PHP 4.3.8 and variables - help!

ppowell

 $stuff .= 'Hello World'; // BREAKS IN PHP 4.3.8 WHY???
 $data = array('hello', 'world');
 echo $data[2]; // ALSO BREAKS IN PHP 4.3.8

I do not understand why, all of a sudden, the following lines are breaking when I moved these lines from a PHP 4.3.2 environment, working just fine and dandy, suddenly spewing "undefined offset, undefined variable, undefined everything else" errors all over the place using PHP 4.3.8.

Is there a setting I can use to switch this off? I'll have to change several hundred lines of code otherwise.

Thanx
Phil

tsinka

Hi,

set error_reporting to E_ALL & ~E_WARNING in php.ini

The problem here is that $stuff wasn't defined before and you want to concatenate a string to this variable. The array only has indexes 0 and 1. This is also a reason for such a warning.

The better way would be to check your code why these warnings appear.

Thomas

ppowell

That's just it, I can't change php.ini as I do not have permissions to do so.

I see the warnings I am getting, but I don't understand why all of a sudden this is happening.

In several environments I've worked in from PHP 4.1.2 to 4.3.2 there has never been a problem with concatenating a string to a non-existent variable, but if I have to instantiate variables every time now in PHP 4.3.8 that would require for me to change literally hundreds of lines of code because I've gotten away with it for years!

Phil

tsinka

Hmmm ... there might be a solution.

You can modify this setting by placing e.g.

php_value error_reporting 1535
or
php_value error_reporting 511

in a .htaccess file (apache) placed into DOCUMENT_ROOT

This can also be done with registry entries when running IIS.

You might need to play around with the values a little bit. read about this here

Thomas

ppowell

Thanx.. I am having to learn all about "+=" and ".=" and variables and objects and, frankly, none of it makes an ounce of sense.

I don't understand why you can do ".=" to a nonexistent variable and then you can't do ".=" to a nonexistent variable in PHP 4.3.8.

Phil

tsinka

It doesn't really depend on the PHP version. It depends on the settings. The default setting might have changed or something like that. You can change the setting of former PHP versions so that they give you that error messages, too.

The best way to get around that is to always have error_reporting set to E_ALL while developing applications. This will make sure that the application will not produce any warnings regardless of the PHP settings.

Thomas

ppowell

I am not sure how to locally change default settings as I do not know what they are. Oh well my head hurts too much at this point.

Thanx
Phil

tsinka

Ok,

which os/web server do you have on that server ? UNIX/Linux or Windows ... Apache or IIS ?

The default settings are not very important in this case ... as long as you're allowed/able to change the settings locally. How to change them depends on the OS/web server version(s).

Weedpacket

You can use error_reporting() on every page you don't want to fix.

I don't understand why you can do ".=" to a nonexistent variable and then you can't do ".=" to a nonexistent variable in PHP 4.3.8.
It's nothing to do with the version of PHP. It's everything to do with the error reporting level.