MySQL-Abfrage über 4 Tabellen auf verschiedenen Kriterien beitreten

Question

Ich habe 4 Tabellen (abgespeckte zu relevanten Spalten der Kürze halber):

CREATE TABLE `papers` (
    `paper_id` int(11) NOT NULL DEFAULT '0', 
    PRIMARY KEY (`paper_id`) 
); 
INSERT INTO papers (paper_id) VALUES(1001); 
INSERT INTO papers (paper_id) VALUES(1002); 
INSERT INTO papers (paper_id) VALUES(1003); 
INSERT INTO papers (paper_id) VALUES(1004); 
INSERT INTO papers (paper_id) VALUES(1005); 
INSERT INTO papers (paper_id) VALUES(1006); 

CREATE TABLE `questions` (
    `question_id` int(11) NOT NULL AUTO_INCREMENT, 
    `type_id` int(11) NOT NULL, 
    PRIMARY KEY (`question_id`) 
); 
INSERT INTO questions (type_id) VALUES(1); 
INSERT INTO questions (type_id) VALUES(2); 
INSERT INTO questions (type_id) VALUES(1); 
INSERT INTO questions (type_id) VALUES(3); 

CREATE TABLE `question_depends` (
    `question_id` int(11) NOT NULL, 
    `depends_question_id` int(11) NOT NULL, 
    `depends_answer_val` int(11) NOT NULL, 
    PRIMARY KEY (`question_id`,`depends_question_id`) 
); 
INSERT INTO question_depends (question_id, depends_question_id, depends_answer_val) VALUES(3, 1, 0); 
INSERT INTO question_depends (question_id, depends_question_id, depends_answer_val) VALUES(2, 1, 1); 
INSERT INTO question_depends (question_id, depends_question_id, depends_answer_val) VALUES(3, 1, 1); 

CREATE TABLE `answers` (
    `paper_id` int(11) NOT NULL, 
    `question_id` int(11) NOT NULL, 
    `answer_val` int(2) NOT NULL, 
    PRIMARY KEY (`paper_id`,`question_id`) 
); 
INSERT INTO answers (paper_id, question_id, answer_val) VALUES(1002, 1, 1); 
INSERT INTO answers (paper_id, question_id, answer_val) VALUES(1002, 4, 0); 
INSERT INTO answers (paper_id, question_id, answer_val) VALUES(1004, 1, 0); 
INSERT INTO answers (paper_id, question_id, answer_val) VALUES(1004, 3, 1); 
INSERT INTO answers (paper_id, question_id, answer_val) VALUES(1005, 1, 1); 

ich mit einer einzigen Abfrage zu kommen versuche, die alle Daten für alle möglichen zeigt Kombinationen:

1. Alle Papiere IDs sollte eine Ausgabe mindestens einmal
2. A gegeben paper_id sein kann oder auch nicht Antworten haben und kann oder auch nicht haben Abhängigkeiten
3. Die ultim Ziel ist es, zu sehen, ob jede Abhängigkeitsfrage beantwortet wird und wenn ja, ob die Antwort val mit den Abhängigkeiten antwortet, val für jede Papier-ID, und ob paper_id Abhängigkeitsfragen hat oder nicht und ob Fragen nicht beantwortet werden Abhängigkeiten)
4. I anpassen können Tabellen/Daten benötigt, wenn

ich kam in der Nähe mit:

select P.paper_id as P_PID, 
    A.paper_id as A_PID, 
    A.question_id as A_QID, 
    A.answer_val as A_VAL, 
    QD.question_id as QD_QID, 
    QD.depends_question_id AS QD_DQID, 
    QD.depends_answer_val AS QD_VAL, 
    Q.type_id AS Q_TYPE 
from papers P 
left join answers A on A.paper_id = P.paper_id 
left join question_depends QD on QD.depends_question_id = A.question_id 
left join questions Q on Q.question_id = QD.question_id 
UNION 
select NULL AS P_PID, 
    NULL AS A_PID, 
    A.question_id as A_QID, 
    A.answer_val as A_VAL, 
    QD.question_id as QD_QID, 
    QD.depends_question_id AS QD_DQID, 
    QD.depends_answer_val AS QD_VAL, 
    Q.type_id AS Q_TYPE 
from question_depends QD 
left join answers A on QD.depends_question_id = A.question_id 
left join questions Q on Q.question_id = QD.question_id 
    where A.question_id IS NULL 

