如何在Oracle SQL中选择子字符串，直到第4个字符类型

Question

我试图在oracle SQL中找到一种方法来获取第4个＆＃34; /＆＃34;之后的所有文本。并计算有多少匹配。我需要在一个SQL语句中执行此操作..

table name: access_log
col name: download
col name: time-stamp
value: Download file:/webdocs/data/groupXXX/case/03_28_54_9_0000011856.pdf

我想结束

case/03_28_54_9_0000011856.pdf

然后计算多少场比赛？可以这样做吗？

Answer 1

我会这样做：

SELECT
    SUBSTR(DOWNLOAD, INSTR(DOWNLOAD, '/', 1, 4) + 1) AS FILENAME,
COUNT(SUBSTR(DOWNLOAD, INSTR(DOWNLOAD, '/', 1, 4) + 1)) AS DOWNLOADS
FROM ACCESS_LOG
GROUP BY SUBSTR(DOWNLOAD, INSTR(DOWNLOAD, '/', 1, 4) + 1)
ORDER BY DOWNLOADS DESC;

返回此结果集：

|                       FILENAME | DOWNLOADS |
|--------------------------------|-----------|
| case/03_28_54_9_0000011856.pdf |         3 |
| case/04_28_54_9_0000011856.pdf |         2 |

有关演示，请参阅此处：SQL Fiddle

Answer 2

试

SELECT SUBSTR(download, INSTR(download, '/', 1, 4)+1)  FROM access_log

获得总尝试次数：

SELECT COUNT(*) , T1.file_name  
FROM (SELECT SUBSTR(download, INSTR(download, '/', 1, 4)+1) as file_name  
           FROM access_log) T1
GROUP BY T1.file_name  ;