ledger/parse.cc
2003-09-30 03:22:38 +00:00

216 lines
4.8 KiB
C++

#include <cstring>
#include <ctime>
#include <cctype>
#include "ledger.h"
#include <pcre.h> // Perl regular expression library
namespace ledger {
//////////////////////////////////////////////////////////////////////
//
// Ledger parser
//
char * next_element(char * buf, bool variable = false)
{
char * p;
if (variable)
p = std::strstr(buf, " ");
else
p = std::strchr(buf, ' ');
if (! p)
return NULL;
*p++ = '\0';
while (std::isspace(*p))
p++;
return p;
}
static int linenum = 0;
inline void finalize_entry(entry * curr) {
if (curr) {
if (! curr->validate()) {
std::cerr << "Failed to balance the following transaction, "
<< "ending on line " << (linenum - 1) << std::endl;
curr->print(std::cerr);
} else {
ledger.push_back(curr);
}
}
}
bool parse_ledger(std::istream& in)
{
static std::time_t now = std::time(NULL);
static struct std::tm * now_tm = std::localtime(&now);
static int current_year = now_tm->tm_year + 1900;
static char line[1024];
static struct std::tm moment;
memset(&moment, 0, sizeof(struct std::tm));
entry * curr = NULL;
// Compile the regular expression used for parsing amounts
static pcre * entry_re = NULL;
if (! entry_re) {
const char *error;
int erroffset;
static const std::string regexp =
"^(([0-9]{4})[./])?([0-9]{2})[./]([0-9]{2})\\s+(\\*\\s+)?"
"(\\(([^)]+)\\)\\s+)?(.+)";
entry_re = pcre_compile(regexp.c_str(), 0, &error, &erroffset, NULL);
}
while (! in.eof()) {
in.getline(line, 1023);
linenum++;
if (in.eof()) {
break;
}
else if (std::isdigit(line[0])) {
static char buf[256];
int ovector[60];
int matched = pcre_exec(entry_re, NULL, line, std::strlen(line),
0, 0, ovector, 60);
if (! matched) {
std::cerr << "Failed to parse, line " << linenum << ": "
<< line << std::endl;
continue;
}
if (curr)
finalize_entry(curr);
curr = new entry;
// Parse the date
int mday, mon, year = current_year;
if (ovector[1 * 2] >= 0) {
pcre_copy_substring(line, ovector, matched, 2, buf, 255);
year = std::atoi(buf);
}
if (ovector[3 * 2] >= 0) {
pcre_copy_substring(line, ovector, matched, 3, buf, 255);
mon = std::atoi(buf);
}
if (ovector[4 * 2] >= 0) {
pcre_copy_substring(line, ovector, matched, 4, buf, 255);
mday = std::atoi(buf);
}
moment.tm_mday = mday;
moment.tm_mon = mon - 1;
moment.tm_year = year - 1900;
curr->date = std::mktime(&moment);
if (ovector[5 * 2] >= 0)
curr->cleared = true;
if (ovector[6 * 2] >= 0) {
pcre_copy_substring(line, ovector, matched, 7, buf, 255);
curr->code = buf;
}
if (ovector[8 * 2] >= 0) {
int result = pcre_copy_substring(line, ovector, matched, 8, buf, 255);
assert(result >= 0);
curr->desc = buf;
}
}
else if (std::isspace(line[0])) {
transaction * xact = new transaction();
char * p = line;
while (std::isspace(*p))
p++;
// The call to `next_element' will skip past the account name,
// and return a pointer to the beginning of the amount. Once
// we know where the amount is, we can strip off any
// transaction note, and parse it.
char * cost_str = next_element(p, true);
char * note_str;
// If there is no amount given, it is intended as an implicit
// amount; we must use the opposite of the value of the
// preceding transaction.
if (! cost_str || *cost_str == ';') {
if (cost_str) {
while (*cost_str == ';' || std::isspace(*cost_str))
cost_str++;
xact->note = cost_str;
}
xact->cost = curr->xacts.back()->cost->copy();
xact->cost->negate();
}
else {
note_str = std::strchr(cost_str, ';');
if (note_str) {
*note_str++ = '\0';
while (std::isspace(*note_str))
note_str++;
xact->note = note_str;
}
for (char * t = cost_str + (std::strlen(cost_str) - 1);
std::isspace(*t);
t--)
*t = '\0';
xact->cost = create_amount(cost_str);
}
account * current = NULL;
for (char * tok = std::strtok(p, ":");
tok;
tok = std::strtok(NULL, ":")) {
if (! current) {
accounts_iterator i = accounts.find(tok);
if (i == accounts.end()) {
current = new account(tok);
accounts.insert(accounts_entry(tok, current));
} else {
current = (*i).second;
}
} else {
account::iterator i = current->children.find(tok);
if (i == current->children.end()) {
current = new account(tok, current);
current->parent->children.insert(accounts_entry(tok, current));
} else {
current = (*i).second;
}
}
}
xact->acct = current;
curr->xacts.push_back(xact);
}
else if (line[0] == 'Y') {
current_year = std::atoi(line + 2);
}
}
if (curr)
finalize_entry(curr);
return true;
}
} // namespace ledger