I wanna know how php interprets this code. Is it something of syntax error or matching priorty? I appreciate any help.
Code:
// php 5.3
class Man {
var $arr = array('name'=>'me');
}
$key = 'name';
$man = new Man();
echo $man->arr['name']; // output me
echo $man->$key['name']; // output nothing along with warning and notice
// output:
PHP Warning: Illegal string offset 'name' in php shell code on line 1
PHP Notice: Undefined property: Man::$n in php shell code on line 1
Well, do you mean $key['name'] or $man->$key and ['name'] thereof? It's ambiguous syntax and PHP is interpreting it the former way, trying to get index 'name' of string 'name', resulting in a warning being output and the interpretation being 'name'[0], i.e. 'n'.
Disambiguate the syntax:
$man->{$key}['name']
'name' of string 'n' not string 'name'. When you don't provide brackets it only considers the first character of the string.
['name'] of a string, clearly meaning it's trying $key['name'] where $key is the string. Since 'name' is an invalid/illegal offset of a string, it produces this warning. This is then falling back to interpreting the offset 'name' as the next best numeric offset, which ends up being 0 ((int)'name' → 0). So it results in $key[0]. Which then translates to $man->n and "Undefined property: Man::$n". See 3v4l.org/MFmgf.This doesn't translate to anything because you're trying to access a property with a tag of a variable. It's not a variable, it won't work:
echo $man->$key['name'];
This on the other hand will work:
echo $man->{$key}['name'];
This lets php know the $key you're about to used is actually a property wrapped in {}
There is no name property define in your class. Because echo $man->$key['name']; convert to echo name['name];
class Man {
var $arr = array("name"=>"me");
}
$key = 'arr';
$man = new Man();
echo $man->arr['name']; // output me
echo $man->{$key}['name']; // output nothing along with warning and notice
Look how i used square brackets around $key. When you don't provide brackets it interprets to $key[0] :
echo $man->a but you can not access array values.
$key = 'arr' And $arr already has index name but still php interprets $key to $a. just run my example without brackets and see the results. I hope you get it this time. just tell me if anything is unclear.Because $man is object of class if you want to print some it should be in class try this code :
class Man{
var $arr=array("name"=>"me");
}
$key='name';
$man=new Man;
echo $man->arr[$key]; //output me
$keyisname. There is no propertynameinManclass. That's why there's a warning. Also the above looks like PHP 4 syntax. please try to avoid it, your co-workers will be grateful.Notice: Undefined property: Man::$name): 3v4l.org/o5sgS