Question

我有一个输出这样的程序：

stuff=Some text with spaces and other $pecial characters stuff2=1 stuff3=0

我试图在变量中填充所有字符串，直到stuff2 = 1，因此变量将是：

stuff=Some text with spaces and other $special characters

我试过这个：

for word in $output
while [ $word != "stuff2=1" ]
do
   var+=" "$word
done
done

但我得到的只是“东西=一些”一遍又一遍。

Answer 1

这个怎么样？

stuff='Some text with spaces and other $pecial characters stuff2=1 stuff3=0'

var=""
for word in $stuff
do
  if [ "$word" == "stuff2=1" ]
     then
    break
  fi
  if [ "$var" != "" ]
  then
    var="${var} "
  fi
  var="${var}${word}"
done
echo "$var"

Answer 2

您可以通过简单的bash参数扩展实现此目的：

line='stuff=Some text with spaces and other $pecial characters stuff2=1 stuff3=0'
truncated=${line% stuff2=1*}
echo "$truncated"

stuff=Some text with spaces and other $pecial characters

当您需要“取消引用”变量时，引用变量至关重要。

Answer 3

可以用bash本身完成

stuff=${stuff/stuff2=1*/}

Answer 4

$ eval $(sed "s/=/='/;s/ [^ ]*=.*/'/")

$ echo $stuff
Some text with spaces and other $pecial characters

将字符串读入变量，直到BASH中的某个单词

4 个答案: