Sir,
I am facing problem in one function , actually in my table
there is two field: id, and password and I am checking the id
field for any user who is added to the friendlist of another user,
like in yahoo messenger.
MYSQL *conn;
MYSQL_RES *res_ptr;
MYSQL_ROW sqlrow;
FILE *fp;
/******************************************************/
int Friend_Result()
{
unsigned int i; int rows;
while( (sqlrow = mysql_fetch_row(res_ptr)) != NULL)
{
for(i = 0; i < mysql_num_fields(res_ptr); i++)
{
if(i > 0) printf("\t");
if(sqlrow[i] != NULL) printf("%s", sqlrow[i]);
else printf("%s", "NULL");
}
printf("\n");
}
if(mysql_errno(conn) != 0)
{
printf("mysql_fetch_row error\n");
return 0;
}
if((unsigned long)mysql_num_rows(res_ptr) == 0)
{
printf("Invalid .... Not have a valid TARA ID...\n");
return 0;
}
return 1;
}
/******************************************************/
int Check_Exist_Friend(char *f)
{
char buf[200];
int result, rows;
printf(" Friend = %s\n", f);
strcpy(buf, "SELECT id from tara_user where id = ");
strcat(buf, "'");
strcat(buf, f);
strcat(buf, "'");
result = mysql_query(conn, buf);
res_ptr = mysql_store_result(conn);
if(res_ptr == NULL) printf("Storing Error\n");
else
{
printf("Res_ptr is not null\n");
if(! Friend_Result() ) return 0;
mysql_free_result(res_ptr);
}
return 1;
}
/************* CHECK THE FRIEND - ID IN DATABASE ***********/
void Add_New_Friend(char *a, char *gp, char *f, int sd)
{
int result;
char buf[200];
printf("Friend name = %s\n", f);
if(! Check_Exist_Friend(f) ) return;
else
{
strcpy(buf, "INSERT INTO flist values(");
strcat(buf, "'");
strcat(buf, a);
strcat(buf, "'");
strcat(buf, ",");
strcat(buf, "'");
strcat(buf, gp);
strcat(buf, "'");
strcat(buf, ",");
strcat(buf, "'");
strcat(buf, f);
strcat(buf,"'");
strcat(buf,")");
result = mysql_query(conn, buf);
if(!result) printf("Inserted %lu rows\n",
(unsigned long)mysql_affected_rows(conn));
else fprintf(stderr, "Insert error %d: %s\n",
mysql_errno(conn),
mysql_error(conn));
}
}
I am calling function Check_Exist_Friend(f) from
Add_New_Friend() but each time where the friend name is existing
in the table or not it give me the same result. I have tried a
lot but couldn't succeed.
I don't know where is fault , so please reply me ..... so that I
can do my rest of work.
If there are other option to check the existence of any user
whether the user is present or not , please let me know....
Hoping positive response from your side.
Thanking you
ASHOK KUMAR
M.C.A. 5TH SEM.
COCHIN UNIVERSITY OF SCIENCE AND TECHNOLOGY
KOCHI : 682 022, KERALA (INDIA).
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