SELECT last_insert_id(); 

-- создаем тестовую табличку
create table test (
 id int not null auto_increment,
 info char not null,
 primary key(id));

-- наполняем ее тестовыми данными
insert into test (info)
 select 'A' union all
 select 'B' union all
 select 'C' union all
 select 'D' union all
 select 'E' union all
 select 'F' union all
 select 'G' union all
 select 'H' union all
 select 'X' union all
 select 'Y' union all
 select 'Z';

-- удалим некоторые записи, чтобы появились "разрывы"
delete from test 
 where id in ( 2 , 3 , 7 , 8 , 10 )

-- после удаления
select * from test

+----+------+
| id | info |
+----+------+
|   1  | A    |
|   4  | D    |
|   5  | E    |
|   6  | F    |
|   9  | X    |
|  11  | Z    |
+----+------+

-- теперь мы хотим определить предшествующую и предыдущую запись 
-- для id с номером 4

select 
 id current,
 IfNULL((select id from test where id=(select max(id) from test where id< 4 ) limit  0 , 1 ),  0 ) prev,
 IfNULL((select id from test where id=(select min(id) from test where id> 4 ) limit  0 , 1 ),  0 ) next
from test
where id= 4 

+---------+------+------+
| current | prev | next |
+---------+------+------+
|        4  |     1  |     5  |
+---------+------+------+

-- теперь сделаем тоже самое для граничной точки (например id=1)

select 
 id current,
 IfNULL((select id from test where id=(select max(id) from test where id< 1 ) limit  0 , 1 ),  0 ) prev,
 IfNULL((select id from test where id=(select min(id) from test where id> 1 ) limit  0 , 1 ),  0 ) next
from test
where id= 1 

+---------+------+------+
| current | prev | next |
+---------+------+------+
|        1  |     0  |     4  |
+---------+------+------+

+----------------+-------+
| D              | P     |
+----------------+-------+
| 01.01.2004     | 0     |
| 02.01.2004     | 0     |
| 03.01.2004     | 1     |
| 04.01.2004     | 1     |
| 06.01.2004     | 0     |
| 07.01.2004     | 1     |
| 08.01.2004     | 0     |
| 09.01.2004     | 1     |
| 11.01.2004     | 1     |
| 14.01.2004     | 0     |
+----------------+-------+

select distinct\\n min(t2.d),\\n t2.p\\nfrom ttt t1, ttt t2\\nwhere t1.p= 0 \\n and t2.p= 1 \\n and t1.d<t2.d\\ngroup by t1.d

\nset @vid =  4 ;\nselect\n(select id from test where id<@vid order by id desc limit  1  ) as min_id,\n(select id from test where id>@vid order by id asc limit  1  ) as max_id;\n

\\ndrop table if exists t_1\\n\\n( 0  row(s) affected)\\n\\n-- создаем тестовую табличку\\ncreate table t_1 (\\n uid int not null auto_increment primary key,\\n uname varchar( 20 ),\\n vote int not null default  0 )\\n\\n-- заливаем инфу\\ninsert into t_1 \\n(uname, vote) \\nvalues \\n(\\\'Smith\\\',  1 ), (\\\'Joe\\\',  1 ), (\\\'Peter\\\',  4 ), (\\\'Kate\\\',  3 ), (\\\'Mike\\\',  10 ), (\\\'BOSS\\\',  8 ), (\\\'John\\\',  0 );\\n\\n( 7  row(s) affected)\\n

\\nselect uid, uname "Best users", vote  \\nfrom t_1\\norder by vote desc\\nlimit  5 \\n\\n-- то получим\\n\\n   uid  Best users  vote  \\n------  ----------  ------\\n      5   Mike             10 \\n      6   BOSS              8 \\n      3   Peter             4 \\n      4   Kate              3 \\n      1   Smith             1 \\n

\\nselect * from t_1 where vote= 1 \\n\\n   uid  uname   vote  \\n------  ------  ------\\n      1   Smith         1 \\n      2   Joe           1 \\n

\\nselect *\\nfrom t_1\\nwhere vote>=(select vote \\n\\t     from t_1\\n\\t     order by vote desc\\n\\t     limit  4 , 1 )\\norder by vote desc\\n\\n   uid  uname   vote  \\n------  ------  ------\\n      5   Mike         10 \\n      6   BOSS          8 \\n      3   Peter         4 \\n      4   Kate          3 \\n      1   Smith         1 \\n      2   Joe           1 \\n

-- установливаем новый формат даты для удобства
set global date_format='%d.%m.%Y'

-- создаем тестовую табличку
create table t ( d date)

( 0  row(s) affected)
( 0  ms taken)

-- заливаем даты
insert into t (d) values (str_to_date('12.06.2005', '%d.%m.%Y'));
insert into t (d) values (str_to_date('13.06.2005', '%d.%m.%Y'));
insert into t (d) values (str_to_date('14.06.2005', '%d.%m.%Y'));
insert into t (d) values (str_to_date('20.06.2005', '%d.%m.%Y'));
insert into t (d) values (str_to_date('21.06.2005', '%d.%m.%Y'));

( 5  row(s) affected)

select date_format(d, '%d.%m.%Y') dates from t

dates     
----------
 12 . 06 . 2005 
 13 . 06 . 2005 
 14 . 06 . 2005 
 20 . 06 . 2005 
 21 . 06 . 2005 

select x1.ones+x2.tens mdays
from (
    select  0  ones union all
    select  1  ones union all
    select  2  ones union all
    select  3  ones union all
    select  4  ones union all
    select  5  ones union all
    select  6  ones union all
    select  7  ones union all
    select  8  ones union all
    select  9  ones) x1,
    (select  0  tens union all
    select  10  tens union all
    select  20  tens union all
    select  30  tens) x2
where x1.ones+x2.tens between  1  and  30 

-- в итоге получим числа
mdays
---------
         1 
         2 
...
        30 
	   
-- сохраним это множество во временной таблице
create temporary table t_1
as
select x1.ones+x2.tens mday
from (
    select  0  ones union all
    select  1  ones union all
    select  2  ones union all
    select  3  ones union all
    select  4  ones union all
    select  5  ones union all
    select  6  ones union all
    select  7  ones union all
    select  8  ones union all
    select  9  ones) x1,
    (select  0  tens union all
    select  10  tens union all
    select  20  tens union all
    select  30  tens) x2
where x1.ones+x2.tens between  1  and  30 

( 30  row(s) affected)

-- теперь все готово, чтобы создать отчет

select concat(t_1.mday, '.', Month('2005-06-00'), '.', YEAR('2005-06-00')) as dates
from t_1 left join t
on t_1.mday=DAYOFMONTH(t.d)
where t.d is null

dates    
---------
 1 . 6 . 2005  
 2 . 6 . 2005  
 3 . 6 . 2005  
 4 . 6 . 2005  
 5 . 6 . 2005  
 6 . 6 . 2005  
 7 . 6 . 2005  
 8 . 6 . 2005  
 9 . 6 . 2005  
 10 . 6 . 2005 
 11 . 6 . 2005 
 15 . 6 . 2005 
 16 . 6 . 2005 
 17 . 6 . 2005 
 18 . 6 . 2005 
 19 . 6 . 2005 
 22 . 6 . 2005 
 23 . 6 . 2005 
 24 . 6 . 2005 
 25 . 6 . 2005 
 26 . 6 . 2005 
 27 . 6 . 2005 
 28 . 6 . 2005 
 29 . 6 . 2005 
 30 . 6 . 2005