string 如何从 Bash 变量中修剪空格？

How to trim whitespace from a Bash variable?

提问by too much php

I have a shell script with this code:

我有一个带有此代码的shell脚本：

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

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

但是条件代码总是会执行，因为hg st总是至少打印一个换行符。

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

• 有没有一种简单的方法来去除空格$var（比如trim()在PHP 中）？

or

或者

• Is there a standard way of dealing with this issue?

• 有没有标准的方法来处理这个问题？

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

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。

回答by MattyV

Let's define a variable containing leading, trailing, and intermediate whitespace:

让我们定义一个包含前导、尾随和中间空格的变量：

FOO=' test test test '
echo -e "FOO='${FOO}'"
# > FOO=' test test test '
echo -e "length(FOO)==${#FOO}"
# > length(FOO)==16

---

---

How to remove all whitespace (denoted by [:space:]in tr):

如何删除所有空格（用[:space:]in表示tr）：

FOO=' test test test '
FOO_NO_WHITESPACE="$(echo -e "${FOO}" | tr -d '[:space:]')"
echo -e "FOO_NO_WHITESPACE='${FOO_NO_WHITESPACE}'"
# > FOO_NO_WHITESPACE='testtesttest'
echo -e "length(FOO_NO_WHITESPACE)==${#FOO_NO_WHITESPACE}"
# > length(FOO_NO_WHITESPACE)==12

---

---

How to remove leading whitespace only:

如何仅删除前导空格：

FOO=' test test test '
FOO_NO_LEAD_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//')"
echo -e "FOO_NO_LEAD_SPACE='${FOO_NO_LEAD_SPACE}'"
# > FOO_NO_LEAD_SPACE='test test test '
echo -e "length(FOO_NO_LEAD_SPACE)==${#FOO_NO_LEAD_SPACE}"
# > length(FOO_NO_LEAD_SPACE)==15

---

---

How to remove trailing whitespace only:

如何仅删除尾随空格：

FOO=' test test test '
FOO_NO_TRAIL_SPACE="$(echo -e "${FOO}" | sed -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_TRAIL_SPACE='${FOO_NO_TRAIL_SPACE}'"
# > FOO_NO_TRAIL_SPACE=' test test test'
echo -e "length(FOO_NO_TRAIL_SPACE)==${#FOO_NO_TRAIL_SPACE}"
# > length(FOO_NO_TRAIL_SPACE)==15

---

---

How to remove both leading and trailing spaces--chain the seds:

如何删除前导和尾随空格 - 链接seds：

FOO=' test test test '
FOO_NO_EXTERNAL_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_EXTERNAL_SPACE='${FOO_NO_EXTERNAL_SPACE}'"
# > FOO_NO_EXTERNAL_SPACE='test test test'
echo -e "length(FOO_NO_EXTERNAL_SPACE)==${#FOO_NO_EXTERNAL_SPACE}"
# > length(FOO_NO_EXTERNAL_SPACE)==14

Alternatively, if your bash supports it, you can replace echo -e "${FOO}" | sed ...with sed ... <<<${FOO}, like so (for trailing whitespace):

或者，如果您的 bash 支持它，您可以替换echo -e "${FOO}" | sed ...为sed ... <<<${FOO}，像这样（用于尾随空格）：

FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"

回答by makevoid

A simple answer is:

一个简单的答案是：

echo "   lol  " | xargs

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

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！

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.

注意：这不会删除所有内部空间，因此"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，因此"foo bar"会变成"foo bar". 此外，它不会删除行尾字符。

回答by bashfu

There is a solution which only uses Bash built-ins called wildcards:

有一个只使用 Bash 内置插件的解决方案，称为通配符：

var="    abc    "
# remove leading whitespace characters
var="${var#"${var%%[![:space:]]*}"}"
# remove trailing whitespace characters
var="${var%"${var##*[![:space:]]}"}"   
printf '%s' "===$var==="

Here's the same wrapped in a function:

这是封装在函数中的相同内容：

trim() {
    local var="$*"
    # remove leading whitespace characters
    var="${var#"${var%%[![:space:]]*}"}"
    # remove trailing whitespace characters
    var="${var%"${var##*[![:space:]]}"}"   
    printf '%s' "$var"
}

You pass the string to be trimmed in quoted form. e.g.:

您以引号形式传递要修剪的字符串。例如：

trim "   abc   "

A nice thing about this solution is that it will work with any POSIX-compliant shell.

这个解决方案的一个好处是它可以与任何符合 POSIX 的 shell 一起工作。

Reference

参考

• Remove leading & trailing whitespace from a Bash variable(original source)

• 从 Bash 变量中删除前导和尾随空格（原始来源）

回答by bashfu

Bash has a feature called parameter expansion, which, among other things, allows string replacement based on so-called patterns(patterns resemble regular expressions, but there are fundamental differences and limitations). [flussence's original line: Bash has regular expressions, but they're well-hidden:]

Bash 有一个称为参数扩展的特性，除其他外，它允许基于所谓的模式（模式类似于正则表达式，但存在根本差异和限制）的字符串替换。[flussence 的原话：Bash 有正则表达式，但它们隐藏得很好：]

The following demonstrates how to remove allwhite space (even from the interior) from a variable value.

