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".
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
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
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; /
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