如何以特定的方式替换新的换行符和换行符

Question

我在 oracle 中名为“DATA”的列中有以下数据。我试图在每行之后删除新的空行。

• 输入

  This is a text from line 1.
  
  This is text from line 2.
  
  This is a text from line 3.The line 3 ends here .
  
  This is a text from line 4.The line ends here .
  

• 输出

  This is a text from line 1.
  This is text from line 2.
  This is a text from line 3.The line 3 ends here .
  This is a text from line 4.The line ends here .
  

我尝试过使用

查询：

Select regexp_replace(regexp_replace(data,chr(13),null),chr(10),null) from main_data;

当我执行下面的查询时，输出就像一个段落。

这是第 1 行的文本。这是第 2 行的文本。这是第 3 行的文本。第 3 行到此结束。这是第 4 行的文本。行到此结束。

任何人都可以说如何实现这一目标吗？

Answer 1

您可以使用'(' || chr(13) || chr(10) || '?|' || chr(10) || '){2,}' 正则表达式：

select regexp_replace(
    'This is a text from line 1.' || chr(13) || chr(10) || chr(13) || chr(10) || 'This is text from line 2.' || chr(10)  || chr(10) || 'This is a text from line 3.The line 3 ends here .'  || chr(10) || chr(10) || 'This is a text from line 4.The line ends here .',
    '(' || chr(13) || chr(10) || '?|' || chr(10) || '){2,}',
    '\1') as Result from dual

该模式匹配 2 次或更多 ({2,}) 连续重复的 CR 符号，后跟可选的（1 或零，?）LF 符号或（|）LF 符号。

匹配被替换为最后一次匹配的 CRLF、CR 或 LF，因为 \1 是使用捕获组 #1 捕获的值的占位符（模式中的第一个 (...) 构造）。

online demo 的输出：

Answer 2

如果字段为空白或NULL，请使用以下内容。

SELECT DATA
FROM main_data
WHERE DATA IS NOT NULL OR DATA != ''

Answer 3

如何将连续出现的行尾标记替换为一个。

DECLARE

  DATA VARCHAR2(2000);
BEGIN
  DATA := 'This is a text from line 1.' || CHR(13) || CHR(13) ||
          'This is text from line 2.' || CHR(13) || CHR(13) ||
          'This is a text from line 3.The line 3 ends here .' || CHR(13) ||CHR(13) ||
          'This is a text from line 4.The line ends here .';

  dbms_output.put_line(regexp_replace(DATA, chr(13) || CHR(13), CHR(13)));

END;

这会给你

This is a text from line 1.
This is text from line 2.
This is a text from line 3.The line 3 ends here .
This is a text from line 4.The line ends here .