handle.c

#include "handle.h"

void handle_class_statement(char*user_subject, char*user_class) {
   /*
   example: a cat is an animal(value)

   Posibilities:
                                      db action      user action
       --------------------------     ------------   --------------------------
   2   cat is unknown       	        add to db     "I'll take note of that"
   1   animal is unknown                             "what's an animal"
   3   cat is known, class = animal                  "I know"
   4   cat known, class != animal                    "no it's a ____"
   5   info is unknown. Add to db

   // 1 class doesn't exist  	// i dont know what an animal is
   // 2 subject doesnt exist, class exists  	// ill take note of that
   // 3 subject exists, class is the same  	// "i knew that"
   // 4 subject exists, class is different  	// no its not
   */

   char value[20];
   char key[20];
   int result;

   // 1 is "animal" in the database?
   if(db_check(user_class) != FOUND) {
      printf("I'm not familiar with %s\n", user_class);
      return;
   }

   //  preparatiom for 2, 3 & 4
//  ex: lookup "cat"
   sprintf(key, "%s > class", user_subject);
   result = db_lookup(key, value);

//  2)
   if(result == NOT_FOUND) {
      sprintf(key, "%s > class", user_subject);
      db_add_pair(key, user_class);
      printf("I'll take a note of that\n");
      return;
   }

//  Note: At this point, "cat" was found in the database.
//  We just need to check if the info matches or not.

//  3) states what already is known
   if(strcmp(user_class, value) == 0) {
      printf("I already know that\n");
      return;
   }

//  4) contradictory info
   else {
      printf("no, it's a %s\n", value);
      return;
   }
}

//--------------------------------------------------
void handle_color_statement(char* user_subject,char* user_color) {
   /*
   example: grass is green

   Posibilities:
   1   grass is unknown                "I dont know what grass is"
   2   grass can't have a color
   3   grass is known. no color is listed
   4   grass is known and already associated with the given color
   5   grass is known, but a different color is listed

   */

   char key[20];
   char value1[20];
   char value2[20];
   int subject_result;
   char user_class;
   int color_result;

   if(DEBUG) printf("handle_color  ");

   // 1 is "grass" in the database?
   if(db_check(user_subject) != FOUND) {
      printf("I'm not familiar with %s\n", user_subject);
      return;
   }

   // 2 is "grass" something that can have a color?
   if(db_root_check(user_subject, "substance")!=MATCH && db_root_check(user_subject, "object") != MATCH) {
      printf("I don't think %s can have a color\r\n", user_subject);
      return;
   }

   // 3 is grass already associated with that color? if not, add it
   sprintf(key, "%s > color", user_subject);
   color_result = db_lookup(key, value1);
   if(color_result == NOT_FOUND) {
      sprintf(key, "%s > color", user_subject);
      db_add_pair(key, user_color);
      printf("I'll take a note of that\n");
      return;
   }

   // 4 if already associated with the given color
   if(strcmp(user_color, value1) == 0) {
      printf("I already know that\n");
      return;
   }

   // 5 if associated with a different color
   else printf("no, it's %s\n", value1);
}
//--------------------------------------------------

void handle_class_question(char* subject) {

   int result;
   char value[20];
   char key[20];

   sprintf(key, "%s > class", subject);
   result = db_lookup(key, value);
   if(result==FOUND) printf("It's a %s\n", value);
   else printf("I've never heard of ""%s""\n", subject);

}

//------------------------------------------------------

void handle_color_question(char* subject) {
   int result;
   char value[20];
   char key[60];

   sprintf(key, "%s > color", subject);
   result = db_lookup(key, value);
   if(result == FOUND) {
      printf("%s\n", value);
      return;
   } else {
      printf("I don't know\n");
      return;
   }

   //else if(result==3)printf("I've never heard of %s\n",key);
// else printf(" I don't know (result %d)\n", result);
}

//----------------------------------------------------

void handle_color_confirmation_question(char* subject, char*value1) {

   int result;
   char value[20];
   char key[60];

   sprintf(key, "%s > color", subject);
   db_lookup(key, value);
   result = strcmp(value, value1);
   if (result == 0) {
      printf("yes\n");
   } else printf("no\n");
}

//----------------------------------------------------

void handle_rating_statement(char* key, char* s2, char* rating) {
//
// in: firstname, firstname or object or substance, rating:0-9
//
//  examples:
//  i like beer
//     1) i'll take note of that
//     2) what is beer?
//     3) i already knew that
//     4) but I thought you hated beer
//  i like jane
//     1) rating added
//     2) i know nothing of this "jane" you speak of
//  jane likes beer
//     1) rating added
//     2) i know nothing of this "jane" you speak of
//     3) i know nothing of this "beer" you speak of
//     4)
//
// Problem: will make multiple entries

