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  • oracle 行列转换

    1、固定列数的行列转换

    student subject grade
    --------- ---------- --------
    student1 语文     80
    student1 数学     70
    student1 英语     60
    student2 语文     90
    student2 数学     80
    student2 英语     100
    ……
    转换为
        语文 数学 英语
    student1 80 70 60
    student2 90 80 100
    ……
    语句如下:select student,

    sum(decode(subject,'语文', grade,null)) "语文",
    sum(decode(subject,'数学', grade,null)) "数学",
    sum(decode(subject,'英语', grade,null)) "英语"
    from
    table
    group by student;
     

    2、不定列行列转换

    c1 c2
    --- -----------
    1 我
    1 是
    1 谁
    2 知
    2 道
    3 不
    ……
    转换为
    1 我是谁
    2 知道
    3 不

    这一类型的转换可以借助于PL/SQL来完成,这里给一个例子
    CREATE OR REPLACE FUNCTION
    get_c2(tmp_c1 NUMBER)
    RETURN VARCHAR2
    IS
    Col_c2
    VARCHAR2(4000);
    BEGIN
    FOR cur IN (SELECT c2 FROM t WHERE c1=tmp_c1)
    LOOP
    Col_c2 := Col_c2||cur.c2;
    END LOOP;
    Col_c2 :=
    rtrim(Col_c2,1);
    RETURN Col_c2;
    END;

    select distinct c1 ,get_c2(c1)
    cc2 from table;


    或者不用pl/sql,利用分析函数和 CONNECT_BY 实现:

    SELECT c1,
    SUBSTR (MAX (SYS_CONNECT_BY_PATH (c2, ';')), 2) NAME
        FROM (SELECT c1, c2,
    rn, LEAD (rn) OVER (PARTITION BY c1 ORDER BY rn) rn1
                FROM (SELECT
    c1, c2, ROW_NUMBER () OVER (ORDER BY c2) rn
                        FROM
    t))
    START WITH rn1 IS NULL
    CONNECT BY rn1 = PRIOR rn
    GROUP BY
    c1;


    3、列数不固定(交叉表行列转置)
    这种是比较麻烦的一种,需要借助pl/sql:

    原始数据:
    CLASS1    CALLDATE         CALLCOUNT
    1          2005-08-08         40
    1          2005-08-07         6
    2          2005-08-08         77
    3          2005-08-09         33
    3          2005-08-08         9
    3          2005-08-07         21

    转置后:
    CALLDATE     CallCount1 CallCount2 CallCount3
    ------------ ---------- ---------- ----------
    2005-08-09   0          0          33
    2005-08-08   40         77         9
    2005-08-07  6    0          21

    试验如下:
    1). 建立测试表和数据
    CREATE TABLE t(
        class1 VARCHAR2(2
    BYTE),
        calldate DATE,
        callcount INTEGER
    );

    INSERT INTO
    t(class1, calldate, callcount)
    VALUES ('1', TO_DATE ('08/08/2005',
    'MM/DD/YYYY'), 40);

    INSERT INTO t(class1, calldate, callcount)
    VALUES
    ('1', TO_DATE ('08/07/2005', 'MM/DD/YYYY'), 6);

    INSERT INTO t(class1,
    calldate, callcount)
    VALUES ('2', TO_DATE ('08/08/2005', 'MM/DD/YYYY'),
    77);

    INSERT INTO t(class1, calldate, callcount)
    VALUES ('3', TO_DATE
    ('08/09/2005', 'MM/DD/YYYY'), 33);

    INSERT INTO t(class1, calldate,
    callcount)
    VALUES ('3', TO_DATE ('08/08/2005', 'MM/DD/YYYY'),
    9);

    INSERT INTO t(class1, calldate, callcount)
    VALUES ('3', TO_DATE
    ('08/07/2005', 'MM/DD/YYYY'), 21);

    COMMIT ;

    2). 建立ref
    cursor准备输出结果集
    CREATE OR REPLACE PACKAGE pkg_getrecord
    IS
        TYPE
    myrctype IS REF CURSOR;
    END pkg_getrecord;
    /

    3). 建立动态sql交叉表函数,输出结果集

    CREATE OR REPLACE FUNCTION fn_rs
        RETURN
    pkg_getrecord.myrctype
    IS
        s VARCHAR2 (4000);
        CURSOR c1 IS
     
      SELECT ',sum(case when Class1='
                || class1
                || '
    then CallCount else 0 end)'
                || ' "CallCount'
                ||
    class1
                || '"' c2
        FROM t
        GROUP BY class1;
        r1
    c1%ROWTYPE;
        list_cursor pkg_getrecord.myrctype;
    BEGIN
        s :=
    'select CallDate ';
        OPEN c1;
        LOOP
            FETCH c1 INTO
    r1;
            EXIT WHEN c1%NOTFOUND;
            s := s || r1.c2;
        END
    LOOP;
        CLOSE c1;
        s := s || ' from T group by CallDate order by
    CallDate desc ';
        OPEN list_cursor FOR s;
        RETURN
    list_cursor;
    END fn_rs;
    /

    4). 测试在sql plus下执行:
    var results
    refcursor;
    exec :results := fn_rs;
    print results;
    CALLDATE    
       CallCount1 CallCount2 CallCount3
    --------------- ---------- ---------- ----------
    2005-08-09      0          0          33
    2005-08-08      40         77         9
    2005-08-07      6          0          21

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  • 原文地址:https://www.cnblogs.com/baiyixianzi/p/2663783.html
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