如何将Oracle中的varchar列分成三列

Question

我有一个地址字段，它可以容纳120个字符，并且需要将它分成三个不同的列，每个字符长度为40个字符。

实施例：

Table name: Address 
Column name: Street_Address 
Select Street_Address * from Address 

输出： 123 Main St North Pole Factory 44, near the rear entrance cross the street and turn left and keep walking straight.

我需要这个地址分成address_1 address_2 和address_3。

所有这三个地址都是varchar(40)数据类型。

所以结果应该是这样的：

Address_1 
152 Main st North Pole Factory 44, near 

Address_2 
the rear entrance cross the street and 

Address_3 
turn left and keep walking straight. 

请注意，每个地址字段可能需要长达40个字符，必须是整个单词，不能在半截断，左无意义。

我正在使用oracle 11i数据库。

Answer 1

这相当“快速和肮脏”，但我认为它给出了正确的结果。
我用流水线表，但也许它可以没有它做...

Here is a sqlfiddle demo

create table t1(id number, adr varchar2(120)) 
/
insert into t1 values(1, '152 Main st North Pole Factory 44, near the rear entrance cross the street and turn left and keep walking straight.') 
/
insert into t1 values(2, '122 Main st Pole Factory 44, near the rear entrance cross the street and turn left and keep walking straight. asdsa') 
/

create or replace type t is object(id number, phrase1 varchar2(40), phrase2 varchar2(40), phrase3 varchar2(40)) 
/
create or replace type t_tab as table of t 
/

create or replace function split_string(id number, str in varchar2) return t_tab 
    pipelined is 

    v_token varchar2(40); 
    v_token_i number := 0; 
    v_cur_len number := 0; 
    v_res_str varchar2(121) := str || ' '; 
    v_p1  varchar2(40); 
    v_p2  varchar2(40); 
    v_p3  varchar2(40); 
    v_p_i  number := 1; 

begin 

    v_token_i := instr(v_res_str, ' '); 

    while v_token_i > 0 loop 

    v_token := substr(v_res_str, 1, v_token_i - 1); 

     if v_cur_len + length(v_token) < 40 then 

     if v_p_i = 1 then 
      v_p1 := v_p1 || ' ' || v_token; 
     elsif v_p_i = 2 then 
      v_p2 := v_p2 || ' ' || v_token; 
     elsif v_p_i = 3 then 
      v_p3 := v_p3 || ' ' || v_token; 
     end if; 

     v_cur_len := v_cur_len + length(v_token) +1; 
    else 
     v_p_i := v_p_i + 1; 

     if v_p_i = 2 then 
      v_p2 := v_p2 || ' ' || v_token; 
     elsif v_p_i = 3 then 
      v_p3 := v_p3 || ' ' || v_token; 
     end if; 

     v_cur_len := length(v_token); 

    end if; 

    v_res_str := substr(v_res_str, v_token_i + 1); 
    v_token_i := instr(v_res_str, ' '); 

    end loop; 

    pipe row(t(id, v_p1, v_p2, v_p3)); 
    return; 
end split_string; 
/

和查询：

select parts.*, length(PHRASE1), length(PHRASE2), length(PHRASE3) 
from t1, table(split_string(t1.id, t1.adr)) parts 

Answer 2

你可以使用递归子查询分解（递归CTE ）：

with s (street_address, line, part_address, remaining) as (
    select street_address, 0 as line, 
    null as part_address, street_address as remaining 
    from address 
    union all 
    select street_address, line + 1 as line, 
    case when length(remaining) <= 40 then remaining else 
     substr(remaining, 1, instr(substr(remaining, 1, 40), ' ', -1, 1)) end 
     as part_address, 
    case when length(remaining) <= 40 then null else 
     substr(remaining, instr(substr(remaining, 1, 40), ' ', -1, 1) + 1) end 
     as remaining 
    from s 
) 
cycle remaining set is_cycle to 'Y' default 'N' 
select line, part_address 
from s 
where part_address is not null 
order by street_address, line; 

哪个机智^ h你的数据得出：

 LINE PART_ADDRESS       
---------- ---------------------------------------- 
     1 152 Main st North Pole Factory 44, near 
     2 the rear entrance cross the street and 
     3 turn left and keep walking straight.  

SQL Fiddle demo有两个地址。

您也可以将这些部分值转换为列，我认为这是您的最终目标，例如，作为看法：

create or replace view v_address as 
with cte (street_address, line, part_address, remaining) as (
    select street_address, 0 as line, 
    null as part_address, street_address as remaining 
    from address 
    union all 
    select street_address, line + 1 as line, 
    case when length(remaining) <= 40 then remaining else 
     substr(remaining, 1, instr(substr(remaining, 1, 40), ' ', -1, 1)) end 
     as part_address, 
    case when length(remaining) <= 40 then null else 
     substr(remaining, instr(substr(remaining, 1, 40), ' ', -1, 1) + 1) end 
     as remaining 
    from cte 
) 
cycle remaining set is_cycle to 'Y' default 'N' 
select street_address, 
    cast (max(case when line = 1 then part_address end) as varchar2(40)) 
    as address_1, 
    cast (max(case when line = 2 then part_address end) as varchar2(40)) 
    as address_2, 
    cast (max(case when line = 3 then part_address end) as varchar2(40)) 
    as address_3 
from cte 
where part_address is not null 
group by street_address; 

Another SQL Fiddle。

值得注意的是，如果street_address长度接近120个字符，它可能不适合整齐地放入3个40个字符的块中 - 根据缠绕到下一个“行”的单词的长度， 。这种方法会生成超过3行，但视图只使用前三行，因此您可能会丢失地址的末尾。您可能希望使字段更长，或者对于这些情况有address_4 ...