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我有一個難以解決的問題，我不知道該怎麼做。存檔一些繼承的實體
我想歸檔一些實體，然後從初始表中刪除它們。問題是這些實體是連在一起的。我有一個接近工作的代碼，但我認爲這不是一個乾淨的方法。它由SQL查詢組成，這些SQL查詢將具有ID的行復制到新表中。 另一個問題是我不需要一些字段來歸檔，因此歸檔實體與初始實體不完全相同。 由於我的表的大小，我正在使用原始SQL查詢而不是DQL。
我要存檔這些實體：科爾，ColleQC，QC，PasserColle，效應初探，ReponseQC，StatistiqueColle，StatistiqueQuestion，RepartitionColle，RepartitionQuestion，Tuteur TO BanqueColle，BanqueColleQC，BanqueQC，BanquePasserColle，BanqueReponse，BanqueReponseQC，BanqueStatistiqueColle，BanqueStatistiqueQuestion ，BanqueRepartitionColle，BanqueRepartitionQuestion，AncienAdherent。
我會將這些存檔用於我的應用的另一部分。
樣本表結構的：
（Tuteur和AncienAdherent擴展用戶）
這是我做歸檔代碼的一部分，但我不認爲這是一個乾淨的方式來做到這一點：
public function archiveTuteurs() {
$db = $this->em->getConnection();
$query = "INSERT INTO ancien_adherent (id)
SELECT u.id
FROM user u
WHERE discr = 'tuteur'";
$stmt = $db->prepare($query);
$stmt->execute();
$query2 = "UPDATE user
SET user.discr = 'ancien'
WHERE discr = 'tuteur'";
$stmt = $db->prepare($query2);
$stmt->execute();
return true;
}
public function archiveColles() {
$db = $this->em->getConnection();
$query = "INSERT INTO banque_colle (id, typeColle, nom, temps_epreuve, matiere_id, dateCreation, ordre, discr)
SELECT colle.id, colle.typeColle, colle.nom, colle.temps_epreuve, colle.matiere_id, colle.dateCreation, colle.ordre, colle.discr
FROM colle";
$stmt = $db->prepare($query);
$stmt->execute();
$query2 = "INSERT INTO banque_colle_qc (id)
SELECT colle_qc.id
FROM colle_qc";
$stmt = $db->prepare($query2);
$stmt->execute();
return true;
}
public function archiveQC() {
$db = $this->em->getConnection();
$query = "INSERT INTO banque_qc (id, titre, id_colle, ordre, qcPere, enonce, donnees, item1, item2, item3, item4,
item5, corrige_item1, corrige_item2, corrige_item3, corrige_item4, corrige_item5, item1_vrai,
item2_vrai, item3_vrai, item4_vrai, item5_vrai, item1_annule, item2_annule, item3_annule,
item4_annule, item5_annule, multiple_choices, inclu)
SELECT qc.id, qc.titre, qc.id_colle, qc.ordre, qc.qcPere, qc.enonce, qc.donnees, qc.item1, qc.item2,
qc.item3, qc.item4, qc.item5, qc.corrige_item1, qc.corrige_item2, qc.corrige_item3, qc.corrige_item4,
qc.corrige_item5, qc.item1_vrai, qc.item2_vrai, qc.item3_vrai, qc.item4_vrai, qc.item5_vrai,
qc.item1_annule, qc.item2_annule, qc.item3_annule, qc.item4_annule, qc.item5_annule,
qc.multiple_choices, qc.inclu
FROM qc
ORDER BY qc.qcPere ASC";
$stmt = $db->prepare($query);
$stmt->execute();
return true;
}
public function archivePassages() {
$db = $this->em->getConnection();
$query = "INSERT INTO banque_passer_colle (colle_id, dateDebut, note)
SELECT passer_colle.colle_id, passer_colle.dateDebut, passer_colle.note
FROM passer_colle";
$stmt = $db->prepare($query);
$stmt->execute();
return true;
}
public function archiveReponses() {
$db = $this->em->getConnection();
$query = "INSERT INTO banque_reponse (id, discr)
SELECT reponse.id, reponse.discr
FROM reponse
WHERE discr='reponseQC'";
$stmt = $db->prepare($query);
$stmt->execute();
$query2 = "INSERT INTO banque_reponse_qc (id, question, A, B, C, D, E, note)
SELECT reponse_qc.id, reponse_qc.question, reponse_qc.A, reponse_qc.B, reponse_qc.C, reponse_qc.D,
reponse_qc.E, reponse_qc.note
FROM reponse_qc";
$stmt = $db->prepare($query2);
$stmt->execute();
return true;
}
public function archiveStats() {
$db = $this->em->getConnection();
$query = "INSERT INTO banque_statistiquecolle (id, colle_id, effectif, moyenne, mediane, note100, major, minor)
SELECT sc.id, sc.colle_id, sc.effectif, sc.moyenne, sc.mediane, sc.note100, sc.major, sc.minor
FROM statistiquecolle_groupe scg
LEFT JOIN statistiquecolle sc ON sc.id = scg.statistiquecolle_id
WHERE scg.groupe_id = 1
AND sc.id NOT IN (SELECT sc1.id
FROM statistiquecolle_groupe scg1
LEFT JOIN statistiquecolle sc1 ON sc1.id = scg1.statistiquecolle_id
WHERE scg1.groupe_id != 1)";
$stmt = $db->prepare($query);
$stmt->execute();
$query2 = "INSERT INTO banque_statistiquequestion (id, question_id, moyenne, nbReponseTot, nbReponseA, nbReponseB,
nbReponseC, nbReponseD, nbReponseE)
SELECT sq.id, sq.question_id, sq.moyenne, sq.nbReponseTot, sq.nbReponseA, sq.nbReponseB, sq.nbReponseC,
sq.nbReponseD, sq.nbReponseE
FROM statistiquequestion_groupe sqg
LEFT JOIN statistiquequestion sq ON sq.id = sqg.statistiquequestion_id
WHERE sqg.groupe_id = 1
AND sq.id NOT IN (SELECT sq1.id
FROM statistiquequestion_groupe sqg1
LEFT JOIN statistiquequestion sq1 ON sq1.id = sqg1.statistiquequestion_id
WHERE sqg1.groupe_id != 1)";
$stmt = $db->prepare($query2);
$stmt->execute();
$query3 = "INSERT INTO banque_repartitioncolle (id, statColle_id, note, nombre, percentOfEffectif)
SELECT rc.id, rc.statColle_id, rc.note, rc.nombre, rc.percentOfEffectif
FROM repartitioncolle rc
WHERE rc.statColle_id IN (SELECT bsc.id
FROM banque_statistiquecolle bsc)";
$stmt = $db->prepare($query3);
$stmt->execute();
$query4 = "INSERT INTO banque_repartitionquestion (id, statQuestion_id, note, nombre, percentOfEffectif)
SELECT rq.id, rq.statQuestion_id, rq.note, rq.nombre, rq.percentOfEffectif
FROM repartitionquestion rq
WHERE rq.statQuestion_id IN (SELECT bsq.id
FROM banque_statistiquequestion bsq)";
$stmt = $db->prepare($query4);
$stmt->execute();
return true;
}
你看過物化視圖嗎？ http://www.fromdual.com/mysql-materialized-views – Iceman
我不認爲這會解決它。如果我理解得很好，這是加快查詢速度的一種方法。 這裏的主要問題是以一種乾淨的方式將一些具有繼承性的數據轉移到其他表中，我找不到如何去做。 –