Sed-删除字符串前的x个字符 [英] Sed - Delete x numbers of characters before a string

问题描述

我有一个txt文件，我想删除特定字符串前的x个字符

I have a txt file and I want to delete x number of characters before a specific string

示例:

52.451878252 foo 845.851878212 foo
98.444854655 foo 458.344854656 foo

我想删除字符串'foo'之前的8个字符以获取此信息:

I want to delete 8 characters before the string ' foo' in order to get this :

52.45 foo 845.66 foo
98.44 foo 458.66 foo

我应该如何使用sed命令(或其他方法)?

How should I do with the sed command (or something else) ?

谢谢:)

推荐答案

在这里:

sed -e 's/........foo/ foo/g' file.txt

或等效地:

sed -e 's/.\{8\}foo/ foo/g' file.txt

甚至:

sed -e 's/.\{8\}\(foo\)/ \1/g' file.txt

或者采用另一种方法，为您的样本提供相同的输出:

Or a different approach, with the same output for your sample:

sed -e 's/[0-9]\{7\} / /g' file.txt

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