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BaFe
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BaFe
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/help says: $reptok(text,token,new,N,C) Replaces the Nth matching token in text with a new token. What about if there are more than 1 words that match the "token", and I want to replace everyone? Nice upgrade would be something like if the N is 0 or null, it would replace every word that matches. (like $replace) while loop is'nt a good trick, it might be infinite in some case. If I'm blind or stupid and there is already something like this; please let me know 
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Joined: Jan 2003
Posts: 2,125
Hoopy frood
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Hoopy frood
Joined: Jan 2003
Posts: 2,125 |
I totally agree with the suggestion, however, a while loop workaround is perfectly possible and doesn't lead to infinite loops. Here's a scripted version that works fine: alias repalltok {
var %i = $findtok($1,$2,0,$4), %a = $1
while %i {
%a = $reptok(%a,$2,$3,1,$4)
dec %i
}
return %a
}Syntax is $repalltok(text,token,new,C)
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BaFe
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BaFe
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hehe, thx.. script you gave kinda works.. Thats right, no infinite loops when using DEC no INC (you clever man) this still bugs a bit eg: $repalltok(one hehe this bugs one really one,one,ONE ONE,32) So it would surely be nice if Kamek could update the $reptok Thx for your help qwerty o/
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Joined: Jan 2003
Posts: 2,973
Hoopy frood
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Hoopy frood
Joined: Jan 2003
Posts: 2,973 |
I'm sure an integrated version wouldn't do any better than qwerty's example. You are afterall adding another token into the sequence when checking. The intent is to replace one token with another token, not a series of them.
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Joined: Jan 2003
Posts: 2,125
Hoopy frood
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Hoopy frood
Joined: Jan 2003
Posts: 2,125 |
Ah, you're right, it was a mistake on my part, which is easily fixed though: alias repalltok {
var %i = $findtok($1,$2,0,$4), %a = $1
while %i {
%a = $reptok(%a,$2,$3,[color:red]%i[/color],$4)
dec %i
}
return %a
}Here's another alternative that doesn't use loops at all (which makes it faster if the string contains many replaceable tokens) alias repalltok2 {
var %x = \x $+ $base($4,10,16,2)
!.echo -q $regsub($gettok($1,1-,$4),/(?<=^| %x )\Q $+ $&
$replacexcs($2,\E,\E\\E\Q) $+ \E(?=$| %x )/gix,$replace($3,\,\\,$,\$),%x)
return %x
}
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BaFe
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BaFe
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ok, points to you ;> however, i should have figured the missing %i by myself  thx again, that'll do!
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