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My googlefu is failing me. I have a bunch of variables I read it from a CSV where I want to strip whitespace. I could 

variable1="$(echo -e "${variable1}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
variable2="$(echo -e "${variable2}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
variable3="$(echo -e "${variable3}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"

but then I'm copying in and editing this for every variable change. It seems to me, from other languages, there should be a way to do a simple for loop that applies to the variables as named and strips the whitespace, but I can't figure out exactly how to do that.

for $i in variable1 variable2 variable3
do
$i=$(echo -e "${$i}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
done

but this doesn't work as far as I can tell. What I want to do is loop over all the variables, strip whitespace, but not have to manually type out each variable name in the script.

4

1 回答 1

3

尝试以下操作：

# Sample values that need trimming.
variable1='  a  '
variable2='      b  c  '
variable3='        sadf sdf sdf     '

for name in ${!variable*}; do
  # Trim leading and trailing whitespace.
  read -r $name <<<"${!name}"
  # Print result
  echo "Variable $name now contains: [${!name}]"
done

该解决方案依赖于变量间接，即通过另一个包含目标变量名称的变量间接引用变量的能力。

请注意，变量间接有其缺陷（请参阅链接），并且通常有更好的解决方案，例如，通过使用数组。

  • ${!variable*}扩展为名称以 . 开头的所有已定义变量的名称variable。
  • ${!name}扩展为名称存储在 variable 中的变量的值$name。
  • read -r，通过读入单个变量，隐式修剪前导和尾随空格。
    • -r确保\字符。在输入中留下单独的，这通常是你想要的。
于 2015-05-13T18:46:54.233 回答