我有一个用逗号分隔的字符串文件。我正试着用新的一行替换逗号。我试过了:
sed 's/,/\n/g' file
但这并不奏效。我错过了什么?
我有一个用逗号分隔的字符串文件。我正试着用新的一行替换逗号。我试过了:
sed 's/,/\n/g' file
但这并不奏效。我错过了什么?
当前回答
只是为了澄清:OSX上sed的手册页(10.8;达尔文内核版本12.4.0)说:
[…]
Sed正则表达式
The regular expressions used in sed, by default, are basic regular expressions (BREs, see re_format(7) for more information), but extended
(modern) regular expressions can be used instead if the -E flag is given. In addition, sed has the following two additions to regular
expressions:
1. In a context address, any character other than a backslash (``\'') or newline character may be used to delimit the regular expression.
Also, putting a backslash character before the delimiting character causes the character to be treated literally. For example, in the
context address \xabc\xdefx, the RE delimiter is an ``x'' and the second ``x'' stands for itself, so that the regular expression is
``abcxdef''.
2. The escape sequence \n matches a newline character embedded in the pattern space. You cannot, however, use a literal newline charac-
ter in an address or in the substitute command.
[…]
所以我想一个人必须使用tr -如上所述-或者nitty
sed "s/,/^M
/g"
注意:你必须输入<ctrl>-v,<return>才能在vi编辑器中得到'^M'
其他回答
请改用tr:
tr , '\n' < file
虽然我写这篇文章晚了,只是更新我的发现。这个答案只适用于Mac OS X。
$ sed 's/new/
> /g' m1.json > m2.json
sed: 1: "s/new/
/g": unescaped newline inside substitute pattern
在上面的命令中,我尝试用Shift+Enter添加新行,但没有工作。所以这一次我尝试了“转义”的“未转义换行”,因为错误告诉。
$ sed 's/new/\
> /g' m1.json > m2.json
成功了!(适用于Mac OS X 10.9.3)
如果sed的使用倾向于完全使用替换表达式(就像我的一样),您也可以使用perl -pe来代替
$ echo 'foo,bar,baz' | perl -pe 's/,/,\n/g'
foo,
bar,
baz
FWIW,下面的行工作在窗口和替换分号在我的路径变量换行。我正在使用安装在我的git bin目录下的工具。
echo %path% | sed -e $'s/;/\\n/g' | less
我发现了另一个同样有效的命令。
find your_filename.txt -type f -exec sed -i 's/,/\n/g' {} \;