Counting number of words (only count words which have at least one alphabetic character) in a File using C?

Question

Words may be separated by any amount of whitespace characters. There can be integers in a file, but the program should only count words which have at least one alphabetic character.

Answer 1

#include<stdio.h>
#include<string.h>

int FindWord(char *);

main(int argc,char **argv)
{
    File *fp;
    char str[100];
    int count=0;

    fp = fopen(argv[1],"r");
    if(fp==NULL)
    {
        printf("Unable to open a file\n");
        return;
    }

    while(fgets(str,sizeof(str),fp)!=NULL)
    {
        count=count+FindWord(str);
        bzero(str,sizeof(str));
    }
    printf("Total Word are %d\n",count);
}

int FindWord(char *p)
{
    int count=1;
    int i,j;
    for(i=0;p[i];i++)
    {
        if(p[i]==' ')
        {
            for(j=i+1;p[j];j++)
            {
                if(p[j]!=' ')
                    break;
            }
            count+=1;
            i=j;
        }
    }

    return count;
}

Comments

This may not be covering all the cases.
1. If line has spaces in the end it may count the wrong numbers
2. If a word has only numbers say 1111 then it should not be counted which is not considered.

Do you like to correct the program :)