sql - 选择分隔字符串作为 Oracle sql 中的表

Question

我有一个像这样的字符串:

"Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4"

我正在编写一个选择查询来选择它作为表格，例如:

Width | Height | Length
-----------------------
10    |  25    |   35  
7     |  5     |   6   
20    |  6     |   3  
45    |  45    |   4  

如果您需要更多信息，请发表评论。

Answer 1

此解决方案适用于任意数量的列(宽度、高度……)和值。

-- your test data  
with data(val) as
 (select 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' from dual),

-- split by ,
cols as
 (select regexp_substr(str, '[^,]+', 1, level) val
    from (select val as str from data)
  connect by regexp_substr((select val as str from data),
                           '[^,]+',
                           1,
                           level) is not null),

-- split by :
hdr_and_cols as
 (select substr(val, 1, instr(val, ':') - 1) as hdr,
         substr(val, instr(val, ':') + 1) as val
    from cols),

-- split by |
hdr_lvl_vals as
 (select distinct x.hdr,
                  level as entry,
                  regexp_substr(x.val, '[^|]+', 1, level) as val
    from hdr_and_cols x
  connect by regexp_substr(x.val, '[^|]+', 1, level) is not null)

select * from hdr_lvl_vals;

结果:

hdr     entry   value
---------------------
Height  1       25
Height  2       5
Height  3       6
Height  4       45
Length  1       35
Length  2       6
Length  3       3
Length  4       4
Width   1       10
Width   2       7
Width   3       20
Width   4       45

您可以按照您喜欢的方式格式化结果，例如

-- your test data  
with data(val) as
 (select 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' from dual),

-- split by ,
cols as
 (select regexp_substr(str, '[^,]+', 1, level) val
    from (select val as str from data)
  connect by regexp_substr((select val as str from data),
                           '[^,]+',
                           1,
                           level) is not null),

-- split by :
hdr_and_cols as
 (select substr(val, 1, instr(val, ':') - 1) as hdr,
         substr(val, instr(val, ':') + 1) as val
    from cols),

-- split by |
hdr_lvl_vals as
 (select distinct x.hdr,
                  level as entry,
                  regexp_substr(x.val, '[^|]+', 1, level) as val
    from hdr_and_cols x
  connect by regexp_substr(x.val, '[^|]+', 1, level) is not null)

-- format output
select w.val as width, h.val as heigth, l.val as length
  from (select entry, val from hdr_lvl_vals where hdr = 'Width') w,
       (select entry, val from hdr_lvl_vals where hdr = 'Height') h,
       (select entry, val from hdr_lvl_vals where hdr = 'Length') l,
       (select level as entry
          from dual
        connect by level <= (select max(entry) from hdr_lvl_vals)) r
 where r.entry = w.entry
   and r.entry = h.entry
   and r.entry = l.entry;

输出:

WIDTH   HEIGTH  LENGTH
--------------------
10      25      35
7       5       6
20      6       3
45      45      4