Hi all, after writing my program, I typed: bash-3.1# gcc -lsqlite3 CreaDB.c -o creadb CreaDB.c: In function 'main': CreaDB.c:35: error: too few arguments to function 'sqlite3_bind_text' CreaDB.c:36: error: too few arguments to function 'sqlite3_bind_text' CreaDB.c:37: error: too few arguments to function 'sqlite3_bind_text'
But I don't see the problem, my program is: int main(int argc, char** argv[]) { int rc, i; sqlite3* db; sqlite3_stmt* stmt; char* sql; const char* tail; rc = sqlite3_open("prova.db", &db); if (rc) { fprintf(stderr, "E' impossibile aprire il file %s\n", sqlite3_errmsg(db)); sqlite3_close(db); exit(1); } sql = "create table modulo(id);"; rc = sqlite3_prepare(db, sql, strlen(sql), &stmt, &tail); if (rc != SQLITE_OK) { fprintf(stderr, "Errore SQL: %s\n", sqlite3_errmsg(db)); } rc = sqlite3_step(stmt); sqlite3_reset(stmt); sql = "insert into modulo(id, nome, classe, istanza) values(?,?,?,?)"; sqlite3_prepare(db, sql, strlen(sql), &stmt, &tail); sqlite3_bind_int(stmt, 1, 1); sqlite3_bind_text(stmt, 2, "nome1"); sqlite3_bind_text(stmt, 3, "classe1"); sqlite3_bind_text(stmt, 4, "istanza1"); sqlite3_step(stmt); while (rc == SQLITE_ROW) { for (i = 0; i < sqlite3_column_count(stmt); i++) fprintf(stderr, "'%s' ", sqlite3_column_text(stmt, i)); fprintf(stderr, "\n"); rc = sqlite3_step(stmt); } sqlite3_finalize(stmt); sqlite3_close(db); return 0; } Thank you very much in advance for answers, savio --------------------------------- Inviato da Yahoo! Mail. La casella di posta intelligente. _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users