如何从字符串中删除空格？ [重复]

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How to trim whitespace from a Bash variable? 41 answers
如何从Bash变量中修剪空格？ 41个答案

In ubuntu bash script how to remove space from one variable

在ubuntu bash脚本中如何从一个变量中删除空间

string will be

字符串将是

   3918912k

Want to remove all blank space.

想要删除所有空白区域。

5 个解决方案

#1

Try doing this in a shell:

尝试在shell中执行此操作：

s="  3918912k"
echo ${s//[[:blank:]]/}

That uses parameter expansion (it's a non posix feature)

这使用参数扩展（这是一个非posix功能）

[[:blank:]] is a POSIX regex class (remove spaces, tabs...), see http://www.regular-expressions.info/posixbrackets.html

[[：blank：]]是一个POSIX正则表达式类（删除空格，制表符......），请参阅http://www.regular-expressions.info/posixbrackets.html

#2

The tools sed or tr will do this for you by swapping the whitespace for nothing

sed或tr工具可以通过交换空白来为你做这件事

sed 's/ //g'

sed's / // g'

tr -d ' '

tr -d''

Example:

例：

$ echo "   3918912k " | sed 's/ //g'
3918912k

#3

You can also use echo to remove blank spaces, either at the beginning or at the end of the string, but also repeating spaces inside the string.

您还可以使用echo来删除字符串开头或结尾处的空格，还可以在字符串中重复空格。

$ myVar="    kokor    iiij     ook      "
$ echo "$myVar"
    kokor    iiij     ook      
$ myVar=`echo $myVar`
$
$ # myVar is not set to "kokor iiij ook"
$ echo "$myVar"
kokor iiij ook

#4

Since you're using bash, the fastest way would be:

由于你使用bash，最快的方法是：

shopt -s extglob # Allow extended globbing
var=" lakdjsf   lkadsjf "
echo "${var//+([[:space:]])/}"

It's fastest because it uses built-in functions instead of firing up extra processes.

它是最快的，因为它使用内置函数而不是激发额外的进程。

However, if you want to do it in a POSIX-compliant way, use sed:

但是，如果要以符合POSIX的方式执行此操作，请使用sed：

var=" lakdjsf   lkadsjf "
echo "$var" | sed 's/[[:space:]]//g'

#5

A funny way to remove all spaces from a variable is to use printf:

从变量中删除所有空格的一种有趣方法是使用printf：

$ myvar='a cool variable    with   lots of   spaces in it'
$ printf -v myvar '%s' $myvar
$ echo "$myvar"
acoolvariablewithlotsofspacesinit

It turns out it's slightly more efficient than myvar="${myvar// /}", but not safe regarding globs (*) that can appear in the string. So don't use it in production code.

事实证明它比myvar =“$ {myvar // /}”稍微有效，但对于可以出现在字符串中的globs（*）不安全。所以不要在生产代码中使用它。

If you really really want to use this method and are really worried about the globbing thing (and you really should), you can use set -f (which disables globbing altogether):

如果你真的真的想要使用这个方法而且真的很担心全局事物（你真的应该），你可以使用set -f（它完全禁用了globbing）：

$ ls
file1  file2
$ myvar='  a cool variable with spaces  and  oh! no! there is  a  glob  *  in it'
$ echo "$myvar"
  a cool variable with spaces  and  oh! no! there is  a  glob  *  in it
$ printf '%s' $myvar ; echo
acoolvariablewithspacesandoh!no!thereisaglobfile1file2init
$ # See the trouble? Let's fix it with set -f:
$ set -f
$ printf '%s' $myvar ; echo
acoolvariablewithspacesandoh!no!thereisaglob*init
$ # Since we like globbing, we unset the f option:
$ set +f

I posted this answer just because it's funny, not to use it in practice.

我发布这个答案只是因为它很有趣，而不是在实践中使用它。

#1