给个时间区间，返回该时间区间所包含的日期

仅能得到时间垮幅为一个月的，当然可以自己更改（用for循环）。

/**
	 * 时间区间所包含的日期.
	 * @param startDate
	 * @param endDate
	 */
	public String[] formatDate(String startDate, String endDate) {
		String[] date = null;
		if (startDate.equals(endDate)) {// 同一天.
			date = new String[] { endDate };
		} else if (startDate.substring(5, 7).equals(endDate.substring(5, 7))) {// 不同天,同一月.
			int a = Integer.parseInt(endDate.substring(8))
					- Integer.parseInt(startDate.substring(8));
			date = new String[a + 1];
			for (int i = 0; i <= a; i++) {
				date[i] = startDate.substring(0, 8)
						+ (Integer.parseInt(startDate.substring(8)) + i);
			}
		} else {// 不同天,不同月.
			String d[] = endDate.toString().split("-");
			int yy = Integer.parseInt(d[0]);
			int mm = Integer.parseInt(d[1]);
			int dd = Integer.parseInt(d[2]);
			mm = mm - 1;
			if (mm == 2) {
				dd = 28;
			} else if (mm == 1 || mm == 3 || mm == 5 || mm == 7 || mm == 8
					|| mm == 10 || mm == 12) {
				dd = 31;
			} else {
				dd = 30;
			}
			// 判断闰年
			boolean r = yy % 4 == 0 && yy % 100 != 0 || yy % 400 == 0;
			if (r && mm == 2)
				dd++;

			String temp = String.valueOf(yy) + "-"
					+ String.valueOf(mm < 10 ? "0" + mm : mm) + "-"
					+ String.valueOf(dd);
			int a = Integer.parseInt(temp.substring(8))
					- Integer.parseInt(startDate.substring(8));
			int b = Integer.parseInt(endDate.substring(8)) - 1;
			date = new String[a + b + 2];
			for (int i = 0; i <= a; i++) {
				date[i] = startDate.substring(0, 8)
						+ (Integer.parseInt(startDate.substring(8)) + i);
			}
			for (int i = 0; i <= b; i++) {
				date[a + i + 1] = endDate.substring(0, 8)
						+ ((i + 1) < 10 ? "0" + (i + 1) : (i + 1));
			}
		}
		for (int i = 0; i < date.length; i++) {
			System.out.println(date[i]);
		}
		return date;
	}

2010-02-25 和 2010-03-02 返回结果：
2010-02-25
2010-02-26
2010-02-27
2010-02-28
2010-03-01
2010-03-02
2010-02-25
2010-02-26
2010-02-27
2010-02-28
2010-03-01
2010-03-02的数组