如何截取所需要的字段，请看

Question

有如下字符串   a=badfd+dfddc+ddasd+sfd 
现在要截取符号之间的字段，也就是badfd；dfddc；ddasd；sfd这四个结果，现提供一下PLSQL语言，
大侠帮助一下，修改修改
declare
v_sql  long;
i      number :=1;
j      number :=1;
begin
 
v_sql:='a=badfd+dfddc+ddasd+sfd';
loop
select  instr(v_sql,'+',1,j)  into i from dual;
j:=j+1;
exit when i=0;
dbms_output.put_line(i); 
end loop;
end;

Answer 1

--这样取吧
select regexp_substr(regexp_replace('a=badfd+dfddc+ddasd+sfd ','[a-zA-Z]+=',''), 
'[^+]+',1,rownum) result from dual   
connect by rownum<=length(regexp_replace('a=badfd+dfddc+ddasd+sfd ','[a-zA-Z]+=',''))
-length(replace(regexp_replace('a=badfd+dfddc+ddasd+sfd ','[a-zA-Z]+=',''), '+', ''))+1;
/*
RESULT                                                                          
----------badfd                                                                           
dfddc                                                                           
ddasd                                                                           
sfd   
*/

Answer 2

--按你的写法
declare
v_sql long;
i number :=1;
j number :=1;
begin
v_sql:='a=badfd+dfddc+ddasd+sfd';
v_sql:=regexp_replace(v_sql,'[a-zA-Z]+=','');
loop
select instr(v_sql,'+',1,j) into i from dual;
dbms_output.put_line(regexp_substr(v_sql,'[^+]+',1,j));
j:=j+1; 
exit when i=0;
end loop;
end;
/
/*
badfd                                                                           
dfddc                                                                           
ddasd                                                                           
sfd    
*/

Answer 3

regexp_replace
regexp_substr