Question

我有以下查询。它返回的结果是非常不受欢迎的：

select stato, (count(1) *100) / sum(1)
from LOG_BONIFICA
group by stato;

它为所有行返回100

Answer 1

在您的情况下，

count(1)等于sum(1)。

尝试这样的事情：

18:39:36 SYSTEM@dwal> ed
Wrote file S:\\tools\buffer.sql

  1  select owner,
  2         count(*) group_cnt,
  3         sum(count(*)) over() total_cnt,
  4         round(100*(count(*) / sum(count(*)) over ()),2) perc
  5    from dba_objects
  6   group by owner
  7*  order by 4 desc
18:39:57 SYSTEM@dwal> /

OWNER                           GROUP_CNT  TOTAL_CNT       PERC
------------------------------ ---------- ---------- ----------
SYS                                 31609      59064      53.52
PUBLIC                              24144      59064      40.88
XDB                                  1660      59064       2.81
SYSTEM                                597      59064       1.01
WMSYS                                 332      59064        .56
EXFSYS                                312      59064        .53
IRKAZDATA                             158      59064        .27
STRMADMIN                              92      59064        .16
DBSNMP                                 55      59064        .09
RI                                     25      59064        .04
PASS                                   16      59064        .03
POTS                                   19      59064        .03
TI                                     11      59064        .02
STRMODS                                11      59064        .02
OUTLN                                  10      59064        .02
APPQOSSYS                               5      59064        .01
ORACLE_OCM                              8      59064        .01

17 rows selected.

Elapsed: 00:00:00.16

使用ratio_to_report

进行

更新：甚至更简单

18:53:36 SYSTEM@dwal> ed
Wrote file S:\\tools\buffer.sql

  1  select owner,
  2         round(100*ratio_to_report(count(*)) over (), 2) perc
  3    from dba_objects
  4   group by owner
  5*  order by 2 desc
18:54:03 SYSTEM@dwal> /

OWNER                                PERC
------------------------------ ----------
SYS                                 53.52
PUBLIC                              40.88
XDB                                  2.81
SYSTEM                               1.01
WMSYS                                 .56
EXFSYS                                .53
IRKAZDATA                             .27
STRMADMIN                             .16
DBSNMP                                .09
RI                                    .04
PASS                                  .03
POTS                                  .03
TI                                    .02
STRMODS                               .02
OUTLN                                 .02
APPQOSSYS                             .01
ORACLE_OCM                            .01

17 rows selected.

Elapsed: 00:00:00.20

Answer 2

您指的是以下内容吗？你需要单独找到一些。

参考SQLFIDDLE

表：

ID  NAME    AMOUNT  STATE
1   john    1000    fl
2   jane    5000    ga
3   james   2000    ca
4   tom     6000    ga
5   tim     8000    fl
6   jim     2000    ga
7   kate    8000    fl
8   jack    3000    tx

结果1：

STATE   SUM(AMOUNT)
ca      2000
fl      17000
ga      13000
tx      3000

最终查询：

select t1.state, concat(
  round(((sum(t1.amount)/t2.total))*100,2),
'%') as pct
from t1, 
(select sum(amount) as total from t1) as t2
group by t1.state
;

结果：

   STATE    COUNT(T1.ID)    SUM(T1.AMOUNT)  TOTAL   PCT
   ca       1               2000           35000    5.71%
   fl       3               17000          35000    48.57%
   ga       3               13000          35000    37.14%
   tx       1               3000           35000    8.57%

Answer 3

使用ratio_to_report分析函数：

SELECT STATO, 
    COUNT(1) STATO_COUNT, 
    RATIO_TO_REPORT(COUNT(1)) OVER() * 100 STATO_PERCENT
FROM LOG_BONIFICA
GROUP BY STATO

所以你不需要自己计算总行数和比率。

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