--С помощью регулярного выражения
select REGEXP_REPLACE('ДГН-04-13601/21-53', '-\d{1,}/', '-********/') from dual;
--Если строка всегда одной длины, то можно с помощью substr
select substr('ДГН-04-13601/21-53', 0, 7) || '********' || substr('ДГН-04-13601/21-53', 13) from dual;Каким способом можно к этому подойти?
sql.execute('UPDATE users SET cash = cash - ' + de)Select ФИО
From Список_специалистов
INNER JOIN Журнал_учёта_работ ON Журнал_учёта_работ.id_Специалиста = Список_специалистов.idSELECT E.FIRST_NAME,
E.LAST_NAME,
L.STREET_ADDRESS
FROM HR.EMPLOYEES E
LEFT JOIN DEPARTMENTS D ON D.DEPARTMENT_ID = E.DEPARTMENT_ID
LEFT JOIN LOCATIONS L ON L.LOCATION_ID = D.LOCATION_ID;
SELECT E.FIRST_NAME,
E.LAST_NAME,
L.STREET_ADDRESS
FROM HR.EMPLOYEES E
LEFT JOIN DEPARTMENTS D USING (DEPARTMENT_ID)
LEFT JOIN LOCATIONS L USING (LOCATION_ID);select pd.product_id,
pd.name
pa.price
from product_description pd
left join product_article pa on pa.product_id = pd.product_id
select sum(rashody)/count(distinct data_prikaza_) over(partition by gosb) from erp_trips
SELECT * FROM Table
WHERE PublishDate between '20201006' AND '20201014'SELECT * FROM Table
WHERE PublishDate >= DATEADD(day,-7, GETDATE())$sql 'UPDATE `_spi`
SET `w_st`=`w_t`+`w_s`+`w_z`+`w_p` + CASE WHEN `p_m` = \'IV\' THEN `w_t`+`w_s` ELSE 0 END';INSERT INTO MyTABLE (Name, CapitalCity, LargestCity, Admission, Population)
select Name, CapitalCity, LargestCity, Admission, Population from OpenJson('
[
{
"name": "Alabama",
"capitalCity": "Montgomery",
"largestCity": "Birmingham",
"admission": "Dec 14, 1819",
"population": "4,903,185"
}, ...
')
with (
Name nvarchar(20) '$.name',
CapitalCity nvarchar(30) '$.capitalCity',
LargestCity nvarchar(20) '$.largestCity',
Admission nvarchar(20) '$.admission',
Population nvarchar(20) '$.population'
)select translate (str, lower(str) || upper(str), upper(str) || lower(str)) from dualсложение 1 и 4 записи, 3 запись. вот так получается.
грубо говоря, мы игнорируем все записили при повторе id_j. из-за этого везде, где в записях есть 2 в id_j игнорируются. и остаются три записи, которые суммируем
SELECT A.ID_H,
SUM(A.COST) COST
FROM (SELECT ID_H,
ID_J,
COST,
RANK() OVER(PARTITION BY ID_J ORDER BY ID_H ASC) RNK
FROM MYTABLE
) A WHERE A.RNK = 1
GROUP BY A.ID_H
SELECT r.id,
c.relation_cpu_id cpu_hash,
LISTAGG(m.relation_monitor_id, ', ') WITHIN GROUP (ORDER BY m.id) AS monitor_hash
FROM report r
LEFT JOIN v_cpu_relation c ON r.id = c.relation_report_id
LEFT JOIN v_monitor_relation m ON r.id = m.relation_report_id
GROUP BY r.id,
c.relation_cpu_id
select l.prod, l.id list_id, l.list_val from list l left join main m on <Дальше сам>SELECT * FROM `objects`
WHERE `bid_type`='$bid_type'
AND `object_type`='$object_type'
AND (`address` LIKE '%$address%'
OR `okrug` LIKE '%$address%')begin
insert into ..... returning id into :id;
-- ...
commit;
end;