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I recently asked for a recursive query in postgreSQL. The answer fit well, but as soon as I add a JOIN element, it gives me the following error:

ERROR: missing FROM-clause entry for table "n": WITH RECURSIVE chain AS ( SELECT n.pordnr, pz.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion FROM pord n LEFT JOIN pordnrzu pz ON pz.pordnr = n.pordnr WHERE n.pordnr = 112 UNION ALL SELECT n.pordnr, pz2.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion FROM chain c LEFT JOIN pordnrzu pz2 ON pz2.pordnr = n.pordnr INNER JOIN pord n ON (c.pordnrzu = n.pordnr) WHERE c.abschl IS NULL ) SELECT * FROM chain c WHERE c.abschl IS NOT NULL

I tried several modifications but I can't get it to work. All I want is, that pordnrzu is being implemented to look for another pordnr.

So: pordnr -> pordnrzu -> pordnr

Is there anything I just missed?

This is the sample table:

CREATE TABLE pord (
  pordnr integer primary key,
  abschl text,
  stg text,
  kzfa text,
  pversion text
);

INSERT INTO pord (pordnr, abschl, stg, kzfa, pversion)
VALUES
(112,     NULL, NULL, NULL, NULL),
(140,     NULL, NULL, NULL, NULL),
(200,     NULL, NULL, NULL, NULL),
(210,     'f2', '140', 'H', '2011'),
(220,     'f2222', '140000', 'HHH', '201111');

CREATE TABLE pordnrzu (
  pordnr integer primary key,
  pordnrzu integer
);

INSERT INTO pordnrzu (pordnr, pordnrzu)
VALUES
(112,     140),
(140,     210),
(200,     220),
(210,     220),
(220,     NULL);

And this my query by now:

WITH RECURSIVE chain AS
(
  SELECT n.pordnr, pz.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion
  FROM pord n
  LEFT JOIN pordnrzu pz
  ON pz.pordnr = n.pordnr
  WHERE n.pordnr = 112
  UNION ALL
  SELECT n.pordnr, pz2.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion
  FROM chain c
  LEFT JOIN pordnrzu pz2
  ON pz2.pordnr = n.pordnr
  INNER JOIN pord n ON (c.pordnrzu = n.pordnr)
  WHERE c.abschl IS NULL
)
SELECT *
FROM chain c
WHERE c.abschl IS NOT NULL;

My aim is that I start with pordnr=112, which then leads per pordnrzu to other pordnr until it finds the first entry with abschl NOT NULL

In this example, the solution would be this row:

pordnr   pordnrzu   abschl   stg   kzfa   pversion
200      210        f2       140   H      2011

SELECT version();

PostgreSQL 9.2.6 on x86_64-unknown-linux-gnu, compiled by gcc (SUSE Linux) 4.3.4 [gcc-4_3-branch revision 152973], 64-bit

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1 回答 1

1

JOIN子句被评估left-to-right。向上移动第二个连接以使这项工作：

WITH RECURSIVE chain AS (
  SELECT n.pordnr, pz.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion
  FROM   pord n
  LEFT   JOIN pordnrzu pz ON pz.pordnr = n.pordnr
  WHERE  n.pordnr = 112

  UNION ALL
  SELECT n.pordnr, pz.pordnrzu, n.abschl, n.stg, n.kzfa, n.pversion
  FROM   chain c
  JOIN   pord  n ON n.pordnr = c.pordnrzu
  LEFT   JOIN pordnrzu pz ON pz.pordnr = n.pordnr
  WHERE  c.abschl IS NULL
)
SELECT *
FROM   chain c
WHERE  c.abschl IS NOT NULL;

或者，您可以在 2nd SELECT 中重写 1st JOIN：

LEFT   JOIN pordnrzu pz ON pz.pordnr = c.pordnrzu
于 2014-02-18T15:27:22.157 回答