string - 如何从Bash变量中修剪空格？(How to trim whitespace from a Bash variable?)

Question

I have a shell script with this code:

(我有一个带有以下代码的shell脚本：)

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

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

(但是条件代码总是执行，因为hg st总是打印至少一个换行符。)

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

  (有没有一种简单的方法可以从$var删除空格（如PHP中的 trim() ）？)

or

(要么)

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

  (有没有解决此问题的标准方法？)

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

(我可以使用sed或AWK ，但是我想认为有一个更优雅的解决方案。)

  ask by community wiki translate from so

Answer 1

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（在tr以[:space:]表示）：)

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:

(如何仅删除尾随空白：)

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 sed s:

(如何同时删除前导空格和尾随空格-链接sed s：)

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 ...

(或者，如果您的bash支持它，则可以替换echo -e "${FOO}" | sed ...)

echo -e "${FOO}" | sed ... with sed ... <<<${FOO} , like so (for trailing whitespace):

(echo -e "${FOO}" | sed ... with sed ... <<<${FOO} ，就像这样（用于尾随空格）：)

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