// second parameter
//
//  known person?    add info
//  unknown person?  dont know this person
//  known object?    add info
//  known substance? add info
//  unknown

   char pn[20]="rating-";
// char value[20];
   int result, result2;
   char id1[20],id2[20];
   int person = FALSE;

// So, take example 2. At this point we know that neither of the words have been checked.
// We'll start by seeing if the first word is the first name of someone we know.
// We can use the get id function for that
// If known, take note that this is a person and continue on, if not, send a message and exit

   result = db_get_id(key);

   if(result != FOUND) {

      result = db_root_check(key,"firstname");

      if(result==FOUND) printf("I don't know%s\n", key);
      else printf("%s?\n", key);

      return;
   }
// Next we'll look at the 2nd parameter and see if it's a known person
//
   result = db_get_id(s2);  // a known entity?
   result2 = db_root_check(s2,"firstname");  // a person?

   if(result==FOUND && result2==FOUND) {
      person = TRUE;
      goto add_info;
   }

   if(result!=FOUND && result2==FOUND) {
      printf("I don't know %s\n", s2);
      return;
   }

   //  Check if the 2nd parameter a object or substance?
   result = db_root_check(s2, "object");
   result2 = db_root_check(s2, "substance");

   if(result == NO_MATCH) {
      printf("I don't know what %s is\n", s2);
      return;
   }
   printf("*%d*\n", result);
   if(result != FOUND && result2 != FOUND) {
      printf("I dont see how %s makes sense\n", s2);
      return;
   }
//---
// If we've made if this far we have a valid sentence and we can store
// the info

add_info:

   if(person==TRUE)strcat(pn, id2);
   else strcat(pn, s2);

//  result = db_lookup(id1, pn, value);

//if(result == SUBJECT_NOT_FOUND){
//     db_add_subject_and_property(id1,pn,rating);
//     db_add_property(id1,"firstname",key);
//     printf("name and rating added\n");
//  }
// else
   if(result==5) {

//     db_add_property(id1, pn, rating); // change this
      printf("rating added\n");
   } else if(result == FOUND) printf("I already know that\n");
   else printf("error, result code:%d\n",result);
}

//--------------------------------------------------


void handle_location_question(char* subject) {

   int result;
   char value[20];
   char key[60];

   // DO WE KNOW THE ANSWER?
//    strcpy(key, subject);
//    strcat(key, " > location");
   sprintf(key, "%s > location", subject);
   result = db_lookup(key, value);
   if(result==FOUND) {
      printf("%s\n", value);
      return;
   }

   // DO WE KNOW THE SUBJECT BUT NOT THE ANSWER?
   //strcpy(key, subject);
   //strcat(key, " > class");
   sprintf(key, "%s > class", subject);
   result = db_lookup(key, value);
   if(result==FOUND) {
      printf(" I don't know where %s is\n", subject);
      return;
   }

   // WE DON'T KNOW THE SUBJECT
   printf(" I never heard of %s\n", key);
   return;

}

//-------------------------------------------------------

void handle_ability_question(char* key, char* value) {

   int result;
   char* temp[40];
   char value2[20];

   result = db_root_check(key,"object");
   if(result == NOT_FOUND) {
      printf(" I never heard of %s\n", key);
      return;
   }

   if(result != MATCH) {
      printf("%s is not an object\n", key);
      return;
   }

   if(db_root_check(value,"action")!= MATCH) {
      printf("%s is not an action\n",key);
      return;
   }

   sprintf(temp,"%s > ability > %s", key, value);
   result = db_lookup(temp, value2);
   printf("RESULT: %d\r\n", result);
}

//------------------------------------------------------
// key is someones first name
// p is what the rating is of
//
void handle_rating_question(char* key, char* p) {

   int result;
   char value[20];
   char id[20];
   char temp[20]="rating-";

   result = db_get_id(key);
   //printf("result %d %s %s %s      ", result, key, p, id);
   //return;

   if(result !=0) {
      printf("I dont know %s\n", key);
      return;
   }

//printf("result %d     ", result);
//  return;

//check if the subject of rating is known

//   result = db_subject_search(p);
   if(result!=0) {
      printf("What is %s?\n", key);
      return;
   }

// get rating if any
   strcat(temp, p);
//  result = db_search(key,temp,value);

   if(result==0) printf("%s\n", value);
   else printf("I dont know (code:%d)\n", result);

}

//----------------------------------------

void handle_list_question(char* subject) {

}

//--------------------------------------------------

void handle_attribute_statement(char* user_subject,char* user_attribute) {
   /*
   example: grass is green

