SELECT companies.name, max(polls.created_at)
FROM companies
JOIN teams
ON teams.company_id = companies.id
LEFT
JOIN polls
ON polls.team_id = teams.id
GROUP BY companies.name
ORDER BY 2
import unicodedata
list_texts = ["ᴰᴼᴺ'ᵀ ᴮᴱ ᴬᶠᴿᴬᴵᴰ ᵀᴼ ᶠᴬᴵᴸ ᴮᴱ ᴬᶠᴿᴬᴵᴰ ᴺᴼᵀ ᵀᴼ ᵀᴿʸ",
" ",
" ", "̈", "ᶠᴿᴬᴺᴷᶠᵁᴿᵀ",
"ʙᴇ ᴏɴ ʏᴏᴜʀ ᴡᴀʏ ᴛᴀᴋᴇ ᴡʜᴀᴛ ʏᴏᴜ ɴᴇᴇᴅ"]
for s in list_texts:
row_str = map(lambda x: x if len(x) == 1 else ' ', (str(unicodedata.name(_)).split(' ')[::-1][0] for _ in s))
print(''.join(list(row_str)))
if ($currDate != $animal['day']){
$currDate = $animal['day']
echo '<li> <p class="timetable-date">'.$currDate.'<p class="timetable-doctors"><span>'.$animal['clerk_name'].'</span></p><li>'}
else {
echo '<li><p class="timetable-doctors"><span>'.$animal['clerk_name'].'</span></p><li>'
}
SELECT act.user_id, act.action_name AS last_action, act.dttm AS last_action_date
FROM actions act
LEFT JOIN actions act1
ON act1.user_id = act.user_id
AND act1.dttm > act.dttm
WHERE act1.user_id IS NULL
select fullname, from_unixtime(startdate) FROM course
declare @a bigint
select @a = 20091203
SELECT convert(datetime, convert(varchar(10), @a), 103)
SELECT fullname, convert(datetime, convert(varchar(10), startdate), 103)
FROM course
WHERE drop.name='name'
Вам подойдёт
WHERE drop.name like 'name%'
Вместо name введите имя вашего предмета.
Если нужно выводить только первые 20, то попробуйте такой вариант
SELECT top 20 drop.id, drop.name, MIN(drop_sell.price) AS min_price, MAX(drop_buy.price) AS max_price
И секцию group by вот так напишите:
group by drop.id, drop.name