How to parse datetime from a file into a class in C++

Question

I am trying to parse a Class data from a txt file in C++. The Class data looks like this, contains some datetime string, some integer and some int.

The data in the txt file looks like this:

2014-08-01 02:06:53,3.9070,9

How to parse the data from the file into the Class?

Problems faced: 1) getting the input from the file for the datetime 2) separating the datetime input from the other data input.

If you're lucky you are on a system with a [`strptime`](http://man7.org/linux/man-pages/man3/strptime.3.html) function. Otherwise there are [alternatives](http://stackoverflow.com/questions/321849/strptime-equivalent-on-windows). — Some programmer dude, Aug 15 '14 at 05:15
And you don't need the `!in.eof()` in your loop condition, it's redundant. — Some programmer dude, Aug 15 '14 at 05:16
Or am I misunderstanding you? The way you read the date-time string from the file does not follow the format you give in the question at all. Instead you construct a string of that format and store in the class. My question is of course why you want to store a date-time stamp internally *as text*? You should read the data-time stamp as text, then parse it into a binary time-stamp than can be used by the C or C++ standard functions, preferably in a [C++11 `chrono` object](http://en.cppreference.com/w/cpp/chrono). — Some programmer dude, Aug 15 '14 at 05:19
And considering that you store the data-time as a string in the structure, what does `Datetime::Parse` do? Parse from a string into a binary time-stamp, then convert that back into a string again? — Some programmer dude, Aug 15 '14 at 05:34
As for the format in the file, please show a *complete* and *unmodified* record. Does it contain newlines? Is it comma-separated between fields? And what is the *actual* format for the date-time? — Some programmer dude, Aug 15 '14 at 05:35
date-time format is locale dependend, that's the whole problem. If everyone used the same date-time - you will never get such a question. For example my date time is: $ date Пт авг 15 10:44:34 UZT 2014 — Tanuki, Aug 15 '14 at 05:44

Some programmer dude · Accepted Answer · 2014-08-15T06:04:13.803

You have multiple problems, most of them regarding your reading and parsing:

Each record (each line) contains three values: A date-time stamp, and two numeric values. You read the date-time stamp (wrongly, I'll come back to that) and then you read six more values.
When you read the numeric values, you read them into a temporary buffer, but then you use the input stream to assign to the structure fields.
Nowhere in the reading you handle the comma that separates the values.
As with the comma, you read the date-time without consideration of the separators in the date-time format.
You read the date-time field into separate values, then create a string of the exact same format as in the file. That's not really needed (I will tell you why below).
The date-time is stored in the structure as a string, so why are you trying to parse it, if you're just storing it in a string anyway? What, if anything, does the DateTime::Parse function really do?

Now to help you with your problem there are a couple of things in the C++ standard library that can help you. The first is the std::getline function, the second is std::istringstream.

First of all I suggest you read each line from the file complete (using std::getline), put the line into a std::istringstream and use that for the parsing, again using std::getline.

Something like

std::string line;
while (std::getline(in, line))
{
    std::istringstream iss(line);

    // Parse each line using the input string stream
}

The std::getline by default uses the newline as separator, but it can be told to use any single character, for example ',': Also note that I put the date-time string directly into your structure.

std::getline(iss, item->datetime, ',');

The above two lines will put the complete date-time string into the datetime member variable. You can then get the two other values easily the same way:

std::string temp;

std::getline(iss, temp, ',');
item->open = std::stod(temp);

std::getline(iss, temp);  // Last item in input, no need for separator
item->volume = std::stoi(temp);

(Read about std::stod and std::stoi.)

I randomly picked two member fields from your structure to initialize, because there are only two numeric fields per record in the file.

How to parse datetime from a file into a class in C++

1 Answers1