print "$counter times 2 is ".($counter*2)."<br>"

ok; sirkent and all the php experts out there!!!

I need some help with this statement which I read in a php code sample...

From what I know, print prints what is in the parentheses " and ".

The confusing part is the .( and the ).

Do the .( and ). tell php that what is inside has to be evaluated ( ie as an expression )???

It seems that its somethng like this...

"print without evaluating here" .(evaluatethecodeandprintoutput)."continue printing without evaluating as xpression"

Ok, PHP is capable of printing variables from within strings and from without, namely because it uses the dollar sign to denote them. But if you were to print the whole statement:

print "$counter times 2 is ($counter*2)
"

in PHP without leaving ($counter*2) outside of the quotes, then you would print:

3 times 2 is 3*2

(for example)

Thus, ($counter*2) needs to be done seperately. But when it is evaluated, you want it to be part of the string you are printing. Thus you need to concatenate your values into one string. The full stop (or period) sign is PHP's concatenation operator. In most other languages it is + or & or something similiar. PHP distinguishes between addition + and concatenation . quite nicely.

So, to summarise (!):

print "$counter times 2 is ".($counter*2)."
"

prints, $counter times 2 is (the result of ($counter*2) added to the string)


Hope this clears that up!

Yes! Thanks . That cleared pretty much everything except the bracket ( or the parentheses ).. Anything that is inside a bracket is to be calculated??? Actually, I've heard of the concatenation operator somewhere but didn't bother to read further

Yes, anything inside a bracket is done first. In this case, the bracket is used partly for that and also because they are both numbers. The bracket lets PHP know that, if the two are of the same type, then it can carry out an operation, in this case, addition.

you mean multiplication?

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Errrm, yes. Sorry!

thanks a lot sirkent! here's your shrine!