如何在bash中从文件中提取和删除版本字符串

I have a simple text file ./version containing a version number. Unfortunately, sometimes the version number in the file is followed by whitespaces and newlines like

我有一个包含版本号的简单文本文件./version。不幸的是,有时文件中的版本号后面跟着空格和换行符

1.1.3[space][space][newline]
[newline]
[newline]

What is the best, easiest and shortest way to extract the version number into a bash variable without the trailing spaces and newlines? I tried

在没有尾随空格和换行符的情况下将版本号提取到bash变量中的最佳,最简单和最短的方法是什么?我试过了

var=`cat ./version | tr -d ' '`

which works for the whitespaces but when appending a tr -d '\n' it does not work.

这适用于空格,但当附加tr -d'\ n'时它不起作用。

Thanks, Chris

4 个解决方案

#1

$ echo -e "1.1.1  \n\n" > ./version
$ read var < ./version
$ echo -n "$var" | od -a
0000000   1   .   1   .   1
0000005

#2

Pure Bash, no other process:

Pure Bash,没有其他过程:

echo -e "1.2.3  \n\n" > .version

version=$(<.version)
version=${version// /}

echo "'$version'"

result: '1.2.3'

#3

I still do not know why, but after deleting and recreating the version file this worked:

我仍然不知道为什么,但删除并重新创建版本文件后,这有效:

var=`cat ./version | tr -d ' ' | tr -d '\n'`

I'm confused... what can you do different when creating a text file. However, it works now.

我很困惑......创建文本文件时你能做些什么呢?但是,它现在有效。

#4

I like the pure bash version from fgm's answer.

我喜欢fgm答案的纯粹bash版本。

I provide this one-line perl command to remove also other characters if any:

我提供这个单行perl命令来删除其他字符,如果有的话:

perl -pe '($_)=/([0-9]+([.][0-9]+)+)/'

The extracted version number is trimmed/stripped (no newline or carriage return symbols):

修剪/剥离提取的版本号(没有换行符或回车符号):

$> V=$( bash --version | perl -pe '($_)=/([0-9]+([.][0-9]+)+)/' )
$> echo "The bash version is '$V'"
The bash version is '4.2.45'

I provide more explanation and give other more sophisticated (but still short) one-line perl commands in my other answer.

我在其他答案中提供了更多解释并给出了其他更复杂(但仍然很短)的单行perl命令。

#1

$ echo -e "1.1.1  \n\n" > ./version
$ read var < ./version
$ echo -n "$var" | od -a
0000000   1   .   1   .   1
0000005

#2