diff options
Diffstat (limited to 'format.cc')
| -rw-r--r-- | format.cc | 67 |
1 files changed, 67 insertions, 0 deletions
@@ -293,4 +293,71 @@ void format_t::format_elements(std::ostream& out, } } +void format_transaction::operator()(transaction_t * xact) +{ + if (inverted) { + xact->total.quantity += - xact->amount; + xact->total.cost += - xact->cost; + } else { + xact->total += *xact; + } + xact->index = index++; + + // This makes the assumption that transactions from a single entry + // will always be grouped together. + + if (last_entry != xact->entry) + first_line_format.format_elements(output_stream, details_t(xact)); + else + next_lines_format.format_elements(output_stream, details_t(xact)); + + last_entry = xact->entry; +} + +void format_account::operator()(const account_t * account, + const unsigned int max_depth, + const bool report_top) +{ + // Don't output the account if only one child will be displayed + // which shows the exact same amount. jww (2004-08-03): How do + // compute the right figure? It should a value expression specified + // by the user, to say, "If this expression is equivalent between a + // parent account and a lone displayed child, then don't display the + // parent." + + if (bool output = report_top || account->parent != NULL) { + int counted = 0; + bool display = false; + + for (accounts_map::const_iterator i = account->accounts.begin(); + i != account->accounts.end(); + i++) { + if (! (*i).second->total) + continue; + + if ((*i).second->total != account->total || counted > 0) { + display = true; + break; + } + counted++; + } + + if (counted == 1 && ! display) + output = false; + + if (output) { + unsigned int depth = account->depth; + if (max_depth == 0 || depth <= max_depth) { + for (const account_t * acct = account; + depth > 0 && acct && acct != last_account; + acct = acct->parent) + depth--; + + format.format_elements(output_stream, details_t(account, depth)); + last_account = account; + } + } + } +} + } // namespace ledger |