Is there a way to debugging PHP to trap errors on property of non-object?

Question

I'm using the Google Map API V3 sending it the Lat/Lng and it returns the full address. I'm parsing the address that's returned by using the Object. It works for many addresses but for some of them it is generating the PHP error of about the property being a non-object:

PHP Notice:  Trying to get property of non-object in get5.php on line 47
PHP Notice:  Trying to get property of non-object in get5.php on line 48
PHP Notice:  Trying to get property of non-object in get5.php on line 49

Here are the lines of code for 47-49:

$city_google_api = $country->AdministrativeArea->Locality->LocalityName;
$state_google_api = $country->AdministrativeArea->AdministrativeAreaName;
$zip_code_google_api = $country->AdministrativeArea->Locality->PostalCode->PostalCodeNumber;

I would like to debug this further, but I don't know which record this is causing these messages regarding the property of non-object. Is there a way to trap on the error or check for "property of non-object" so that I can dump the variables? I don't want to dump this for each record because that's too much output. In general, I would like to know if there is a way to trap errors to dump variables? Thanks!

Lightness Races in Orbit · Accepted Answer · 2012-10-14 22:44:25Z

Take:

$city_google_api = $country->AdministrativeArea->Locality->LocalityName;

and temporarily re-write it:

$area            = $country->AdministrativeArea;
$locality        = $area->Locality;
$city_google_api = $locality->LocalityName;

Do the same for the other three lines and, presto, the line number in the error will tell you precisely where the problem is.

Divide and conquer — your debug technique friend.

You could also program defensively from the start so that the error is not just better reported, but "caught" at runtime:

if (!is_object($country))
   throw SomeException();

$area = $country->AdministrativeArea;

if (!is_object($area))
   throw SomeException();

$locality = $area->Locality;

if (!is_object($locality))
   throw SomeException();

$city_google_api = $locality->LocalityName;

or go further and examine the actual type of each object. This is usually overkill, though — once you have fixed your code, you will know what type each variable has, and the original three lines will suffice.

goat · Accepted Answer · 2012-10-14 22:44:23Z

0

if (!is_object($country)) {
    //debug
}
if (!is_object($country->AdministrativeArea)) {
    //debug
}
//etc...

answered Oct 14, 2012 at 22:44

goat

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Comments

Community · Accepted Answer · 2023-11-17 19:32:19Z

0

Actually there is not, but you can check if

an object has a certain property using property_exists()
check if it actually is an object using is_object()
or has a certain type using instanceof.

edited Nov 17, 2023 at 19:32

CommunityBot

11 silver badge

answered Oct 14, 2012 at 22:44

Sven

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