CODEWITHSUNDEEP
java
user179516
user179516

Reputation: 337

Get the newest file based on timestamp in Java

I have a directory which contains files in the following format. The files are in a diretory called /incoming/external/data

ABC_20100806.csv
ABC_20100807.csv
ABC_20100808.csv
ABC_20100809.csv
ABC_20100810.csv
ABC_20100811.csv
ABC_20100812.csv

As you can see the filename of the file includes a timestamp. i.e. [RANGE]_[YYYYMMDD].csv

What i need to do is find out which of these files has the newest date using the timestamp on the filename not the system timestamp and store the filename in a variable and move it to another directory and move the rest to a different directory in java.

Upvotes: 0

Views: 3044

Answers (1)

helderdarocha
helderdarocha

Reputation: 23637

You can read the filenames into an array using:

File directory = new File("/incoming/external/data");
String[] fileNames = directory.list(new FilenameFilter() {
    public boolean accept(File dir, String fileName) {
         return fileName.endsWith(".csv");
    }
});

And from there simply sort the array if your files always have the same prefix:

Arrays.sort(fileNames);

One way you can remove the prefix and suffix of each fileName to extract the date is:

int underline = fileName.indexOf("_");
int dot = fileName.indexOf(".");
String datePart = fileName.substring(underline, dot);

And then you can add that string to an array and sort (lexically).

If for some other reason you want to convert the dates into Java Dates, you can use:

SimpleDateFormat dt = new SimpleDateFormat("yyyymmdd");
Date date = dt.parse(datepart);

And you'll have a Java date, which you can also sort in an array or list.

Upvotes: 3

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