Bash 循环获取文件中的行，跳过注释和空行
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Bash loop to get lines in a file, skipping comments and blank lines
提问by Cat5InTheCradle
I want to get lines from a file that might look like this
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!
The code I have for handling the comments is this:
我处理评论的代码是这样的：
#!/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 $line
done
But that doesn't skip the whitespace. Not sure what I need to add. Is there a better way to do this?
但这并没有跳过空格。不确定我需要添加什么。有一个更好的方法吗？
回答by anubhava
You can tweak loop like this:
您可以像这样调整循环：
while read -r line; do
[[ -n "$line" && "$line" != [[:blank:]#]* ]] && echo "$line"
done < file
回答by Jordan Running
This ought to do. No loop needed unless you need to do further operations on the remaining lines.
这是应该的。除非您需要对剩余的行做进一步的操作，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*$|#)' $hosts
回答by Tripp Kinetics
How about:
怎么样：
#!/bin/bash
hosts = inventory/hosts
# List hosts we're going to try to connect to
sed -e '/^\s*$/ d' -e '/^#/ d' $hosts | while read line; do
echo $line
done
回答by jimm-cl
If you just want to skip comments and blank lines, then maybe a simple sedcommand would be enough:
如果您只想跳过注释和空行，那么也许一个简单的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!
Try something like sed '/^#/d'to get rid of the comments, and sed '/^$/d'to get rid of blank lines.
尝试诸如sed '/^#/d'删除注释和sed '/^$/d'删除空白行之类的方法。
$ sed '/^#/d' test | sed '/^$/d'
But this line will not
Nor this one
but not this one!
