Question

我想从一个看起来像这样的文件中获取行

# This will be skipped
But this line will not
Nor this one

# this and the above blank line will be skipped
but not this one!

我处理评论的代码是：

#!/bin/bash

hosts = inventory/hosts

# List hosts we're going to try to connect to
cat $hosts | while read line; do
  case "$line" in \#*) continue ;; esac
        echo $line
done

但这并没有跳过这个空白。不确定我需要添加什么。有更好的方法吗？

Answer 1

你可以像这样调整循环：

while read -r line; do
    [[ -n "$line" && "$line" != [[:blank:]#]* ]] && echo "$line"
done < file

Answer 2

这应该做的。除非您需要对其余行进行进一步操作，否则不需要循环。

#!/bin/bash

hosts="inventory/hosts"

# List hosts we're going to try to connect to
grep -vE '^(\s*$|#)' $hosts

Answer 3

怎么样：

#!/bin/bash

hosts = inventory/hosts

# List hosts we're going to try to connect to
sed -e '/^\s*$/ d' -e '/^#/ d' $hosts | while read line; do
    echo $line
done

Answer 4

如果你只是想跳过评论和空行，那么简单的sed命令就足够了：

$ cat test
# This will be skipped
But this line will not
Nor this one

# this and the above blank line will be skipped
but not this one!

尝试使用类似sed '/^#/d'的内容来删除注释，并sed '/^$/d'删除空白行。

$ sed '/^#/d' test | sed '/^$/d'
But this line will not
Nor this one
but not this one!

使用Bash循环获取文件中的行，跳过注释和空行

4 个答案: