I'm receiving a JSON object with date value like this:
{"PostingDate":"\/Date(1325134800000-0500)\/"}
And I want to parse it in Java code to Date
or getting it as a String
.
I want to know what is the easy way of doing it.
I take it the first number (1325134800000
) is the number of milliseconds since epoch, and -0500
is the time zone. This appears to be the case given the sample code below, which seems to do what you want.
The following code parses the JSON input using Jackson, which I recommend if you don't have a JSON parsing library of choice yet. It lacks error checking etc.
Sample code:
public final class Foo
{
public static void main(final String... args)
throws IOException
{
// What the JSON value must match exactly
// Not anchored since it will be used with the (misnamed) .matches() method
final Pattern pattern
= Pattern.compile("\\\\/Date\\((\\d+)(-\\d+)?\\)\\\\/");
final ObjectMapper mapper = new ObjectMapper();
// Parse JSON...
final JsonNode node = mapper.readTree(
"{\"PostingDate\": \"\\/Date(1325134800000-0500)\\/\"}");
if (!node.has("PostingDate")) {
System.err.println("Bad JSON input!");
System.exit(1);
}
// Get relevant field
final String dateSpec = node.get("PostingDate").getTextValue();
// Try and match the input.
final Matcher matcher = pattern.matcher(dateSpec);
if (!matcher.matches()) {
System.err.println("Bad pattern!"); // Yuck
System.exit(1);
}
// The first group capture the milliseconds, the second one the time zone
final long millis = Long.parseLong(matcher.group(1));
String tz = matcher.group(2);
if (tz.isEmpty()) // It can happen, in which case the default is assumed to be...
tz = "+0000";
// Instantiate a date object...
final Date date = new Date(millis);
// And print it using an appropriate date format
System.out.printf("Date: %s %s\n",
new SimpleDateFormat("yyyy/MM/dd HH:MM:ss").format(date), tz);
}
}
Output:
Date: 2011/12/29 06:12:00 -0500
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