提交 0a1f5dcf authored 作者: Stéphane Eintrazi's avatar Stéphane Eintrazi

update code style

上级 72e7bbd8
...@@ -1508,7 +1508,7 @@ public class Function extends Expression implements FunctionCall { ...@@ -1508,7 +1508,7 @@ public class Function extends Expression implements FunctionCall {
result = ValueInt.get(getDatePart(v1, field)); result = ValueInt.get(getDatePart(v1, field));
}else { } else {
// Case where we retrieve the EPOCH time. // Case where we retrieve the EPOCH time.
// First we retrieve the dateValue and his time in nanoseconds. // First we retrieve the dateValue and his time in nanoseconds.
...@@ -1534,7 +1534,7 @@ public class Function extends Expression implements FunctionCall { ...@@ -1534,7 +1534,7 @@ public class Function extends Expression implements FunctionCall {
// number of days and multiply it by the number of seconds in a day. // number of days and multiply it by the number of seconds in a day.
result = ValueDecimal.get(numberOfDays.multiply(secondsPerDay)); result = ValueDecimal.get(numberOfDays.multiply(secondsPerDay));
}else if(v1 instanceof ValueTimestampTimeZone) { } else if(v1 instanceof ValueTimestampTimeZone) {
// Case where the value is a of type ValueTimestampTimeZone ('2000:01:01 10:00:00+05). // Case where the value is a of type ValueTimestampTimeZone ('2000:01:01 10:00:00+05).
// We retrieve the time zone offset in minute // We retrieve the time zone offset in minute
...@@ -1546,7 +1546,7 @@ public class Function extends Expression implements FunctionCall { ...@@ -1546,7 +1546,7 @@ public class Function extends Expression implements FunctionCall {
.add(numberOfDays.multiply(secondsPerDay)) .add(numberOfDays.multiply(secondsPerDay))
.subtract(timeZoneOffsetSeconds)); .subtract(timeZoneOffsetSeconds));
}else { } else {
// By default, we have the date and the time ('2000:01:01 10:00:00) if no type is given. // By default, we have the date and the time ('2000:01:01 10:00:00) if no type is given.
// We just have to sum the time in nanoseconds and the total number of days in seconds. // We just have to sum the time in nanoseconds and the total number of days in seconds.
......
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