Ale*_*sio 9 sql postgresql quoted-identifier
我试着从postgres控制台这个命令:
select sim.id as idsim,
num.id as idnum
from main_sim sim
left join main_number num on (FK_Numbers_id=num.id);
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我得到了这样的答复:
ERROR: column "fk_numbers_id" does not exist LINE 1: ...m from main_sim sim left join main_number num on (FK_Numbers...
但如果我只是检查我的表:
dbMobile=# \d main_sim
id | integer | not null default
Iccid | character varying(19) | not null
...
FK_Device_id | integer |
FK_Numbers_id | integer |
Indexes:
"main_sim_pkey" PRIMARY KEY, btree (id)
"main_sim_FK_Numbers_id_key" UNIQUE, btree ("FK_Numbers_id")
"main_sim_Iccid_key" UNIQUE, btree ("Iccid")
"main_sim_FK_Device_id" btree ("FK_Device_id")
Foreign-key constraints:
"FK_Device_id_refs_id_480a73d1" FOREIGN KEY ("FK_Device_id") REFERENCES main_device(id) DEFERRABLE INITIALLY DEFERRED
"FK_Numbers_id_refs_id_380cb036" FOREIGN KEY ("FK_Numbers_id") REFERENCES main_number(id) DEFERRABLE INITIALLY DEFERRED
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......因为我们可以看到该列存在.
可能是语法错误,但我无法看到...
任何帮助将不胜感激.阿莱西奥
a_h*_*ame 19
不,该列FK_Numbers_id不存在,只"FK_Numbers_id"存在一列
显然,您使用双引号创建了表,因此所有列名称现在都区分大小写,您必须始终使用双引号:
select sim.id as idsim,
num.id as idnum
from main_sim sim
left join main_number num on ("FK_Numbers_id" = num.id);
Run Code Online (Sandbox Code Playgroud)
列foo和FOO是相同的,所述列"foo"和"FOO"都没有.
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