Trabalhando com datas

13 de abril de 2006 4 respostas

kurokibr 13 de abril de 2006

Amigos, como fazer para subtrair um dia de uma determinada data(hoje)?
estou tentando da seguinte maneira mas o resultado é a mesma data:

Calendar cl = Calendar.getInstance();
SimpleDateFormat sdf = new SimpleDateFormat(“dd/MM/yyyy”);

System.out.println("Antes: "+ sdf.format(cl.getTime()) );

cl.getTime().setTime(cl.getTime().getTime() - 24<em>60</em>60*1001);

System.out.println("Depois: "+ sdf.format(cl.getTime()) );

valeu galera

4 Respostas

Cristiano_Pires1 13 de abr. de 2006

Calendar calendarA = Calendar.getInstance();

Calendar calendarB = Calendar.getInstance();

int multiplier;

if (b.getTime() - a.getTime() > 0) {

multiplier = -1;

calendarA.setTime(b);

calendarB.setTime(a);

} else {

multiplier = 1;

calendarA.setTime(a);

calendarB.setTime(b);

}

int years = calendarA.get(Calendar.YEAR) - calendarB.getCalendar.YEAR);
    int months = calendarA.get(Calendar.MONTH) - calendarB.get(Calendar.MONTH);
    int days = calendarA.get(Calendar.DAY_OF_MONTH) - calendarB.get(Calendar.DAY_OF_MONTH);
    if (years > 0 && (months < 0 || (months == 0 && days < 0))) {
        years -= 1;
    }
    return years * multiplier;

velo 13 de abr. de 2006

dah uma olhada em GregorianCalendar

VELO

Mantu 13 de abr. de 2006

Tenta usar o método add da classe Calendar. Isso aqui é um texto que eu tirei da API desta classe:

add(f, delta) adds delta to field f. This is equivalent to calling set(f, get(f) + delta) with two adjustments:
Add rule 1. The value of field f after the call minus the value of field f before the call is delta, modulo any overflow that has occurred in field f. Overflow occurs when a field value exceeds its range and, as a result, the next larger field is incremented or decremented and the field value is adjusted back into its range.

Add rule 2. If a smaller field is expected to be invariant, but   it is impossible for it to be equal to its prior value because of changes in its minimum or maximum after field f is changed, then its value is adjusted to be as close as possible to its expected value. A smaller field represents a smaller unit of time. HOUR is a smaller field than DAY_OF_MONTH. No adjustment is made to smaller fields that are not expected to be invariant. The calendar system determines what fields are expected to be invariant.
In addition, unlike set(), add() forces an immediate recomputation of the calendar’s milliseconds and all fields.

Example: Consider a GregorianCalendar originally set to August 31, 1999. Calling add(Calendar.MONTH, 13) sets the calendar to September 30, 2000. Add rule 1 sets the MONTH field to September, since adding 13 months to August gives September of the next year. Since DAY_OF_MONTH cannot be 31 in September in a GregorianCalendar, add rule 2 sets the DAY_OF_MONTH to 30, the closest possible value. Although it is a smaller field, DAY_OF_WEEK is not adjusted by rule 2, since it is expected to change when the month changes in a GregorianCalendar.

Veja a sintaxe em
http://java.sun.com/j2se/1.5.0/docs/api/java/util/Calendar.html#add(int,%20int)

thiago.correa 13 de abr. de 2006

http://www.guj.com.br/posts/list/16631.java

Criado 13 de abril de 2006

Ultima resposta 13 de abr. de 2006

Respostas 4

Participantes 5

4 Respostas

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