如何从 Bash 变量中修剪空格? [英] How to trim whitespace from a Bash variable?

问题描述

我有一个带有此代码的 shell 脚本:

I have a shell script with this code:

var=`hg st -R "$path"`
if [ -n "$var" ]; then
    echo $var
fi

但条件代码总是会执行，因为hg st 总是至少打印一个换行符.

But the conditional code always executes, because hg st always prints at least one newline character.

• 是否有一种简单的方法可以从 $var 中去除空格(例如 PHP)?

• Is there a simple way to strip whitespace from $var (like trim() in PHP)?

或

• 是否有处理这个问题的标准方法?

我可以使用 sed 或 AWK，但我想有一个更优雅的解决方案来解决这个问题.

I could use sed or AWK, but I'd like to think there is a more elegant solution to this problem.

推荐答案

一个简单的答案是:

echo "   lol  " | xargs

Xargs 将为您进行修剪.这是一个命令/程序，没有参数，返回修剪后的字符串，就这么简单！

Xargs will do the trimming for you. It's one command/program, no parameters, returns the trimmed string, easy as that!

注意:这不会删除所有内部空格，所以 "foo bar" 保持不变；它不会变成 "foobar".但是，多个空格会被压缩为单个空格，所以 "foo bar" 会变成 "foo bar".此外，它不会删除行尾字符.

Note: this doesn't remove all internal spaces so "foo bar" stays the same; it does NOT become "foobar". However, multiple spaces will be condensed to single spaces, so "foo bar" will become "foo bar". In addition it doesn't remove end of lines characters.

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