如何解决在postgres中不存在
我正在使用postgres db,我试图在sql下面执行以插入不存在的表中
我想先将数据插入表中,然后再检查是否存在,否则不要插入。
我希望简化查询,因为我将在带有pycopg2库的python代码中使用它
INSERT INTO T1 (a1,a2,a3,a4)
VALUES (123,'20e16411-b8f7',4,(SELECT u1 from T2 where u2 = 'test@test.com'))
WHERE NOT EXISTS (SELECT a3
FROM TI
WHERE a1 =123
AND a2 = '20e16411-b8f7'
AND a3 = 4
AND a4 = (SELECT u1 from T2 where u2 = 'test@test.com'))
sql Error [42601]: ERROR: Syntax error at or near "WHERE" Position: 236
我在哪里语法上遇到了问题,但是,如果不能很好地纠正我的查询,我不确定这是否是实现我的逻辑的正确查询
If not exist (select * from t1 where <check>)
Begin
Insert
End
解决方法
我认为您想要<html>
<head>
</head>
<body>
<div class="survery-questions">
<form action="">
<div class="price-form">
<input name="price-1" type="text" placeholder="price 1" class="home-price-footer" id="input" required>
</div>
<div class="phone-form">
<input name="price-2" type="text" placeholder="price 2" class="home-price-footer" id="input" required>
</div>
<div class="email-form">
<input type="price-3" placeholder="price 3" class="home-price-footer" id="input" required>
</div>
<div class="price-form">
<input name="price-4" type="text" placeholder="price 4" class="home-price-footer" id="input" required>
</div>
<div class="phone-form">
<input name="price-5" type="text" placeholder="price 5" class="home-price-footer" id="input" required>
</div>
<div class="email-form">
<input type="price-6" placeholder="price 6" class="home-price-footer" id="input" required>
</div>
<div class="price-form">
<input name="price-7" type="text" placeholder="price 7" class="home-price-footer" id="input" required>
</div>
<div class="phone-form">
<input name="price-8" type="text" placeholder="price 8" class="home-price-footer" id="input" required>
</div>
<div class="email-form">
<input type="price-9" placeholder="price 9" class="home-price-footer" id="input" required>
</div>
</form>
</div>
</body>
</html>
:
insert ... select
,
让数据库验证唯一性。在您关注的列上定义唯一约束:
create unique index unq_t1_a1_a2_a3_a4 on t1(a1,a2,a3,a4);
然后使用on conflict
:
insert into T1 (a1,a4)
select 123,'20e16411-b8f7',4,t2.u1
from T2
where t2.u2 = 'test@test.com'
on conflict (a1,a4) do nothing;
,
您不能在VALUES之后使用WHERE,这就是为什么会出现错误,在此查询中,您必须使用SELECT而不是VALUES,类似这样;
INSERT INTO T1 (a1,a4)
SELECT 123,(SELECT u1 from T2 where u2 = 'test@test.com')
WHERE NOT EXISTS (SELECT a3
FROM TI
WHERE a1 =123
AND a2 = '20e16411-b8f7'
AND a3 = 4
AND a4 = (SELECT u1 from T2 where u2 = 'test@test.com'))
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