oracle - 如何删除或跳过字符串中的多个新行？

Question

DECLARE
   v_string VARCHAR2(4000) := 'A database is an organized collection of data, 
   generally stored and accessed electronically 
   from a computer system. Where databases are more
  
  
   
  complex they are often developed 
  using formal design and modeling techniques.';
   v_sql_code_new VARCHAR2(4000);
BEGIN
END;
 /  

  Desired output: 
     A database is an organized collection of data, 
     generally stored and accessed electronically 
     from a computer system. Where databases are more
     complex they are often developed 
     using formal design and modeling techniques.

我正在尝试从字符串中删除所有换行符。我尝试使用 regexp_replace 但无法获得所需的结果。 先谢谢你

Answer 1

虽然您也可以将 '^\s*$' 与 modifier=>'mn' 一起使用，但最简单的方法是替换多个 chr(10):

DECLARE
   v_string VARCHAR2(4000) := 'A database is an organized collection of data, 
   generally stored and accessed electronically 
   from a computer system. Where databases are more
  
  
   
  complex they are often developed 
  using formal design and modeling techniques.';
   v_sql_code_new VARCHAR2(4000);
BEGIN
   dbms_output.put_line(regexp_replace(v_string,chr(10)||'(\s*'||chr(10)||')+',chr(10)));
END;
/

\s* 这里只是删除只包含空格字符的行。