Question

假设您有两个带有替换的表，这些替换必须保持原样，而另一个表包含一个名称主体。我如何获得所有可能的替代品？

Substitution Table
--------------------------------------
word        subs_list
MOUNTAIN    MOUNTAIN, MOUNT, MT, MTN
HOUSE       HAUS, HOUSE
VIEW        VU, VIEW

Synonyms table
-------------------------------------------------
EDUCATION   SCHOOL, UNIVERSITY, COLLEGE, TRAINING
FOOD        STORE, FOOD, CAFE
STORE       FOOD, STORE, MARKET
REFRIGERATION   FOODLOCKER, FREEZE, FRIDGE

names table
------------------------------------------------
MOUNT VU FOOD USA 
MOUNTAIN VU STORE CA

注意：我知道只有一个替换表是可取的，但是两个替换表都必须保留，因为它们除了用于上述目的外，还用于其他目的，这些表已经被使用。此外，两个表中的替换列表只是一个varchar，其字符串用逗号分隔

考虑到前一个问题，问题是生成可能通过替换派生的名称。例如，名称MOUNT VU FOOD USA应当分解为MOUNTAIN VIEW FOOD USA和MOUNTAIN VIEW STORE USA，第二种方法将采用相同的方式。

我已经能够以错误的顺序获得替换品，并且全部起作用，有一种方法获得替换后生成的具有不同名称的输出数组？到目前为止，我已经创建了该函数进行替换：

create or replace function replace_companies_array(i_sentence IN VARCHAR) returns VARCHAR[] AS $p_replaced$
DECLARE
  p_replaced VARCHAR[];
  subs RECORD;
  flag boolean:= True;
  cur_s CURSOR(i_sentence VARCHAR)
    FOR SELECT w.input, coalesce(x.word, w.input) as word, count(*) OVER (PARTITION BY w.input) as counter
    FROM regexp_split_to_table(trim(i_sentence), '\s') as w(input) 
     LEFT JOIN (
      select s.word, trim(s1.token) as token
      from subs01 s
       cross join unnest(string_to_array(s.subs_list, ',')) s1(token)
      union
      select sy.word, trim(s2.token) as token
      from syns01 sy
       cross join unnest(string_to_array(sy.syn_list, ',')) s2(token)
     ) as x on lower(trim(w.input)) = lower(x.token)
   order by counter;
BEGIN
  OPEN cur_s(i_sentence);

  LOOP
   --fetch row into the substitutions
     FETCH cur_s INTO subs;

   --Exit when no more rows to fetch
     EXIT WHEN NOT FOUND;

     SELECT REGEXP_REPLACE(i_sentence,'(^|[^a-z0-9])' || subs.input || '($|[^a-z0-9])','\1' || UPPER(subs.word) || '\2','g')
     INTO i_sentence; 

  END LOOP;
  p_replaced:=array_append(p_replaced, i_sentence);

  RETURN p_replaced;

END;
$p_replaced$ LANGUAGE plpgsql;

非常感谢您的贡献

Answer 1

我没有得到最终结果，但是我很接近它！

从句子MOUNT VU FOOD USA中，我得到{"MOUNTAIN VIEW MARKET USA","MOUNTAIN VIEW STORE USA","MOUNTAIN VIEW CAFE USA","MOUNTAIN VIEW FOOD USA"}

这是我所有用于重新创建同义词和替代表的脚本：

DROP TABLE IF EXISTS subs01;
DROP TABLE IF EXISTS syns01;
CREATE TABLE subs01 (word VARCHAR(20), subs_list VARCHAR(200));
CREATE TABLE syns01 (word VARCHAR(20), syn_list VARCHAR(200));

INSERT INTO subs01 (word, subs_list) VALUES ('MOUNTAIN', 'MOUNTAIN, MOUNT, MT, MTN'),('HOUSE', 'HAUS, HOUSE'),('VIEW', 'VU, VIEW');
INSERT INTO syns01 (word, syn_list) VALUES ('EDUCATION', 'SCHOOL, UNIVERSITY, COLLEGE, TRAINING'),('FOOD', 'STORE, FOOD, CAFE'),('STORE', 'FOOD, STORE, MARKET'),('REFRIGERATION', 'FOODLOCKER, FREEZE, FRIDGE');

我决定将工作分为两个阶段：

用以下单词代替：

CREATE OR REPLACE function substitute_words (i_sentence IN VARCHAR) returns VARCHAR AS $p_substituted$
DECLARE
  --p_substituted VARCHAR;
  subs_cursor CURSOR FOR select su.word, trim(s2.token) as token from subs01 su cross join unnest(string_to_array(su.subs_list, ',')) s2(token);
  subs_record record;
BEGIN
  OPEN subs_cursor;
  LOOP
    FETCH subs_cursor INTO subs_record;
    EXIT WHEN NOT FOUND;
    RAISE NOTICE 'INFO : TOKEN (%) ',subs_record.token ;

    IF i_sentence LIKE '%'|| subs_record.token || '%' THEN
      RAISE NOTICE '-- FOUND : TOKEN (%) ',subs_record.token ;
      SELECT replace (i_sentence, subs_record.token, subs_record.word) INTO i_sentence;
    END IF;
  END LOOP;
  CLOSE subs_cursor;
  RETURN i_sentence;
END
$p_substituted$ LANGUAGE plpgsql;

用公理代替已知单词：

CREATE OR REPLACE function synonymize_sentence (i_sentence IN VARCHAR) returns TABLE (sentence_result VARCHAR) AS $p_syn$
DECLARE
  syn_cursor CURSOR FOR select su.word, trim(s2.token) as token from syns01 su cross join unnest(string_to_array(su.syn_list, ',')) s2(token);
  syn_record record;
BEGIN
  CREATE TEMPORARY TABLE record_syn (result VARCHAR(200)) ON COMMIT DROP;
  INSERT INTO record_syn (result) SELECT i_sentence;
  OPEN syn_cursor;
  LOOP
    FETCH syn_cursor INTO syn_record;
    EXIT WHEN NOT FOUND;
    RAISE NOTICE 'INFO : WORD (%) ',syn_record.word ;

    INSERT INTO record_syn (result) SELECT replace (result, syn_record.word, syn_record.token) FROM record_syn where result LIKE '%'||     syn_record.word || '%';

  END LOOP;
  CLOSE syn_cursor;
  RETURN QUERY SELECT distinct result FROM record_syn;
END;
$p_syn$ LANGUAGE plpgsql;

然后，要生成结果数组，请执行以下语句：

SELECT ARRAY(SELECT synonymize_sentence (substitute_words ('MOUNT VU FOOD USA')));

Postgresql中的字符串替换产生附加字符串数组

1 个答案: