Checking past c->final (somewhat)

[Dark Nexus]

Arlgitht, i think you have to parse throught string or something to figure it out, but how would i do

 

/!bleh bloh blah

 

how would it recognize blah?

[50% Packetloss]

c->final is the "bloh blah" part of your example. Its the input after the command. The simple way to seperate them would be to

String Blah=c->final; //string object found in datatypes.h

Blah.split(' '); //split the string where you find the 1st ' ' (space) charater

 

Thus the string would now be seperated, and Blah.msg=="blah"

If you wanted to get the left side of the string too then do

String Bloh=Blah.split(' ');

and Bloh.msg=="bloh"

[Dark Nexus]

quick to answer thanx, i thought it would be a little bit more challenging than that. Is there a different way that requieres only temporary variables?

[50% Packetloss]

You can always do the same task by hand. Increment through the charater array until you find the space and then store the rest in another charater array. Doing everything manuelly makes the program much faster but may be more time consuming. Something such as a mervbot plugin should be easy to read for noobie programmers or if you need to look back at your code after a long period of time. Speed isnt a problem with a small merv plugin, especially with the cpu power that is in wide spread use nowadays

[Guest AKA. Versetti]

If you do it that way how would you get the middle all alone without adding the end vaule. Example:

/!test var1 var2

 

now how Would I set var1 into it's onw char and var2 in it's own also. I'll need this info for more then merv bot since I'm Adding a Msg Board too the SSG Biller.

This would help me alot my alias is Versetti but I'm at the schools pc now and don't feel like typing my password in and all that good stuff.

[50% Packetloss]

String Var2=c->final;

String Var1=Var2.split(' ');

 

var1 will have the left side and var2 will have the right side