tic tac toe in c, trying to determine who wins - c

Question

Trying to determine who wins in a game of tic tac toe,I am new to programming. Currently getting user has won after only entering 1 X or O input. Inputs must be entered with 2 int, row and coloumn. Any help is greatly appreciated!
#include <stdio.h>
#include <stdlib.h>
void drawBoard(char board[][3])
{
int rows, columns;
for ( rows = 0 ; rows < 3 ; rows++ )
{
for ( columns = 0 ; columns < 3 ; columns++ )
{
if(board[rows][columns]){
printf( "|%c", board[rows][columns] );
}else{
printf("| ");
}
}
printf("|\n");
}
}
int main()
{
char game[3][3]={{0}};
int totalEntry =0,row,column;
char value;
while(totalEntry<=9){
printf("Please choose x or o: ");
scanf("%c",&value);
getchar();
printf("Enter row number: ");
scanf("%d",&row);
getchar();
printf("Enter Column number: ");
scanf("%d",&column);
getchar();
game[row][column] = value;
drawBoard(game);
if((game[0][0] == game[0][1]) && (game[0][1] == game[0][2]) && game[0][0] != 'x')
if((game[1][0] == game[1][1]) && (game[1][1] == game[1][2]) && game[1][0] != 'x')
if((game[2][0] == game[2][1]) && (game[2][1] == game[2][2]) && game[2][0] != 'x')
if((game[0][0] == game[1][0]) && (game[1][0] == game[2][0]) && game[0][0] != 'x')
if((game[0][1] == game[1][1]) && (game[1][1] == game[2][1]) && game[0][1] != 'x')
if((game[0][2] == game[1][2]) && (game[1][2] == game[2][2]) && game[0][2] != 'x')
if((game[0][0] == game[1][1]) && (game[1][1] == game[2][2]) && game[0][0] != 'x')
if((game[2][0] == game[1][1]) && (game[1][1] == game[0][2]) && game[2][0] != 'x')
printf("User x has won!");
if((game[0][0] == game[0][1]) && (game[0][1] == game[0][2]) && game[0][0] != 'o')
if((game[1][0] == game[1][1]) && (game[1][1] == game[1][2]) && game[1][0] != 'o')
if((game[2][0] == game[2][1]) && (game[2][1] == game[2][2]) && game[2][0] != 'o')
if((game[0][0] == game[1][0]) && (game[1][0] == game[2][0]) && game[0][0] != 'o')
if((game[0][1] == game[1][1]) && (game[1][1] == game[2][1]) && game[0][1] != 'o')
if((game[0][2] == game[1][2]) && (game[1][2] == game[2][2]) && game[0][2] != 'o')
if((game[0][0] == game[1][1]) && (game[1][1] == game[2][2]) && game[0][0] != 'o')
if((game[2][0] == game[1][1]) && (game[1][1] == game[0][2]) && game[2][0] != 'o');
printf("User o has won!");
break;
}
return 0;
}

Answer 1

I would have used something like following, using a loop. Note that since the board is initialized to zero, assigning winner to a board position which still has zero means that a winner isn't found even if the test for all equal passes.
//declare this outside the main while-loop:
char winner=0;
for(i=0;i<3 && !winner;i++) {
if(game[i][0]==game[i][1] && game[i][1]==game[i][2])
winner=game[i][0]; // across
else if(game[0][i]==game[1][i] && game[1][i]==game[2][i])
winner=game[0][i]; // down
}
if( !winner && ( game[0][0]==game[1][1] && game[1][1]==game[2][2] ||
game[0][2]==game[1][1] && game[1][1]==game[2][0] ) )
winner = game[1][1]; // diagonal
if(winner) {
printf( "Winner is %c!\n",winner );
break;
}

Answer 2

IckIckIck.
First, it'll make you life much easier (at least for determining the winner), if you use a one-dimenisional array of 9 elements.
Then, write a function which test one possible win and returns true or false.
bool Test(char board[9], char player, int a, int b, int c)
{
return board[a] == player
&& board[b] == player
&& board[c] == player;
}
Then Test(game, 'x', 0,3,6); -- tests of x wins down left side.

Answer 3

The logic in your conditional is incorrect. Imagine the board looks like this. 0 denotes an empty spot.
0 0 0
0 0 x
0 0 0
The statement if((game[0][0] == game[0][1]) && (game[0][1] == game[0][2]) && game[0][0] != 'x') will evaluate to true because game[0][0] is empty game[0][1] is empty and game[0][2] is empty so they are all equal.
Also since game[0][0] is empty, it is not equal to 'x'.
So changing your != to == will help in that regard.