下面演示了如何从变量值中删除所有空白（甚至从内部）。

$ var='abc def'
$ echo "$var"
abc def
# Note: flussence's original expression was "${var/ /}", which only replaced the *first* space char., wherever it appeared.
$ echo -n "${var//[[:space:]]/}"
abcdef

回答by rkachach

In order to remove all the spaces from the beginning and the end of a string (including end of line characters):

为了删除字符串开头和结尾的所有空格（包括行尾字符）：

echo $variable | xargs echo -n

This will remove duplicate spaces also:

这也将删除重复的空格：

echo "  this string has a lot       of spaces " | xargs echo -n

Produces: 'this string has a lot of spaces'

产生：'这个字符串有很多空格'

回答by Brian Cain

Strip one leading and one trailing space

去掉一个前导和一个尾随空格

trim()
{
    local trimmed=""

    # Strip leading space.
    trimmed="${trimmed## }"
    # Strip trailing space.
    trimmed="${trimmed%% }"

    echo "$trimmed"
}

For example:

例如：

test1="$(trim " one leading")"
test2="$(trim "one trailing ")"
test3="$(trim " one leading and one trailing ")"
echo "'$test1', '$test2', '$test3'"

Output:

输出：

'one leading', 'one trailing', 'one leading and one trailing'

Strip allleading and trailing spaces

去除所有前导和尾随空格

trim()
{
    local trimmed=""

    # Strip leading spaces.
    while [[ $trimmed == ' '* ]]; do
       trimmed="${trimmed## }"
    done
    # Strip trailing spaces.
    while [[ $trimmed == *' ' ]]; do
        trimmed="${trimmed%% }"
    done

    echo "$trimmed"
}

For example:

例如：

test4="$(trim "  two leading")"
test5="$(trim "two trailing  ")"
test6="$(trim "  two leading and two trailing  ")"
echo "'$test4', '$test5', '$test6'"

Output:

输出：

'two leading', 'two trailing', 'two leading and two trailing'

回答by GuruM

From Bash Guide section on globbing

来自 Bash 指南关于globbing 的部分

To use an extglob in a parameter expansion

在参数扩展中使用 extglob

 #Turn on extended globbing  
shopt -s extglob  
 #Trim leading and trailing whitespace from a variable  
x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}  
 #Turn off extended globbing  
shopt -u extglob  

Here's the same functionality wrapped in a function (NOTE: Need to quote input string passed to function):

这是封装在函数中的相同功能（注意：需要引用传递给函数的输入字符串）：

trim() {
    # Determine if 'extglob' is currently on.
    local extglobWasOff=1
    shopt extglob >/dev/null && extglobWasOff=0 
    (( extglobWasOff )) && shopt -s extglob # Turn 'extglob' on, if currently turned off.
    # Trim leading and trailing whitespace
    local var=
    var=${var##+([[:space:]])}
    var=${var%%+([[:space:]])}
    (( extglobWasOff )) && shopt -u extglob # If 'extglob' was off before, turn it back off.
    echo -n "$var"  # Output trimmed string.
}

Usage:

用法：

string="   abc def ghi  ";
#need to quote input-string to preserve internal white-space if any
trimmed=$(trim "$string");  
echo "$trimmed";

---

---

If we alter the function to execute in a subshell, we don't have to worry about examining the current shell option for extglob, we can just set it without affecting the current shell. This simplifies the function tremendously. I also update the positional parameters "in place" so I don't even need a local variable

如果我们将函数更改为在子 shell 中执行，我们不必担心检查 extglob 的当前 shell 选项，我们可以设置它而不影响当前 shell。这极大地简化了功能。我还“就地”更新了位置参数，所以我什至不需要局部变量

trim() {
    shopt -s extglob
    set -- "${1##+([[:space:]])}"
    printf "%s" "${1%%+([[:space:]])}" 
}

so:

所以：

$ s=$'\t\n \r\tfoo  '
$ shopt -u extglob
$ shopt extglob
extglob         off
$ printf ">%q<\n" "$s" "$(trim "$s")"
>$'\t\n \r\tfoo  '<
>foo<
$ shopt extglob
extglob         off

回答by VAmp

You can trim simply with echo:

你可以简单地修剪echo：

foo=" qsdqsd qsdqs q qs   "

# Not trimmed
echo \'$foo\'

# Trim
foo=`echo $foo`

# Trimmed
echo \'$foo\'

回答by Paul Tomblin

I've always done it with sed

我一直都是用 sed 做的

  var=`hg st -R "$path" | sed -e 's/  *$//'`

If there is a more elegant solution, I hope somebody posts it.

如果有更优雅的解决方案，我希望有人发布它。

回答by Adam Rosenfield

You can delete newlines with tr:

您可以使用以下命令删除换行符tr：

var=`hg st -R "$path" | tr -d '\n'`
if [ -n $var ]; then
    echo $var
done