如何从猛砸变量修剪空白？ [英] How to trim whitespace from a Bash variable?

问题描述

我有这个code 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code总是执行，因为汞ST 始终打印至少有一个换行符。

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

• 是否有从 $ VAR 剥离空白（如修剪（）的简单方式中的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式？

我可以使用 SED 或的 AWK ，但我想，认为还有一个更优雅的解决了这个问题。

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

让我们来定义包含变量前置，和中间的空格：

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

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

echo命令使用的选项 -n 来避免增加返回字符，并造成 WC -m 算一笔附加的字符。

The echo command used the option -n to avoid adding a return character and causing wc -m count an additional character.

如何删除所有空格（由表示[[：空间：]] ）

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

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

---

如何删除前导空格只：

---

How to remove leading whitespace only:

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

---

如何删除尾随空格只：

---

How to remove trailing whitespace only:

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)==$(echo -ne "${FOO_NO_TRAIL_SPACE}" | wc -m)"
# > length(FOO_NO_TRAIL_SPACE)==15

---

如何删除这两个首尾空格 - 链 SED 取值：

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)==$(echo -ne "${FOO_NO_EXTERNAL_SPACE}" | wc -m)"
# > length(FOO_NO_EXTERNAL_SPACE)==14

另外，如果您的bash支持的话，你可以替换回声-e$ {} FOO| SED ... 与 SED ...＆LT;＆LT;＆LT; $ {FOO} ，像这样（尾部空白）：

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

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

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