... aber die Ausgabe hat die Antwortdaten für jede Zeile von jeder paper_id nicht nur die für diese Papier-ID. Irgendwelche Gedanken geschätzt! Mit Hilfe dieser kleinen Beispieldaten die oben Auswahlausgänge gesetzt:

P_PID A_PID A_QID A_VAL QD_QID QD_DQID QD_VAL Q_TYPE 
1001        
1002 1002 1  1  2  1  1  2 
1002 1002 1  1  3  1  0  1 
1002 1002 4  0  NULL NULL NULL NULL 
1003 NULL NULL NULL NULL NULL NULL NULL 
1004 1004 1  0  2  1  1  2 
1004 1004 1  0  3  1  0  1 
1004 1004 3  1  NULL NULL NULL NULL 
1005 1005 1  1  2  1  1  2 
1005 1005 1  1  3  1  0  1 
1006 NULL NULL NULL NULL NULL NULL NULL 

Der ideale Ausgang (wenn ich nichts Typo hat) wäre:

P_PID A_PID A_QID A_VAL QD_QID QD_DQID QD_VAL Q_TYPE 
1001 NULL NULL NULL 2  1  1  2 
1001 NULL NULL NULL 3  1  0  1 
1002 1002 1  1  2  1  1  2 
1002 NULL NULL NULL 3  1  0  1 
1002 1002 4  0  NULL NULL NULL 3 
1003 NULL NULL NULL 2  1  1  2 
1003 NULL NULL NULL 3  1  0  1 
1004 1004 1  0  2  1  1  2 
1004 NULL NULL NULL 3  1  0  1 
1004 1004 3  1  NULL NULL NULL 1 
1005 1005 1  1  2  1  1  2 
1005 NULL NULL NULL 3  1  0  1 
1006 NULL NULL NULL 2  1  1  2 
1006 NULL NULL NULL 3  1  0  1 

Answer 1

"Divide et impera". Teilen Sie das Problem in drei Fälle auf.

SELECT /*Answers and no depends*/ p.paper_id AS p_pid, 
    a.paper_id AS a_pid, 
    a.question_id AS a_qid, 
    a.answer_val AS a_val, 
    qd.question_id AS qd_qid, 
    qd.depends_question_id AS qd_dqid, 
    qd.depends_answer_val AS qd_val, 
    q.type_id AS q_type 
FROM papers p 
JOIN answers a 
    ON a.paper_id = p.paper_id 
LEFT OUTER JOIN question_depends qd 
    ON a.question_id = qd.depends_question_id 
     AND 
     a.answer_val = qd.depends_answer_val 
LEFT OUTER JOIN questions q 
    ON q.question_id = a.question_id 
WHERE qd.question_id IS NULL 
UNION 
SELECT /*Answers and depends*/ p.paper_id AS p_pid, 
    a.paper_id AS a_pid, 
    a.question_id AS a_qid, 
    a.answer_val AS a_val, 
    qd.question_id AS qd_qid, 
    qd.depends_question_id AS qd_dqid, 
    qd.depends_answer_val AS qd_val, 
    q.type_id AS q_type 
FROM papers p 
JOIN answers a 
    ON a.paper_id = p.paper_id 
LEFT OUTER JOIN question_depends qd 
    ON a.question_id = qd.depends_question_id 
     AND 
     a.answer_val = qd.depends_answer_val 
LEFT OUTER JOIN questions q 
    ON q.question_id = qd.question_id 
WHERE qd.question_id IS NOT NULL 
UNION 
SELECT /*Missing answer*/ p.paper_id AS p_pid, 
    a.paper_id AS a_pid, 
    a.question_id AS a_qid, 
    a.answer_val AS a_val, 
    qd.question_id AS qd_qid, 
    qd.depends_question_id AS qd_dqid, 
    qd.depends_answer_val AS qd_val, 
    q.type_id AS q_type 
FROM papers p 
CROSS JOIN question_depends qd 
JOIN questions q 
    ON q.question_id = qd.question_id 
LEFT OUTER JOIN answers a 
    ON a.paper_id = p.paper_id 
     AND 
     a.question_id = qd.depends_question_id 
     AND 
     a.answer_val = qd.depends_answer_val 
WHERE a.question_id IS NULL 
ORDER BY 1, 7 DESC; 

Sie können das Ergebnis auf SQL Fiddle überprüfen http://sqlfiddle.com/#!9/fbd3a9/3