Build string to reference object

[Hugh Oxford]

I want to build a string to reference an object.

I can reference is manually thus:

print_r($this->struct->parts[0]->parts[1]);

but if I build a string...

$string = "->parts[0]->parts[1]"

....and try to reference it thus...

print_r($this->struct{$string });


I get Fatal error : Cannot use object of type stdClass as array

Any thoughts?

[Jerry Stuckle]

Re: Build string to reference object

Hugh Oxford wrote:

I want to build a string to reference an object.
>
I can reference is manually thus:
>
print_r($this->struct->parts[0]->parts[1]);
>
but if I build a string...
>
$string = "->parts[0]->parts[1]"
>
...and try to reference it thus...
>
print_r($this->struct{$string });
>
>
I get Fatal error : Cannot use object of type stdClass as array
>
Any thoughts?

True. You can't do it that way. You could use eval() - but that's a
bad thing to use.

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[petersprc]

Re: Build string to reference object

I would say, typically this isn't necessary, as you can pass around a
reference to the nested objects that you want to update or you map
strings to objects.

If you have a large app which requires this type of dynamic path-based
access, you might consider using an XML document data structure and
querying it with XPath queries.

In general, your example could be directly implemented using either
eval:

$str = 'parts[0]->parts[1]';
eval("\$val = \$this->struct->$str;");
print_r($val);

Or with a function getNestedProper ty like so:

<?

function getNestedProper ty($obj, $path)
{
{
$val = $obj;
foreach ($path as $part) {
$val = is_object($val) ? $val->$part :
$val[$part];
}
return $val;
}

class c
{
function c()
{
$this->arr = array('a' =array(1, 2, 3));
}
}

print_r($val = getNestedProper ty(new c,
array('arr', 'a', 1)));

?>

Regards,

John Peters

On May 28, 8:51 am, Hugh Oxford <ares...@fas.co mwrote:

I want to build a string to reference an object.
>
I can reference is manually thus:
>
print_r($this->struct->parts[0]->parts[1]);
>
but if I build a string...
>
$string = "->parts[0]->parts[1]"
>
...and try to reference it thus...
>
print_r($this->struct{$string });
>
I get Fatal error : Cannot use object of type stdClass as array
>
Any thoughts?

[Hugh Oxford]

Re: Build string to reference object

petersprc wrote:

I would say, typically this isn't necessary, as you can pass around a
reference to the nested objects that you want to update or you map
strings to objects.
>
If you have a large app which requires this type of dynamic path-based
access, you might consider using an XML document data structure and
querying it with XPath queries.
>
In general, your example could be directly implemented using either
eval:
>
$str = 'parts[0]->parts[1]';
eval("\$val = \$this->struct->$str;");
print_r($val);
>
Or with a function getNestedProper ty like so:
>
<?
>
function getNestedProper ty($obj, $path)
{
{
$val = $obj;
foreach ($path as $part) {
$val = is_object($val) ? $val->$part :
$val[$part];
}
return $val;
}
>
class c
{
function c()
{
$this->arr = array('a' =array(1, 2, 3));
}
}
>
print_r($val = getNestedProper ty(new c,
array('arr', 'a', 1)));
>
?>
>
Regards,
>
John Peters
>
On May 28, 8:51 am, Hugh Oxford <ares...@fas.co mwrote:

>I want to build a string to reference an object.
>>
>I can reference is manually thus:
>>
>print_r($thi s->struct->parts[0]->parts[1]);
>>
>but if I build a string...
>>
> $string = "->parts[0]->parts[1]"
>>
>...and try to reference it thus...
>>
>print_r($thi s->struct{$string });
>>
>I get Fatal error : Cannot use object of type stdClass as array
>>
>Any thoughts?

>

Thanks Peter and for your other replies elsewhere.