今天有人问了个关于oracle字符串截取和拼接的问题，让我帮他写出sql，看了下问题描述还比较清晰就试着解决下，利用午休时间把功能实现了，，问题看似不难，但思路一定要清晰，不然就乱了，关键大量应用了oracle的substr 和instr函数，下面贴出问题和脚本：
问题：sql中一个字段值为：1788987565327、768374872394903、21437238740213483874629、23412341234252345。其中顿号间隔的每一组数字位数和尾数不定，现在要使前面这个字段值中顿号前的数字尾数即7、3、9、5都分别加1，变成8、4、0、6输出成1788987565328、768374872394904、21437238740213483874620、23412341234252346。注意其中第三个数，从9加1后，输出成0，而不是10
脚本：
declare targetstr varchar2(2000); strlength number; position number; maxposition number; retrunstr varchar2(2000); tempstr varchar2(2000); endstr number;begin targetstr := '1788987565327、768374872394903、21437238740213483874629、23412341234252345'; maxposition := 0; select length(targetstr) into strlength from dual; for i in 1..strlength loop select instr(str,'、',1,i) into position from （select targetstr as str from dual）; --dbms_output.put_line(position); if position > 0 then if maxposition = 0 then select substr(str,0,instr(str,'、',1,1)-1) into retrunstr from （select targetstr as str from dual）; select to_number(substr(restr,-1)) into endstr from （select retrunstr as restr from dual）; if endstr = 3 or endstr = 5 or endstr = 7 then endstr := endstr + 1; elsif endstr = 9 then endstr := 0; end if; select substr(str,0,instr(str,'、',1,1)-2)||to_char(endstr) into retrunstr from （select targetstr as str from dual）; elsif maxposition position then --特别处理最后一段 tempstr := ''; select substr(str,maxposition-length(str)) into tempstr from （select targetstr as str from dual）; select to_number(substr(restr,-1)) into endstr from （select tempstr as restr from dual）; if endstr = 3 or endstr = 5 or endstr = 7 then endstr := endstr + 1; elsif endstr = 9 then endstr := 0; end if; select substr(str,1,length(str)-1) ||to_char(endstr) into tempstr from （select tempstr as str from dual）; retrunstr := retrunstr || '、'|| tempstr; end if; exit; end if; end loop; dbms_output.put_line(retrunstr);end;运行结果如下：
原字符串：1788987565327、768374872394903、21437238740213483874629、23412341234252345