複数行のデータを別々の列で同じ行に表示するにはどうすればいいですか？

Question

テーブル名はID_TXNです。ここでRECONCILE_SL列には複数のtxnがあり、1txnには複数のcontra txnがあります。 。

      My Table Data 

| RECONCILE_SL | AMOUNT_DR | AMOUNT_CR | TR_REF_NO | TRN_DATE | 
| ------------ | --------- | --------- | ----------- | ----------| 
| 280874  | 0   | 200000 | 1703700164 | 07-Feb-17 | 
| 280874  | 0   | 200000 | 1703900110 | 09-Feb-17 | 
| 280874  | 0   | 200000 | 1704400070 | 14-Feb-17 | 
| 280874  | 200000 | 0   | AIBIF170137 | 07-Feb-17 | 
| 280874  | 400000 | 0   | AIBIF170149 | 09-Feb-17 | 

          My Desire Result 

| TRN_DATE | TR_REF_NO | AMOUNT_DR | AMOUNT_CR | ContraTRN_DATE | ContraTR_REF_NO | Contra AMOUNT_DR | Contra AMOUNT_CR | RECONCILE_SL | 
|-----------|-------------|-----------|-----------|----------------|-----------------|------------------|------------------|--------------| 
| 07-Feb-17 | AIBIF170137 | 200000 |   | 07-Feb-17  | 1703700164  |     | 200000   | 280874  | 
| 09-Feb-17 | AIBIF170149 | 400000 |   | 09-Feb-17  | 1703900110  |     | 200000   | 280874  | 
|   |    |   |   | 14-Feb-17  | 1704400070  |     | 200000   | 280874  | 

Answer 1

何かのように、あなたのデータを結合する機能row_number()を使用して人工配列（rn）を追加し、自己：

select reconcile_sl, 
     dr.trn_date date_dr, dr.amount_dr, dr.tr_ref_no ref_no_dr, 
     cr.trn_date date_cr, cr.amount_cr, cr.tr_ref_no ref_no_cr 
    from (
     select t.*, row_number() over (partition by reconcile_sl order by tr_ref_no) rn 
     from id_txn t where amount_dr > 0) dr 
    full join (
     select t.*, row_number() over (partition by reconcile_sl order by tr_ref_no) rn 
     from id_txn t where amount_cr > 0) cr 
    using (reconcile_sl, rn) 

テスト：

create table id_txn (RECONCILE_SL number(8), AMOUNT_DR number(8), AMOUNT_CR number(8), 
        TR_REF_NO varchar2(15), TRN_DATE date); 

insert into id_txn values(280874,  0, 200000, '1703700164', date '2017-02-07'); 
insert into id_txn values(280874,  0, 200000, '1703900110', date '2017-02-09'); 
insert into id_txn values(280874,  0, 200000, '1704400070', date '2017-02-14'); 
insert into id_txn values(280874, 200000,  0, 'AIBIF170137', date '2017-02-07'); 
insert into id_txn values(280874, 400000,  0, 'AIBIF170149', date '2017-02-09'); 

結果：

RECONCILE_SL DATE_DR  AMOUNT_DR REF_NO_DR  DATE_CR  AMOUNT_CR REF_NO_CR 
------------ ----------- --------- --------------- ----------- --------- --------------- 
     280874 2017-02-07  200000 AIBIF170137  2017-02-07  200000 1703700164 
     280874 2017-02-09  400000 AIBIF170149  2017-02-09  200000 1703900110 
     280874          2017-02-14  200000 1704400070 

Answer 2

は、ここで私は私の解決策（クエリ）私たちは高品質な質問と回答を気にStackOverflowで

SELECT * 
FROM (SELECT acc_no 
      ,dr.trn_date ledger_trn_date 
      ,dr.tr_ref_no tr_ref_no 
      ,dr.amount_dr 
      ,dr.amount_cr 
      ,cr.trn_date contra_trn_date 
      ,cr.tr_ref_no contra_tr_ref_no 
      ,cr.amount_dr contra_amount_dr 
      ,cr.amount_cr contra_amount_cr 
      ,rec_category 
      ,reconcile_sl 
      ,process_user 
      ,type_of_recons 
     FROM (SELECT t.*, ROW_NUMBER() OVER (PARTITION BY reconcile_sl ORDER BY tr_ref_no) rn 
       FROM ID_TXN t 
       WHERE acc_type = 'S') dr 
       FULL JOIN (SELECT t.*, ROW_NUMBER() OVER (PARTITION BY reconcile_sl ORDER BY tr_ref_no) rn 
          FROM ID_TXN t 
          WHERE acc_type = 'N') cr 
         USING (reconcile_sl 
          ,acc_no 
          ,rec_category 
          ,process_user 
          ,type_of_recons 
          ,rn)) m 
WHERE acc_no = '101974' 
     AND reconcile_sl = 280874 
     AND EXISTS 
       (SELECT 1 
        FROM ID_TXN h 
        WHERE trn_date BETWEEN :day_1 AND :day_2 
        AND m.reconcile_sl = h.reconcile_sl) 
ORDER BY reconcile_sl, ledger_trn_date, contra_trn_date 

Result: 

reconcile_sl ledger_trn_date ledger_tr_ref_no amount_dr amount_cr contra_trn_date contra_tr_ref_no contra_amount_dr contra_amount_dr rec_category process_user type_of_recons 
------------ --------------- ---------------- --------- ---------- --------------- ---------------- ---------------- ---------------- ------------ ------------ -------------- 
     280874  07/feb/17  aibif170137  200000   0  07/feb/17  1703700164    0   200000    s-n   saleh     m 
     280874  09/feb/17  aibif170149  400000   0  09/feb/17   1703900110    0   200000    s-n   saleh     m 
     280874                 14/feb/17   1704400070    0   200000    s-n   saleh     m