I have a PHP Object with an attribute having a dollar ($) sign in it.
How do I access the content of this attribute ?
Example :
echo $object->variable; // Ok
echo $object->variable$WithDollar; // Syntax error :-(
With variable variables:
$myVar = 'variable$WithDollar';
echo $object->$myVar;
With curly brackets:
echo $object->{'variable$WithDollar'};
Thanks to your answers, I just found out how I can do that the way I intended :
echo $object->{'variable$WithDollar'}; // works !
I was pretty sure I tried every combination possible before.
I assume you want to access properties with variable names on the fly. For that, try
echo $object->{"variable".$yourVariable}
You don't.
The dollar sign has a special significance in PHP. Although it is possible to bypass the variable substitution in dereferencing class/object properties you NEVER should be doing this.
Don't try to declare variables with a literal '$'.
If you're having to deal with someoneelse's mess - first fix the code they wrote to remove the dollars then go and chop off their fingers.
C.
There are reflection methods that also allow you to construct method and attribute names that may be built by variables or contain special characters. You can use the ReflectionClass::getProperty ( string $name ) method.
$object->getProperty('variable$WithDollar');
Related
I recently saw this line in a PHP piece of code:
$dbObject = json_decode($jsonString);
$dbObject->{'mysql-5.4'}[0]->credentials
What does this mean? In the PHP docs we can read, that
Both square brackets and curly braces can be used interchangeably for accessing array elements (e.g. $array[42] and $array{42} will both do the same thing in the example above).
But how can the Object $dbObject be defined to allow ->{...}[...]access? Is this code kind of unsafe? Which PHP version does allow this?
Did I miss anything in the PHP docs?
It's to enable access to properties which would be invalid syntax as bare literals. Meaning:
$dbObject->mysql-5.4[0]->credentials
This is invalid/ambiguous syntax. To make clear to PHP that mysql-5.4 is a property and not a property minus a float, you need to use the {'..'} syntax.
To be exact, ->{..} enables you to use any expression as the property name. For example:
$dbObject->{ sprintf('%s-%.1f', 'mysql', 5.4) }
The curly brace syntax allows you to use a string literal or variable as a property or method name.
This is useful for several reasons:
(per #Deceze's answer): It allows you to access property names that would otherwise not be valid PHP syntax -- eg if the property name contains a dot or dash, per your example. Typically properties like this would be accessed via a __get() magic method, since they would also be invalid syntax to define as an actual property within the class.
(per #feela's answer): It allows you to use variables to reference your properties. This is similar to the $$ variable variable syntax. It's generally not really very good practice, but is useful in some cases.
(not mentioned by any other answers yet): It allows you to remove certain potential ambiguities from your code. Consider the following code for example:
$output = $foo->$bar['baz'];
This is ambiguous. You can resolve the ambiguity by adding braces or brackets:
$output = $foo->{$bar['baz']}; //gets the value of a property from $foo which has a name that is defined in $bar['baz']
$output = ($foo->$bar)['baz']; //gets the ['baz'] element of $foo->$bar array.
This is particularly important because the forthcoming relese of PHP 7 will change the default behaviour of code like like. This will make the language behave more consistently, but will break existing code that doesn't have the braces.
See also the PHP documentation for this change here: https://wiki.php.net/rfc/uniform_variable_syntax
But even without the language change to force the issue, code like this should have braces anyway to help with readability -- if you don't put braces, then you may end up struggling to work out what you were trying to do when you come back to the code in six months' time.
The curly braces syntax in the example is not from the array syntax, but comes from a possible syntax to access variables with variable names.
Curly braces may also be used, to clearly delimit the property name.
They are most useful when accessing values within a property that
contains an array, when the property name is made of mulitple parts,
or when the property name contains characters that are not otherwise
valid (e.g. from json_decode() or SimpleXML).
Examples from the PHP docs:
echo $foo->{$baz[1]} . "\n";
echo $foo->{$start . $end} . "\n";
echo $foo->{$arr[1]} . "\n";
See “PHP Variable variables”: http://php.net/manual/en/language.variables.variable.php
(More examples on the usage there…)
As an addition to #deceze's answer:
The -> operator means that you are accessing an object property. In this case the property name should be mysql-5.4 which is not valid PHP identifier (it must contain letters, numbers or underscores, see here). So it is 100% sure that the property name has been created dynamically.
PHP allows overloading properties using magic method called __get(). The body of this method allows you to handle any property you would wish - this can be any string, or any variable, even an object, which is cast to a string.
