Question

我有一个字符串。让我们说：“abcdea” 我必须找出使用匿名pl / sql块的字符串中最常用的字母。

我该怎么做？

Answer 1

也许不是很优雅，但仍在工作：

declare  
 str varchar2(6 char) := 'abcdea';  
 alphabet varchar2(26 char) := 'abcdefghijklmnopqrstuvwxyz';  
 most_frequent char := '';  
 max_count INTEGER := 0;  
 v_count INTEGER := 0;  
 checking CHAR := ''; 
begin   
 for i in 1..26 loop
        checking := SUBSTR(alphabet, i, 1);
        for k in 1..length(str) 
        loop
          if SUBSTR(str, k, 1) = checking THEN
            v_count := v_count + 1; 
          end if;
        end loop;
        if v_count > max_count THEN
          max_count := v_count;
          most_frequent := checking;
        end if;
        v_count :=0;   
end loop;   
 dbms_output.put_line('Most frequent letter is "'|| most_frequent||'"'); 
end;

Answer 2

关联数组非常便于存储计数：

DECLARE
    TYPE t_counts IS TABLE OF PLS_INTEGER INDEX BY VARCHAR2(1);
    --
    v_counts t_counts;
    v_string VARCHAR2(100) := 'the quick brown fox jumped over the lazy dog';
    v_char   VARCHAR2(1);
BEGIN
    FOR i IN 1 .. length( v_string ) LOOP
        v_char := substr( v_string, i, 1 );
        IF NOT v_counts.exists( v_char ) THEN
            v_counts( v_char ) := 0;
        END IF;
        v_counts( v_char ) := v_counts( v_char ) + 1;
    END LOOP;
END;
/

在数组中查找具有最高计数的索引将是微不足道的添加

Answer 3

创建表

create table charcount ( c char(1) )

将字符串拆分为单个字符，并使用PL / SQL将它们插入此表中。答案就是第一行：

select c, count(1)
from charcount
group by c
order by count(1) desc

Answer 4

以下是使用几种SYS类型的解决方案：

declare
  str varchar2(6) := 'abcdea';
  tab AWRDRPT_TEXT_TYPE_TABLE := AWRDRPT_TEXT_TYPE_TABLE();
begin
  for i in 1..length(str) loop
     tab.extend(1);
     tab(i) := AWRDRPT_TEXT_TYPE (substr(str,i,1));
  end loop;
  for r in (select output, count(*) as cnt
            from table(tab)
            group by output
            order by cnt desc)
  loop
     dbms_output.put_line(r.output);
     exit;
  end loop;
end;

Answer 5

在纯SQL上没有PL / SQL就可以了

declare
    l_string varchar2(1000) := 'asddafsbaterbtretbbfdsbgdfsgnrtynrthfhfghntumfyuy';
    l_char   char(1);
begin
    with t as
     (select l_string as str from dual)
    select c
      into l_char
      from (select substr(str, level, 1) as c, count(*) as cnt
              from t
            connect by level <= length(str)
             group by substr(str, level, 1)
             order by 2 desc)
     where rownum < 2;
    dbms_output.put_line('Most used char is ' || l_char);
exception
    when no_data_found then
        dbms_output.put_line('Sorry string is empty');
end;

PL / SQL - 使用字符串！

5 个答案: