我只是在脑海中写下了这个，但是如果您必须使用简单的sql来做的话，这应该是个主意。
insert into restaurant(x, y)values select valuex, valuey from dual where not exists( select 1 from restaurant where name = 'restaurantname')
编辑： 同样，我无法解析它，但您可能可以使用WITH子句：
with validation as( select 1 from restaurant where name = 'restaurantname')insert into restaurant(x, y)values ( select value1x, value1y from dual where validation.v = 1), ( select value2x, value2y from dual where validation.v = 1)
