text Q - GameCreators Forum

Author

Message

t10dimensional

16

Years of Service

User Offline

Joined: 22nd Mar 2009

Location: Code Cave, USA

Posted: 12th Jun 2009 10:30 Edited at: 12th Jun 2009 10:51

Link

How do you grab a word out of a sentance?

or how would you search for a word in a sentance?

RND Q

what are the snapshots for?can I delete them

If at first you don't succeed-Pause-Go to last checkpoint

Back to top

Profile PM

BN2 Productions

21

Years of Service

User Offline

Joined: 22nd Jan 2004

Location:

Posted: 12th Jun 2009 11:54

Link

Quote: "what are the snapshots for?can I delete them"

They are just screen shots, they aren't needed.

Quote: "How do you grab a word out of a sentance?"

There are many ways. For a text parser system a year or two ago, I wrote a system that would go through a string character by character using MID$() and identified where the spaces were. Then it would load the strings between the spaces into an array using a combination of Left$() and RIGHT$(). Then you could just pick whatever word you need. My system would check each one against lists to identify if they were important, looking for verbs, correct words, nouns etc.

Great Quote:
"Time...LINE??? Time isn't made out of lines...it is made out of circles. That is why clocks are round!" -Caboose

Back to top

Profile PM

That1Smart Guy

16

Years of Service

User Offline

Joined: 26th Feb 2009

Location: Somewhere...... yep

Posted: 12th Jun 2009 16:05

Link

ya I did something similar with an equation style calculator, you would enter an equation like you would on a regular calculator, it would cycle through and find any parenthesis or operations using mid$() then compute the result

There are only 10 kinds of people in the world, those who understand binary and those who dont

Back to top

Profile PM Email

t10dimensional

16

Years of Service

User Offline

Joined: 22nd Mar 2009

Location: Code Cave, USA

Posted: 12th Jun 2009 20:54

Link

Thanks,

I'll mess around with it.If I can't figure it out I'll ask.

If at first you don't succeed-Pause-Go to last checkpoint

Back to top

Profile PM

t10dimensional

16

Years of Service

User Offline

Joined: 22nd Mar 2009

Location: Code Cave, USA

Posted: 12th Jun 2009 22:15 Edited at: 12th Jun 2009 22:19

Link

This is what I got so far I'm glad I could work it out on my own

it was fun.I could probally shorten it somehow.Tell me if you see

anything I could make better.

a=1
y=1
z=1
dim line$(5)
dim word$(100)
dim verb$(10)
dim object$(10)
dim letter(20)
dim l$(20)

verb$(1)="run"
verb$(2)="walk"
verb$(3)="use"
verb$(4)="find"
verb$(5)="travle"
verb$(6)="attack"
verb$(7)="ask"
verb$(8)="jump"
verb$(9)="drop"
verb$(10)="defend"

object$(1)="sword"
object$(2)="horse"
object$(3)="armor"
object$(4)="potion"
object$(5)="elixir"
object$(6)="book"
object$(7)="bow"
object$(8)="river"
object$(9)="item"
object$(10)="sheild"

text 1,20, "enter your sentance"
input a$
line$(1)=a$
cls

for x=1 to len(line$(y))
   letter(z)=asc(lower$(mid$(line$(y),x)))
   if letter(z)=32 or x=len(line$(y)) then gosub word:z=0
   z=z+1 
  next x

for e=1 to 20
 print word$(e)
next e

wait key

print word$(1)+word$(2)+word$(3)+word$(4)+word$(5)+word$(6)+word$(7)

wait key

End

word:
for b=1 to z
 l$(b)=chr$(letter(b))
next b

word$(a)=l$(1)+l$(2)+l$(3)+l$(4)+l$(5)+l$(6)+l$(7)+l$(8)+l$(9)+l$(10)+l$(11)+l$(12)+l$(13)+l$(14)+l$(15)+l$(16)+l$(17)+l$(18)+l$(19)+l$(20)

for w=1 to 20
 l$(w)=""
next w

for w=1 to 20
 letter(w)=0
next w

a=a+1
return

+ Code Snippet

a=1
y=1
z=1
dim line$(5)
dim word$(100)
dim verb$(10)
dim object$(10)
dim letter(20)
dim l$(20)

verb$(1)="run"
verb$(2)="walk"
verb$(3)="use"
verb$(4)="find"
verb$(5)="travle"
verb$(6)="attack"
verb$(7)="ask"
verb$(8)="jump"
verb$(9)="drop"
verb$(10)="defend"

object$(1)="sword"
object$(2)="horse"
object$(3)="armor"
object$(4)="potion"
object$(5)="elixir"
object$(6)="book"
object$(7)="bow"
object$(8)="river"
object$(9)="item"
object$(10)="sheild"

text 1,20, "enter your sentance"
input a$
line$(1)=a$
cls

  for x=1 to len(line$(y))
   letter(z)=asc(lower$(mid$(line$(y),x)))
   if letter(z)=32 or x=len(line$(y)) then gosub word:z=0
   z=z+1 
  next x

for e=1 to 20
 print word$(e)
next e

wait key

print word$(1)+word$(2)+word$(3)+word$(4)+word$(5)+word$(6)+word$(7)

wait key

End

word:
for b=1 to z
 l$(b)=chr$(letter(b))
next b

word$(a)=l$(1)+l$(2)+l$(3)+l$(4)+l$(5)+l$(6)+l$(7)+l$(8)+l$(9)+l$(10)+l$(11)+l$(12)+l$(13)+l$(14)+l$(15)+l$(16)+l$(17)+l$(18)+l$(19)+l$(20)

for w=1 to 20
 l$(w)=""
next w

for w=1 to 20
 letter(w)=0
next w

a=a+1
return

If at first you don't succeed-Pause-Go to last checkpoint

Back to top

Profile PM

That1Smart Guy

16

Years of Service

User Offline

Joined: 26th Feb 2009

Location: Somewhere...... yep

Posted: 12th Jun 2009 22:25

Link

i havent tested it but as i was looking thru i realized something small you could do make reading the code easier

you have:

+ Code Snippet

word$(a)=l$(1)+l$(2)+l$(3)+l$(4)+l$(5)+l$(6)+l$(7)+l$(8)+l$(9)+l$(10)+l$(11)+l$(12)+l$(13)+l$(14)+l$(15)+l$(16)+l$(17)+l$(18)+l$(19)+l$(20)

but i was thinking you could shorten that to:

+ Code Snippet

for num=1 to 20
    word$(a)=word$(a)+l$(num)
next num

adding strings basically adds the second (l$()) to the end of the first (wordS(a))

so as long as at the start of the loop word$(a)="" then this will work

There are only 10 kinds of people in the world, those who understand binary and those who dont

Back to top

Profile PM Email

t10dimensional

16

Years of Service

User Offline

Joined: 22nd Mar 2009

Location: Code Cave, USA

Posted: 12th Jun 2009 22:42

Link

thanks,
did'nt think about that.I thought thier would be a better way to do that.

If at first you don't succeed-Pause-Go to last checkpoint

Back to top

Profile PM

Sorry your browser is not supported!

DarkBASIC Discussion / text Q