So in your case somebody created a class with a __get() magic method that handles string mysql-5.4. The curly braces { and } are given to denote that the string should be treated as property name (however there is no property with this name).
most of the time there is a $ after ->{ so it's like ->$variable
for example we have:
$languageCode = 'en';
$obj->{$languageCode};
$obj->{$languageCode} it is equivalent to $obj->en
============
in your case it's because it's not allowed in php write like:
$dbObject->mysql-5.4
because there is a - and there is a . in the mysql-5.4 and - and . makes a word to two words.
so you have to write it like this:
$dbObject->{'mysql-5.4'}[0]->credentials
I'm trying to modify the name of the key dynamically based on which rows are being fetched, but my syntax seems to be slightly off within the query. After moving the quotes around more times than I care to admit, I finally decided it was time to ask for help ;-)
$var = '$foo_row';
$MAX_5A = ${$var . '["MAX_5A"]'};
Instead of
$MAX_5A = $foo_row['MAX_5A'];
Bonus points if someone wants to explain to me the logic behind the correct syntax :-)
This should work for you:
(Just use variable variables with curly quotes to make sure PHP doesn't think this: ${$var["MAX_5A"]}. Also note I removed the dollar sign in the string)
$var = 'foo_row';
//^ dollar sign removed
$MAX_5A = ${$var}["MAX_5A"];
$var = 'foo';
$bar = 'var';
echo $$bar; // foo
Logic: A variable variable takes the value of a variable and treats that as the name of a variable.
i have this line here.. it gives me an error..
Could you please take a look at this?
Thanks
$slideshow-auto2=$this->params->get("slideshow-auto2");
Invalid variable name:
$slideshow-auto2=$this->params->get("slideshow-auto2");
^---can't have this in a var name.
You're trying to do (from PHP's view), $slideshow minus constant "auto2" equals ...
I think you're missing a >:
$slideshow->auto2=$this->params->get("slideshow-auto2");
// ^ Right here
$slideshow-auto2 is not a valid variable name. You can't have hyphens in a variable name (PHP sees it as a minus).
Most of the other answers are guessing that you intended to use the -> syntax. If $slideshow is an object and auto2 is a property of that object, then this is what you want.
However, given the context of the rest of your line of code, my guess is that you want to have an actual variable named $slideshow-auto2. Unfortunately, this just isn't allowed. You'll need to work around it. You could name your variable $slideshowAuto2 or $slideshow_auto2 or various other alternatives, but not $slideshow-auto2.
You're trying to substract a property from an object, I guess you want to access that property so add a '>'
$slideshow->auto2=$this->params->get("slideshow-auto2");
Are you trying to use a hyphen within a variable name? That won't work because it's being interpreted as a minus sign and subtracting a property from an object does not work. You probably want something like this instead:
$slideshow->auto2=$this->params->get("slideshow-auto2");
Edit:
If you don't intend to access the property 'auto2', simply replace the hyphen with a valid character for a variable name.
I understand what the following line does but i don't understand how the brackets are used? I have always used brackets in an if, while and other statements but i have never used them in this fashion.
Are there rules to using them this way, should i not use them in this way? Any help would be appreciated... Thanks
${$key} = $temp;
In that specific case, there is effectively no difference between using brackets and not.
So your code is equivalent to the following:
$$key = $temp;
The brackets are typically used to force PHP to interpolate variables in strings, which isn't necessary in this case.
Using the brackets is very helpful for reducing ambiguity in a statement using array indices:
${$array[0]} = $temp;
As opposed to
$$array[0] = $temp;
The parser will think that you meant ($$array)[0], not $($array[0])
Have a look at:
Variable variables
Today I learned about PHP variable variables; "variable variable takes the value of a variable and treats that as the name of a variable". Also, variable
It seems to be using variable variables.
Otherwise, braces like that are usually used for variable interpolation in strings, where the variable is an object property or array member.
In the example above, they are not necessary. It also seems you can subscript an array with variable variables, with no issues.
I get the basics of variable variables, but I saw a syntax just know, which bogles my mind a bit.
$this->{$toShow}();
I don't really see what those {} symbols are doing there. Do they have any special meaning?
PHP's variable parser isn't greedy. The {} are used to indicate what should be considered part of a variable reference and what isn't. Consider this:
$arr = array();
$arr[3] = array();
$arr[3][4] = 'Hi there';
echo "$arr[3][4]";
Notice the double quotes. You'd expect this to output Hi there, but you actually end up seeing Array[4]. This is due to the non-greediness of the parser. It will check for only ONE level of array indexing while interpolating variables into the string, so what it really saw was this:
echo $arr[3], "[4]";
But, doing
echo "{$arr[3][4]}";
forces PHP to treat everything inside the braces as a variable reference, and you end up with the expected Hi there.
They tell the parser, where a variable name starts and ends. In this particular case it might not be needed, but consider this example:
$this->$toShow[0]
What should the parser do? Is $toShow an array or $this->$toShow ? In this case, the variable is resolved first and the array index is applied to the resulting property.
So if you actually want to access $toShow[0], you have to write:
$this->{$toShow[0]}
These curly braces can be used to use expressions to specify the variable identifier instead of just a variable’s value:
$var = 'foo';
echo ${$var.'bar'}; // echoes the value of $foobar
echo $$var.'bar'; // echoes the value of $foo concatenated with "bar"
$this->{$toShow}();
Break it down as below:
First this is a object oriented programming style as you got to see $this and ->. Second, {$toShow}() is a method(function) as you can see the () brackets.
So now {$toShow}() should somehow be parsed to a name like 'compute()'. And, $toShow is just a variable which might hold a possible function name. But the question remains, why is {} used around.
The reason is {} brackets substitues the value in the place. To clarify,
$toShow="compute";
so,
{$toShow}(); //is equivalent to compute();
but this is not true:
$toShow(); //this is wrong as a variablename is not a legal function name