計數順序匹配的單詞

Question

我想返回字的順序匹配的兩個字符串 例數的查詢：

表

Id column1    column2  result 
1 'foo bar live'  'foo bar'  2 
2 'foo live tele'  'foo tele'  1 
3 'bar foo live'  'foo bar live' 0 

拿到發生的總數我使用：

select id, column1,column2, 
extractvalue(dbms_xmlgen.getxmltype('select cardinality (
    sys.dbms_debug_vc2coll(''' || replace(lower(column1), ' ', ''',''') || ''') multiset intersect 
    sys.dbms_debug_vc2coll('''||replace(lower(column2), ' ', ''',''')||''')) x from dual'), '//text()') cnt 
from table. 

任何人都可以請建議在類似的線路上進行順序匹配查詢，因爲我想要的順序匹配數和出現次數一起顯示。

Answer 1

就個人而言，在這種情況下，我會選擇普通SQL的PL/SQL代碼。喜歡的東西：

包裝規格：

create or replace package PKG is 
    function NumOfSeqWords(
    p_str1 in varchar2, 
    p_str2 in varchar2 
) return number; 
end; 

包體：

create or replace package body PKG is 
    function NumOfSeqWords(
    p_str1 in varchar2, 
    p_str2 in varchar2 
) return number 
    is 
    l_str1  varchar2(4000) := p_str1; 
    l_str2  varchar2(4000) := p_str2; 
    l_res  number default 0; 
    l_del_pos1 number; 
    l_del_pos2 number; 
    l_word1 varchar2(1000); 
    l_word2 varchar2(1000); 
    begin 
    loop 
     l_del_pos1 := instr(l_str1, ' '); 
     l_del_pos2 := instr(l_str2, ' '); 
     case l_del_pos1 
     when 0 
     then l_word1 := l_str1; 
      l_str1 := ''; 
     else l_word1 := substr(l_str1, 1, l_del_pos1 - 1); 
     end case; 
     case l_del_pos2 
     when 0 
     then l_word2 := l_str2; 
      l_str2 := ''; 
     else l_word2 := substr(l_str2, 1, l_del_pos2 - 1); 
     end case; 
     exit when (l_word1 <> l_word2) or 
       ((l_word1 is null) or (l_word2 is null)); 

     l_res := l_res + 1; 
     l_str1 := substr(l_str1, l_del_pos1 + 1); 
     l_str2 := substr(l_str2, l_del_pos2 + 1); 
    end loop; 
    return l_res; 
    end; 
end; 

測試用例：

with t1(Id1, col1, col2) as(
    select 1, 'foo bar live' ,'foo bar'  from dual union all 
    select 2, 'foo live tele' ,'foo tele' from dual union all 
    select 3, 'bar foo live' ,'foo bar live'from dual 
) 
    select id1 
     , col1 
     , col2 
     , pkg.NumOfSeqWords(col1, col2) as res 
    from t1 
    ; 

結果：

 ID1 COL1   COL2    RES 
---------- ------------- ------------ ---------- 
     1 foo bar live foo bar    2 
     2 foo live tele foo tele    1 
     3 bar foo live foo bar live   0 

Answer 2

爲什麼放棄查詢的方法。我知道這是一個有點複雜，我希望有人可以對其進行處理，以改善它，但在我的業餘時間，我能夠調用的一個下午生存這方面的工作...

這裏就SQLFidlle

SELECT Table1.id, 
     Table1.column1, 
     Table1.column2, 
     max(nvl(t.l,0)) RESULT 
FROM (
    SELECT id, 
      column1, 
      column2, 
      LEVEL l, 
      decode(LEVEL, 
        1, 
       substr(column1, 1, instr(column1,' ', 1, LEVEL) -1), 
       substr(column1, 1, (instr(column1,' ', 1, LEVEL))) 
       ) sub1, 
      decode(LEVEL, 
        1, 
       substr(column2, 1, instr(column2,' ', 1, LEVEL) -1), 
       substr(column2, 1, (instr(column2,' ', 1, LEVEL))) 
       ) sub2 

    FROM (SELECT id, 
        column1 || ' ' column1, 
        column2 || ' ' column2 
      FROM Table1) 
    WHERE decode(LEVEL, 
         1, 
        substr(column1, 1, instr(column1,' ', 1, LEVEL) -1), 
        substr(column1, 1, (instr(column1,' ', 1, LEVEL))) 
       ) = 
      decode(LEVEL, 
         1, 
        substr(column2, 1, instr(column2,' ', 1, LEVEL) -1), 
        substr(column2, 1, (instr(column2,' ', 1, LEVEL))) 
       ) 
    START WITH column1 IS NOT NULL 
    CONNECT BY instr(column1,' ', 1, LEVEL) > 0 
) t 
RIGHT OUTER JOIN Table1 ON trim(t.column1) = Table1.column1 
         AND trim(t.column2) = Table1.column2 
         AND t.id = Table1.id 
GROUP BY Table1.id, 
      Table1.column1, 
      Table1.column2 
ORDER BY max(nvl(t.l,0)) DESC