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How can I read comma separated values from a text file in Java?

Tags:

java

string

io

I have got this text file with latitude and longitude values of different points on a map.

How can I split my string into latitudes and longitudes? What is the general way to do these type of things that is with other delimiters like space or tab etc.? Sample file:

28.515046280572285,77.38258838653564
28.51430151808072,77.38336086273193
28.513566177802456,77.38413333892822
28.512830832397192,77.38490581512451
28.51208605426073,77.3856782913208
28.511341270865113,77.38645076751709

This is the code I am using to read from the file:

try(BufferedReader in = new BufferedReader(new FileReader("C:\\test.txt"))) {
    String str;
    while ((str = in.readLine()) != null) {
        System.out.println(str);
    }
}
catch (IOException e) {
    System.out.println("File Read Error");
}
like image 276
rishiag Avatar asked Mar 19 '26 13:03

rishiag


2 Answers

You may use the String.split() method:

String[] tokens = str.split(",");

After that, use Double.parseDouble() method to parse the string value to a double.

double latitude = Double.parseDouble(tokens[0]);
double longitude = Double.parseDouble(tokens[1]);

Similar parse methods exist in the other wrapper classes as well - Integer, Boolean, etc.

like image 55
KV Prajapati Avatar answered Mar 22 '26 03:03

KV Prajapati


Use OpenCSV for reliability. Split should never be used for these kind of things. Here's a snippet from a program of my own, it's pretty straightforward. I check if a delimiter character was specified and use this one if it is, if not I use the default in OpenCSV (a comma). Then i read the header and fields

CSVReader reader = null;
try {
    if (delimiter > 0) {
        reader = new CSVReader(new FileReader(this.csvFile), this.delimiter);
    }
    else {
        reader = new CSVReader(new FileReader(this.csvFile));
    }

    // these should be the header fields
    header = reader.readNext();
    while ((fields = reader.readNext()) != null) {
        // more code
    }
catch (IOException e) {
    System.err.println(e.getMessage());
}
like image 43
Jeffrey Vandenborne Avatar answered Mar 22 '26 01:03

Jeffrey Vandenborne