如何以特定方式替换新行和换行符

Question

我在 oracle.I 中名为 "DATA" 的列中有以下数据，试图删除每行后的新空行。

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;

当我执行以下查询时，输出就像一个段落。

这是第 1.This 行的文本 2.This 是第行的文本。第 3 行到此结束。这是第 4.The 行的文本行结束这里.

有人能说说如何实现吗？

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,}',
    '') as Result from dual

该模式匹配 2 个或更多 ({2,}) 个连续重复的 CR 符号，后跟一个可选的（1 或零，?）LF 符号或（|）一个低频符号。

匹配被单个最后匹配的 CRLF、CR 或 LF 替换，因为 </code> 是捕获组 #1 捕获的值的占位符（第一个 <code>(...) 构造在模式中）。

online demo 的输出：

Answer 2

如果字段为空或NULL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 .

如何以特定方式替换新行和换行符

How to replace the new line and line feed in a specific manner

regex

oracle

oracle11g

regexp-replace