我创建了以下功能
function lob_replace( p_lob in out clob,p_what in varchar2,p_with in clob ) return clob
as
l_temp_number number;
l_temp_number_1 number;
l_temp_clob clob;
l_return_clob clob;
l_temp1_clob clob;
l_temp2_clob clob;
begin
l_temp_number:=dbms_lob.instr(p_lob, p_what);
--Create a lob locator
DBMS_LOB.createtemporary(l_temp_clob,true);
DBMS_LOB.createtemporary(l_temp1_clob,true);
DBMS_LOB.createtemporary(l_temp2_clob,true);
---substract and build the LOBs
l_temp_number_1:=length(p_lob);
l_temp_clob:=dbms_lob.substr(p_lob,l_temp_number-1,1);
l_temp1_clob:=dbms_lob.substr(p_lob,l_temp_number_1-l_temp_number+1 ,l_temp_number +length(p_what) );
--append three diff lob to one
dbms_lob.append(l_temp2_clob,l_temp_clob);
dbms_lob.append(l_temp2_clob,p_with);
dbms_lob.append(l_temp2_clob,l_temp1_clob);
l_return_clob :=l_temp2_clob;
--remove the tmp lob
DBMS_LOB.freetemporary(l_temp_clob);
DBMS_LOB.freetemporary(l_temp1_clob);
DBMS_LOB.freetemporary(l_temp2_clob);
return l_return_clob;
end;
如果我将此函数调用如下
,则会抛出错误declare
temp clob;
begin
temp:='replace this #a#';
temp:=lob_replace(temp,'#a#','with this');
end;
它会抛出错误
ORA-06502: PL/SQL: numeric or value error: invalid LOB locator specified: ORA-22275
ORA-06512: at "SYS.DBMS_LOB", line 639
ORA-06512: at "LOB_REPLACE", line 24
ORA-06512: at line 5
但不会抛出错误
declare
temp clob;
begin
temp:='replace this #a# ';
temp:=lob_replace(temp,'#a#','with this');
end;
请注意temp:='replace this #a# ';
有人知道原因吗?
答案 0 :(得分:0)
当您构建l_temp1_clob
时,如果有任何要添加的内容,则只能添加其余的lob。改变你的代码:
l_temp1_clob:=dbms_lob.substr(p_lob,l_temp_number_1-l_temp_number+1 ,l_temp_number +length(p_what) );
对此:
if l_temp_number +length(p_what) < l_temp_number_1
then
l_temp1_clob := dbms_lob.substr(p_lob, l_temp_number_1 - l_temp_number - 1 , l_temp_number +length(p_what) );
end if;
逻辑是:如果lob substr的开始是在lob的结束之前。
此外,您的测试呼叫中未声明dd
变量:
declare
temp clob;
begin
temp:='replace this #a#';
temp:=lob_replace(dd,'#a#','with this');
end;