Hey guys I can't seem to get the syntax here right, I'm just trying to see if type exists,
I've tried
if($obj->type) /* do something */
if($obj[0]->type) /* do something */
if($obj->type[0]) /* do something */
On this
stdClass::__set_state(
array(
'0' =>
stdClass::__set_state(
array(
'report_number' => '555',
'type' => 'citrus',
'region' => 'Seattle',
'customer_number' => '9757',
'customer' => 'The Name',
'location' => 'West Seattle, WA',
'shipper' => 'Yamato Transport',
'po' => '33215',
'commodity' => 'RARE',
'label' => 'PRODUCE',
)),
))
But just can't seem to get it right, I believe it has something to do with
[0] being an int instead of varchar but I have no idea....
The output in your question is created by var_export
which represents the data of the object as an array
. But don't get mislead by that, it's in fact an object.
Object properties are accessed by this syntax:
$obj->property
In your case the property is named 0
which is hard for PHP to deal with in plain code as it's not a property name you could write right away like this:
$obj->0
Instead, you need to tell the PHP parser what the name is by enclosing it into {}
:
$obj->{0}
You can then traverse further on to access the type
property of 0
:
$obj->{0}->type
Hope this is helpful, the question comes up from time to time, this is a related one with more related links: How do I access this object property?.
The PHP Manual has this documented, but not very obvious on the Variable variables entry:
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.
Have you tried:
<?php
if (property_exists($obj, 'type')) {
// Do something
}
?>