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Joined: Apr 2005
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Hoopy frood
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OP
Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
i have issue with combo box which get me a bit frustrated... in short i have this in INIT:
on *:dialog:name:init:0:{
did -a $dname 25 [0] *!user@host.domain
did -a $dname 25 [1] *!*user@host.domain
did -a $dname 25 [2] *!*@host.domain
did -a $dname 25 [3] *!*user@*.domain
did -a $dname 25 [4] *!*@*.domain
did -a $dname 25 [5] nick!user@host.domain
did -a $dname 25 [6] nick!*user@host.domain
did -a $dname 25 [7] nick!*@host.domain
did -a $dname 25 [8] nick!*user@*.domain
did -a $dname 25 [9] nick!*@*.domain
if (!%host.type) { set %host.type 2 | did -c $dname 25 %host.type }
elseif (%host.type == 0) { did -c $dname 25 1 }
...
...
and this in SCLICK
if ($did = 25) { set %host.type $gettok($remove($did(25),[,]),1,32) }
is that even if i CLICK on host type 0 on combo control (combo value 1) it will always next time on INIT put it on host type 1 (combo value 2) and as seen on end of INIT even code, my "chekcup" doesnt work... and if i set anything else (than host type 0) value in combo dissaperas :P so what should i do?
IceCapped
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Joined: Oct 2004
Posts: 8,330
Hoopy frood
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Hoopy frood
Joined: Oct 2004
Posts: 8,330 |
For your init, why not just use: did -c $dname 25 $calc(%host.type + 1) That will save a lot of elseif stuff. The end of your init will look like this:
if (!%host.type) { set %host.type 2 | did -c $dname 25 %host.type }
else { did -c $dname 25 $calc(%host.type + 1) }
Invision Support
#Invision on irc.irchighway.net
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Joined: Apr 2005
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Hoopy frood
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OP
Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
because when i run dialog it always gives +1 +1
so if i last time set hos type to 0, next time i run dialog it
will be set on host type 1, if i set last time host type on 5,
next time i run dialog it will be set on 6...
and i would like dialog to REMEMBER last taken host type
but for reasons that host type 0 is item 1 on combo control
its all fucked up :P
IceCapped
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Joined: Oct 2004
Posts: 8,330
Hoopy frood
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Hoopy frood
Joined: Oct 2004
Posts: 8,330 |
set %host.type $did(25).sel
or
set %host.type $calc($did(25).sel - 1)
Pick from those based on whether or not you want the line number stored or the actual host type stored (the second one will store the host type instead of the line number). If you use the second one, then use the +1 calculation I showed you above for selecting the correct line.
Invision Support
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Hoopy frood
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OP
Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
this is ok, thanks.
just to ask why doesnt mirc do this correct.
for example what ever value i chose (1-10)
for //echo -a $address($me,(%host.type - 1))
i always get only result for host type 0
why?
IceCapped
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Joined: Oct 2004
Posts: 8,330
Hoopy frood
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Hoopy frood
Joined: Oct 2004
Posts: 8,330 |
You need to use $calc if you're doing any calculations (such as %host.type - 1).
Invision Support
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Hoopy frood
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OP
Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
well i tried
//echo -a $address($me,($calc(%host.type - 1)))
and
//echo -a $address($me,$calc(%host.type - 1))
for both i get * /echo: insufficient parameters
:P
IceCapped
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Joined: Jul 2006
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Vogon poet
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Vogon poet
Joined: Jul 2006
Posts: 107 |
For that echo to work from the command line, you have to evaluate the $calc part. //echo -a $address($me,$($calc(%host.type - 1),2)) It may work in a script without the separate evaluation. I can never keep that straight. Also, you'd have to be online. From the help file: If the Internal Address List doesn't contain a matching nickname, the identifier returns $null.
LonDart
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Joined: Apr 2005
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Hoopy frood
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Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
aww nice, that works.
just to ask why is number 2 in the end of this "code" ?
),2))
Last edited by raZOR; 13/02/07 04:36 AM.
IceCapped
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Posts: 400
Fjord artisan
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Fjord artisan
Joined: Apr 2006
Posts: 400 |
it's all still under $address, so it's $address(......,2) which shows in *!*@ADDRESS.COM
-Kurdish_Assass1n
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Hoopy frood
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Hoopy frood
Joined: Apr 2005
Posts: 1,009 |
no it doesnt
i tested it with different host types and it returns correct ones
IceCapped
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Joined: Oct 2004
Posts: 8,330
Hoopy frood
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Hoopy frood
Joined: Oct 2004
Posts: 8,330 |
The ,2 is part of the $() and causes a double evaluation.
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Hoopy frood
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Hoopy frood
Joined: Sep 2005
Posts: 2,881 |
For that echo to work from the command line, you have to evaluate the $calc part.
You don't. Reasons why $address may return $null is that you're not connected or not on a channel, but you certainly don't need to use $() here.
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