从Linux文件中删除颜色代码 [英] Remove colour code from linux files

问题描述

我有一个来自测试脚本的输出文件(我不能更改)，由于编码，该输出在终端中看起来很棒，它以漂亮的颜色显示了输出.

I have an output file from a testing script (which I cannot alter), the output looks great in the terminal thanks to the encoding, which displays the output in nice colours.

但是，当我对文件进行vim处理时，会得到以下信息:

However when I vim the file, I get the following:

^[[1m0024^[[0m, ^[[36munknown.10^[[0m --> ^[[32mUNKNOWN^[[0m

我希望文件包含:

0024, unknown.10 --> UNKNOWN

关于stackover流有两个类似的问题，但是到目前为止，我还没有找到适合我的解决方案.

There are a couple of similar questions on stackover flow, but so far I have not found a solution that works for me.

任何帮助将不胜感激！

非常感谢！

其他信息:

我不想隐藏颜色字符，我想将它们从文件中删除.

I don't want to conceal the colour characters, I would like to remove them from the file.

输出进入证据文件，然后将该文件上推至GIT供团队审核.使用这些颜色代码的GIT UI很难:(

The output goes into an evidence file, and then that file is pushed up to a GIT for the team to review. It is difficult to the GIT UI with those colour codes :(

推荐答案

要删除颜色控制字符，可以使用以下sed命令:

To remove color control character, you may use the following sed command:

sed 's/\x1b\[[^\x1b]*m//g' file

如此处所示，颜色代码由< Esc> [FormatCodem .

转义字符为十六进制的 \ x1b (有时标记为 \ e 或 \ 033 ).

The escape character is \x1b in hexadecimal (sometimes noted as \e or \033).

该命令查找序列转义符，后跟方括号 \ x1b \ [，直到字符 m (如果找到)将其删除.

The command looks for the sequence escape followed by square bracket \x1b\[ until the character m, if found it deletes it.

除了转义字符本身 [^ \ x1b] * 之外，这两个字符之间的所有内容均被允许.这允许具有最短的正则表达式.

Everything in between these 2 characters is allowed except the escape character itself [^\x1b]*. This allows to have the shortest regex.

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