Postgresql 调用将JSONB作为参数传递的函数-Postgres
下面是我写的函数Postgresql 调用将JSONB作为参数传递的函数-Postgres,postgresql,function,jsonb,postgresql-9.6,Postgresql,Function,Jsonb,Postgresql 9.6,下面是我写的函数 CREATE OR REPLACE FUNCTION upd.insert_testdata(spec jsonb[]) RETURNS void LANGUAGE 'plpgsql' AS $BODY$ BEGIN --Consider all columns in specialist table as character varying and code column as integer.
CREATE OR REPLACE FUNCTION upd.insert_testdata(spec jsonb[])
RETURNS void
LANGUAGE 'plpgsql' AS $BODY$
BEGIN
--Consider all columns in specialist table as character varying and code column as integer.
insert into upd.specialist (to_cd, empid, code, booking_status, availability)
select j.spec->>'to_cd',
j.spec->>'empid',
j.spec->>'code',
j.spec->>'booking_status',
j.spec->>'availability'
from jsonb_populate_record(spec) j;
END;
$BODY$;
我正试图通过以下命令调用函数
SELECT upd.insert_testdata(
'{
"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status":"Y",
"availability":"MTWTFSS"
}'::jsonb[]
);
但我得到的错误是格式不正确的数组文字
详细信息:意外的数组元素。
SQL状态:22P02
另外,我还想知道如何在单个json变量中插入多条记录/传递多行您实际上不必作为
jsonb
发送和接收。您可以在函数中作为文本发送并转换为jsonb
对象
CREATE OR REPLACE FUNCTION upd.insert_testdata(spec text)
RETURNS void LANGUAGE 'plpgsql'
AS $BODY$
declare
your_json jsonb;
begin
your_json := spec::jsonb;
然后您将jsonb
作为文本发送
select upd.insert_testdata(
'{
"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status":"Y",
"availability":"MTWTFSS"
}'
)
不要传递json值数组,传递json数组:
CREATE OR REPLACE FUNCTION upd.insert_testdata( spec jsonb)
RETURNS void
LANGUAGE plpgsql
AS $BODY$
begin
--Consider all columns in specialist table as character varying and code column as integer.
insert into upd.specialist (to_cd, empid, code, booking_status, availability)
select j.spec->>'to_cd',
j.spec->>'empid',
(j.spec->>'code')::int,
j.spec->>'booking_status',
j.spec->>'availability'
from jsonb_array_elements(spec) as j(spec);
end;
$BODY$;
如果您不想手动列出所有键,并且如果您100%确定键名始终与表中的列名匹配,则可以使用jsonb\u populate\u record
CREATE OR REPLACE FUNCTION upd.insert_testdata( spec jsonb)
RETURNS void
LANGUAGE plpgsql
AS $BODY$
begin
--Consider all columns in specialist table as character varying and code column as integer.
insert into upd.specialist (to_cd, empid, code, booking_status, availability)
select (jsonb_populate_record(null::specialist, j.spec)).*
from jsonb_array_elements(spec) as j(spec);
end;
$BODY$;
然后像这样使用它:
select upd.insert_testdata('[
{"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status": "Y",
"availability":"MTWTFSS"}
]'::jsonb);
select upd.insert_testdata('[
{"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status": "Y",
"availability":"MTWTFSS"},
{"to_cd":"CFG",
"empid":"test2",
"code": 456,
"booking_status": "N",
"availability":"MT"}
]'::jsonb);
如果要传递多个元素,可以按如下方式使用:
select upd.insert_testdata('[
{"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status": "Y",
"availability":"MTWTFSS"}
]'::jsonb);
select upd.insert_testdata('[
{"to_cd":"NFG",
"empid":"test",
"code":123,
"booking_status": "Y",
"availability":"MTWTFSS"},
{"to_cd":"CFG",
"empid":"test2",
"code": 456,
"booking_status": "N",
"availability":"MT"}
]'::jsonb);
我的基本目的是在表中插入数据。这将如何将数据插入表中?错误:类型json的输入语法第2行:(“[^DETAIL:Expected:”,但找到“,”。我在您的JSON中的某个地方使用postgres v9.6,您没有使用
:
来分隔键和值。代码是的,它正在工作,最后一个问题不应该是jsonb_populate_记录更适合这种情况。。。???