我需要在句子或字符串中反转单词的位置.
For example : "Hello World! I Love StackOverflow", to be displayed as "StackOverflow Love I World! Hello".
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可以用一个SQL吗?字长不大于VARCHAR2(4000)Oracle VARCHAR2表列中的最大长度支持.
我得到了用于反转字符串的解决方案(字符按相反顺序)
基于XML的版本,以避免定义自己的功能; 需要11克listagg():
select listagg(word, ' ') within group (order by rn desc) as reversed
from (
select word, rownum as rn
from xmltable('for $i in ora:tokenize($STR, " ") return $i'
passing 'Hello World! I Love StackOverflow' as str
columns word varchar2(4000) path '.'
)
);
REVERSED
----------------------------------------
StackOverflow Love I World! Hello
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该XMLTable()做的tokenising,并分配行号:
select rownum as rn, word
from xmltable('for $i in ora:tokenize($STR, " ") return $i'
passing 'Hello World! I Love StackOverflow' as str
columns word varchar2(4000) path '.'
);
RN WORD
---------- --------------------
1 Hello
2 World!
3 I
4 Love
5 StackOverflow
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在listagg()随后片它以相反的顺序重新走到一起.
创建一个函数:
REGEXP_SUBSTR('Your text here','[^ ]+', 1, ?)将使用Space作为分隔符从文本中提取单词.Tt在Exception上返回原始String本身!
CREATE OR REPLACE FUNCTION reverse_words (v_STRING IN VARCHAR2)
RETURN VARCHAR2
IS
L_TEMP_TEXT VARCHAR2(4000);
L_FINAL_TEXT VARCHAR2(4000);
V_LOOPCOUNT NUMBER :=0;
T_WORD VARCHAR2(4000);
BEGIN
L_TEMP_TEXT := regexp_replace(V_STRING,'[[:space:]]+',' '); -- Replace multiple spaces as single
LOOP
v_LOOPCOUNT := v_LOOPCOUNT+1;
T_WORD := REGEXP_SUBSTR(L_TEMP_TEXT,'[^ ]+', 1, V_LOOPCOUNT);
L_final_TEXT := T_WORD||' '||L_final_TEXT;
EXIT WHEN T_WORD IS NULL;
END LOOP;
RETURN(TRIM(L_final_TEXT));
EXCEPTION
WHEN OTHERS THEN
DBMS_OUTPUT.PUT_LINE(sqlerrm||chr(10)||dbms_utility.format_error_backtrace);
RETURN V_STRING;
END reverse_words;
/
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样本结果:
你可以打电话 reverse_words(yourcolumn) from your_table
SQL> select reverse_words('Hello World! I Love StackOverflow') "Reversed" from dual;
Reversed
--------------------------------------------------------------------------------
StackOverflow Love I World! Hello
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