Why did this not give an error?

[jodleren]

This line

while(!adminok && ($i>=0))

was missing a $-sign, should be while(!$adminok && ($i>=0))

Though, it did not give an error, and I do not have any adminok
elsewhere... so how come, that it accepted this, and the programme
failed instead...?

WBR
Sonnich

[Willem Bogaerts]

This line

>
while(!adminok && ($i>=0))

was missing a $-sign, should be while(!$adminok && ($i>=0))

Though, it did not give an error, and I do not have any adminok
elsewhere... so how come, that it accepted this, and the programme
failed instead...?

Probably because error display was off and you did not provide any error
handling. Mind you, not displaying errors is good on a production
machine, as it won't help any hackers by showing internals of your code.

Furthermore, any word that is not defined as a constant and used as a
value is taken as an unquoted string. So your code actually was treated as:

while(!'adminok ' && ($i>=0))

Regards,
--
Willem Bogaerts

Application smith
Kratz B.V.
http://www.kratz.nl/

[jodleren]

On Dec 17, 9:47 am, Willem Bogaerts
<w.bogae...@kra tz.maardanzonde rditstuk.nlwrot e:

This line

while(!adminok && ($i>=0))

was missing a $-sign, should be while(!$adminok && ($i>=0))

Though, it did not give an error, and I do not have any adminok
elsewhere... so how come, that it accepted this, and the programme
failed instead...?

Probably because error display was off and you did not provide any error
handling. Mind you, not displaying errors is good on a production
machine, as it won't help any hackers by showing internals of your code.

Error handling is on...

Furthermore, any word that is not defined as a constant and used as a
value is taken as an unquoted string. So your code actually was treated as:

while(!'adminok ' && ($i>=0))

But this might explain it.

[Willem Bogaerts]

Error handling is on...

>

>Furthermore, any word that is not defined as a constant and used as a
value is taken as an unquoted string. So your code actually was treated as:

while(!'admino k' && ($i>=0))

But this might explain it.

I think such an unquoted string generates a warning and not an error...

--
Willem Bogaerts

Application smith
Kratz B.V.
http://www.kratz.nl/

[Jonas Werres]

Though, it did not give an error, and I do not have any adminok

elsewhere...

Probably because your error level is misconfigured, which - in my opinion -
means "not E_ALL || E_STRICT".
So the warning(?) that the undefined constant was taken as string has not
been displayed.

[Michael Fesser]

..oO(jodleren)

>Error handling is on...

Set error_reporting to E_ALL|E_STRICT in your php.ini.

Micha