i'm workin on some modification of the xdirectory module and i found a problem with the use of the || (or) logical operator on smarty.

i will be verbose, for all that people don't knowing this module :)
xdirectory pass to smarty templates the 'link' array, containing the two variables involved in the problem.
these variables are $link.logourl (containing the name of the logo image for a business entry) and $link.premium (that select some additional displaying features for a business entry).

if i try to set a logical operator like this, it does not work:



if i set only a 'if $link.premium or a 'if $link.logourl' conditional statement, this work fine, but setting both the variables as conditional, lets work only the $link.premium condition
more clearly, a business entry without a logo will show no image because the img tag will result like this:



viceversa if i set a logical operators like this, it will work:



it seems to me very strange, but maybe there is something of smarty i didn't understand...

anyone can tell me if it is a problem of smarty or of the xoop's smarty implementation (o may be a my mistake? )

Many thanx and sorry for my bad english

Max

Only thing I can see is that you write

and not


If this is just a typo in the post try putting in some brackets like this:

Also be sure to check the smarty doc site for proper use of the || operator, my suggestion would be to use OR instead.

http://smarty.php.net/manual/en/language.function.if.php

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thanks for the answers...

Mithrandir:
yes, the Slink.logourl is a typo in the post, i checked many times the syntax :)
before posting i also tried to separate the condition with bracket, without any result :-/

Draven:
i tried also the use of OR instead of ||, again without any result, and i have read the smarty documentation many times searching for any possible idea.

I will try to reproduce this error in another page containing only the $link array or another array, to see it the error happen again.

I have the same problem with wfchannel_index.html.

This is ok
<{if $article.headline != "Saveurs régionales"}>
<{$article.headline}>
<{/if}>

But this lines are not implemented

<{if substr($article.headline,1,4) != "Menu"}>

<{if $article.headline != "Saveurs régionales" or $article.headline != "Menu de saison"}>

Have you a idea ?

Quote:

This works for me. Note that substr(...,1,4) returns four characters starting at the second character of the string. For example, substr("abcdef",1,4) is "bcde".

Quote:

This also works. But it doesn't make much sense, since the if-clause will always be true. I.e., if A != B, as they are in your case, then the expression (C != A or C != B) is a tautology (always true).

Thousand thanks Dave, it's ok now