我有引用其他表的表:如何在postgresql中插入其他表的值?
CREATE TABLE scratch
(
id SERIAL PRIMARY KEY,
name TEXT NOT NULL,
rep_id INT NOT NULL REFERENCES reps,
term_id INT REFERENCES terms
);
CREATE TABLE reps (
id SERIAL PRIMARY KEY,
rep TEXT NOT NULL UNIQUE
);
CREATE TABLE terms (
id SERIAL PRIMARY KEY,
terms TEXT NOT NULL UNIQUE
);
我想補充一個新的記錄劃傷給出的名,則代表和方面值,即我既沒有相應的rep_id也不term_id。
現在是我唯一的想法就是:
insert into scratch (name, rep_id, term_id)
values ('aaa', (select id from reps where rep='Dracula' limit 1), (select id from terms where terms='prepaid' limit 1));
我的問題是這樣的。我試圖使用參數化查詢API(從使用節點postgres的包節點),其中一個插入件的查詢看起來像這樣:
insert into scratch (name, rep_id, term_id) values ($1, $2, $3);
,然後$ 1,$ 2和$ 3個值的陣列作爲傳遞一個單獨的論點。最後,當我對參數化查詢感到滿意時,想法是將它們提升爲準備好的語句,以便利用最有效和最安全的方式來查詢數據庫。
但是,我很困惑我怎麼能用我的例子來做到這一點,在這個例子中,不同的表格必須被子查詢。
請幫忙。
P.S.
我使用的是PostgreSQL 9.2,對於PostgreSQL特定的解決方案沒有問題。
編輯1
C:\Users\markk>psql -U postgres
psql (9.2.4)
WARNING: Console code page (437) differs from Windows code page (1252)
8-bit characters might not work correctly. See psql reference
page "Notes for Windows users" for details.
Type "help" for help.
postgres=# \c dummy
WARNING: Console code page (437) differs from Windows code page (1252)
8-bit characters might not work correctly. See psql reference
page "Notes for Windows users" for details.
You are now connected to database "dummy" as user "postgres".
dummy=# DROP TABLE scratch;
DROP TABLE
dummy=# CREATE TABLE scratch
dummy-# (
dummy(# id SERIAL NOT NULL PRIMARY KEY,
dummy(# name text NOT NULL UNIQUE,
dummy(# rep_id integer NOT NULL,
dummy(# term_id integer
dummy(#);
NOTICE: CREATE TABLE will create implicit sequence "scratch_id_seq" for serial column "scratch.id"
NOTICE: CREATE TABLE/PRIMARY KEY will create implicit index "scratch_pkey" for table "scratch"
NOTICE: CREATE TABLE/UNIQUE will create implicit index "scratch_name_key" for table "scratch"
CREATE TABLE
dummy=# DEALLOCATE insert_scratch;
ERROR: prepared statement "insert_scratch" does not exist
dummy=# PREPARE insert_scratch (text, text, text) AS
dummy-# INSERT INTO scratch (name, rep_id, term_id)
dummy-# SELECT $1, r.id, t.id
dummy-# FROM reps r, terms t
dummy-# WHERE r.rep = $2 AND t.terms = $3
dummy-# RETURNING id, name, $2 rep, $3 terms;
PREPARE
dummy=# DEALLOCATE insert_scratch2;
ERROR: prepared statement "insert_scratch2" does not exist
dummy=# PREPARE insert_scratch2 (text, text, text) AS
dummy-# INSERT INTO scratch (name, rep_id, term_id)
dummy-# VALUES ($1, (SELECT id FROM reps WHERE rep=$2 LIMIT 1), (SELECT id FROM terms WHERE terms=$3 LIMIT 1))
dummy-# RETURNING id, name, $2 rep, $3 terms;
PREPARE
dummy=# EXECUTE insert_scratch ('abc', 'Snowhite', '');
id | name | rep | terms
----+------+-----+-------
(0 rows)
INSERT 0 0
dummy=# EXECUTE insert_scratch2 ('abc', 'Snowhite', '');
id | name | rep | terms
----+------+----------+-------
1 | abc | Snowhite |
(1 row)
INSERT 0 1
dummy=# EXECUTE insert_scratch ('abcd', 'Snowhite', '30 days');
id | name | rep | terms
----+------+----------+---------
2 | abcd | Snowhite | 30 days
(1 row)
INSERT 0 1
dummy=# EXECUTE insert_scratch2 ('abcd2', 'Snowhite', '30 days');
id | name | rep | terms
----+-------+----------+---------
3 | abcd2 | Snowhite | 30 days
(1 row)
INSERT 0 1
dummy=#
EDIT 2
我們可以利用該rep_id
需要的事實,即使terms_id
是可選的,使用INSERT-SELECT以下版本:
PREPARE insert_scratch (text, text, text) AS
INSERT INTO scratch (name, rep_id, term_id)
SELECT $1, r.id, t.id
FROM reps r
LEFT JOIN terms t ON t.terms = $3
WHERE r.rep = $2
RETURNING id, name, $2 rep, $3 terms;
然而,這個版本有兩個問題:
- 沒有區分是否有缺失
terms
值(即, '')和一個無效的值(即完全從術語表中缺失的非空值)。兩者都被視爲失蹤條款。 (但插入有兩個子查詢會遇到同樣的問題) - 該版本取決於需要
rep
的事實。但是如果rep_id
也是可選的呢?
EDIT 3
找到的第2項所述溶液 - 被要求對代表消除依賴性。再加上使用WHERE語句有問題,如果代表無效 - 它只是插入0行,而我想在這種情況下顯式失敗,SQL不會失敗。我的解決方法就是使用虛擬一行CTE:
PREPARE insert_scratch (text, text, text) AS
WITH stub(x) AS (VALUES (0))
INSERT INTO scratch (name, rep_id, term_id)
SELECT $1, r.id, t.id
FROM stub
LEFT JOIN terms t ON t.terms = $3
LEFT JOIN reps r ON r.rep = $2
RETURNING id, name, rep_id, term_id;
如果代表缺失或無效,此SQL將嘗試插入NULL到rep_id領域,因爲該領域是NOT NULL
錯誤將提高 - 正是我需要。如果進一步我決定讓代表可選 - 沒問題,同樣的SQL也適用於此。
你能不能改變你的例子,利用'PREPARE'聲明?我想看看你的例子看起來像$ 1,$ 2,... – mark
由於一些奇怪的原因,這與INSERT插入不插入任何東西給我。你有沒有在一個實際的PostgreSQL數據庫上檢查它? – mark
@mark是的,我執行並檢查。查看更新。 – joop