   Posibilities:
   1   grass is unknown                      "reply: "i dont know what grass is"
   2   grass is not an object or substance
   3   green is unknown
   4   green is not an attribute
   5   info already known
   6   info is contradictory
   7   ok to add to db
   */

   int result;
   int result1;
   int result2;
   char value[20];
   char key[60];
   int subject_result;
   char subject_class[20];
   char db_class[20];
   char attribute_type[20];
   char db_attribute[20];

   // 1) ex: is "grass" in database?
   sprintf(key, "%s > class", user_subject);
   subject_result = db_lookup(key, subject_class);

   // If subject_result === UNKNOWN, I don't know what %s is

   // 2) Is grass an object or substance?
   result1 = db_root_check(user_subject, "object");
   result2 = db_root_check(user_subject, "substance");
   if ((result1 == NOT_FOUND) && (result2 == NOT_FOUND)) {
      printf("I don't think %s can have such an attribute\n", user_subject);
      return;
   }

   // 3) is green in db?
   sprintf(key, "%s > class", user_attribute);
   if(db_lookup(key, db_class) == NOT_FOUND) {
      printf("I'm unfamiliar with %s\n", user_attribute);
      return;
   }

   // 4) is "green" an attribute
   if(db_root_check(user_attribute, "attribute") == NOT_FOUND) {
      printf("%s is not an attribute\n", user_attribute);
      return;
   }

   // Preparation for 5, 6 and 7

   // Get the type of the attribute. ex: "green is a color"
   sprintf(key, "%s > class", user_attribute);  // assemble key, ex: "green > class"
   db_lookup(key, attribute_type);  // ex: returns "color"

   // Get the attribute from db if any
   sprintf(key, "%s > %s", user_subject, attribute_type);  // assemble key, ex: "green > color"
   result = db_lookup(key, db_attribute);  // ex: returns "green"

   // 5) Is the attribute already known?
    if(strcmp(user_attribute, db_attribute) == 0) {   // compare "green"(what the user typed) with green(from db)
      printf("I already know that\n");
      return;
   }

   // 6 Check if the attrubute contradicts what's in the db
   if(result == FOUND) {
      printf("no, it's %s\n", db_attribute);
      return;
   }

   // 7 Since posiblities 1 to 6 have been eliminated, it's ok to add the info to the db
   // Determine attribute type

   // Add to db


   sprintf(key, "%s > %s", user_subject, value);  // assemble key, ex: grass > color
   db_add_pair(key, user_attribute);

   printf("I'll take a note of that\n");
   return;

}

//--------------------------------------------------
// WORK IN PROGRESS
//
// handle hi, hello
//
//
void handle_greetings(void) {
   // if not logged in
   if(current_user_id == 0) {
      printf("hi, what is your name?\r\n");
   }
   // if logged in
   else {
      printf("hi %s\r\n", current_user_name);
   }
}

 /*
void handle_is_statement(char*user_subject, char*user_attribute) {

   UNDER CONSTRUCTION, will likely be abandoned


   example: a cat is an animal(value)

   Posibilities:
                                      db action      user action
       --------------------------     ------------   --------------------------
      cat is unknown       	        add to db     "I'll take note of that"
      animal is unknown                             "what's an animal"
      cat is known, class = animal                  "I know"
      cat known, class != animal                    "no it's a ____"

   //  class doesn't exist  	// i dont know what an animal is
   //  subject doesnt exist, class exists  	// ill take note of that
   //  subject exists, value is the same  	// "i knew that"
   //  subject exists, value is different  	// no its not

   At some point we need to figure out if this is a class or an attribute.
   The syntax tells us what to expect, then we have to verify it.


   class or attribute?
   B)if(root_check(p1) == attribute) deal with it, goto label1
   A)if(p1 is in db) it's a class so deal with it,
   if yes to both,


   label1:





   cat in db?




   char value[20];
   char key[20];
   int result;

   // 1 is "animal" in the database?
   if(db_check(user_class) != FOUND) {
      printf("I'm not familiar with %s\n", user_class);
      return;
   }

   //2,3&4
//  ex: lookup "cat"
   sprintf(key, "%s > class", user_subject);
   result = db_lookup(key, value);

//  2)
   if(result == NOT_FOUND) {
      sprintf(key, "%s > class", user_subject);
      db_add_pair(key, user_class);
      printf("I'll take a note of that\n");
      return;
   }

//  Note: At this point, "cat" was found in the database.
//  We just need to check if the info matches or not.

//  3) states what already is known
   if(strcmp(user_class, value) == 0) {
      printf("I already know that\n");
      return;
   }

//  4) contradictory info
   else {
      printf("no, it's a %s\n", value);
      return;
   }
}
*/