumps for my logs but you could also use something like print_r() or var_export().
This is how I use it:
$foo = 'Elvis';
$bar = 42;
$obj = new SomeFancyObject();
log('Something went wrong! vars='.vlog('foo', 'bar', 'obj'));
And this would print out like this to the logs:
Something went wrong! vars={"foo":"Elvis","bar":42,"obj":{"nestedProperty1":1, "nestedProperty2":"etc."}}
Word of warning though: This will only work for variables declared in the global scope (so not inside functions or classes. In there you need to evoke compact() directly so it has access to that scope, but that's not really that big of a deal since this vlog() is basically just a shortcut for json_encode(compact('foo', 'bar', 'obj')), saving me 16 keystrokes each time I need it.
Nope, not possible. Sorry.
Not elegantly... BUT YOU COULD FAKE IT!
1) Drink enough to convince yourself this is a good idea (it'll take a lot)
2) Replace all your variables with variable variables:
$a = 10
//becomes
$a = '0a';
$$a = 10;
3) Reference $$a in all your code.
4) When you need to print the variable, print $a and strip out the leading 0.
Addendum: Only do this if you are
Never showing this code to anyone
Never need to change or maintain this code
Are crazy
Not doing this for a job
Look, just never do this, it is a joke
I know this is very very late, but i did it in a different way.
It might honestly be a bit bad for performance, but since it's for debugging it shouldn't be a problem.
I read the file where the function is called, on the line it was called and I cut out the variable name.
function dump($str){
// Get the caller with debug backtrace
$bt = debug_backtrace();
$caller = array_shift($bt);
// Put the file where the function was called in an array, split by lines
$readFileStr = file($caller['file']);
// Read the specific line where the function was called
$lineStr = $readFileStr[$caller['line'] -1];
// Get the variable name (including $) by taking the string between '(' and ')'
$regularOutput = preg_match('/\((.*?)\)/', $lineStr, $output);
$variableName = $output[1];
// echo the var name and in which file and line it was called
echo "var: " . $variableName . " dumped in file: " . $caller['file'] . ' on line: ' . $caller['line'] . '<br>';
// dump the given variable
echo '<pre>' . var_export($str, true) . '</pre>';
}
i've had the same thought before, but if you really think about it, you'll see why this is impossible... presumably your debug function will defined like this: function debug($someVar) { } and there's no way for it to know the original variable was called $myvar.
The absolute best you could do would be to look at something like get_defined_vars() or $_GLOBALS (if it were a global for some reason) and loop through that to find something which matches the value of your variable. This is a very hacky and not very reliable method though. Your original method is the most efficient way.
No, the closer you will get is with get_defined_vars().
EDIT: I was wrong, after reading the user comments on get_defined_vars() it's possible with a little hack:
function ev($variable){
foreach($GLOBALS as $key => $value){
if($variable===$value){
echo '<p>$'.$key.' - '.$value.'</p>';
}
}
}
$lol = 123;
ev($lol); // $lol - 123
Only works for unique variable contents though.
Bit late to the game here, but Mach 13 has an interesting solution: How to get a variable name as a string in PHP
You could use eval:
function debug($variablename)
{
echo ($variablename . ":<br/>");
eval("global $". $variablename . ";");
eval("var_dump($" . $variablename . ");");
}
Usage: debug("myvar") not debug($myvar)
This is late post but I think it is possible now using compact method
so the code would be
$a=1;
$b=2;
$c=3
var_dump(compact('a','b','c'));
the output would be
array (size=3)
'a' => int 1
'b' => int 2
'c' => int 3
where variable name a, b and c are the key
Hope this helps
I believe Alix and nickf are suggesting this:
function debug($variablename)
{
echo ($variablename . ":<br/>");
global $$variablename; // enable scope
var_dump($$variablename);
}
I have tested it and it seems to work just as well as Wagger's code (Thanks Wagger: I have tried so many times to write this and the global variable declaration was my stumbling block)