使用 Bash 读取文件中的行并避免使用 # 行
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Reading lines in a file and avoiding lines with # with Bash
提问by nixnotwin
I tried this:
我试过这个：
file="myfile"
while read -r line
do
[[ $line = \#* ]] && continue
"address=$line7.0.0.1"
done < "$file"
This code doesn't avoid the lines that begin with comments. Even if I don't have any comments, dnsmasqtells that there are errors.
此代码不会避免以注释开头的行。即使我没有任何评论，也dnsmasq告诉我有错误。
Its going to be a dnsmasqconf file, and it will read and insert domain names like so: address=\mydomain.com\127.0.0.1.
它将是一个dnsmasqconf 文件，它将像这样读取和插入域名：address=\mydomain.com\127.0.0.1.
EDIT:1
编辑：1
Input file:
输入文件：
domain1.com
domain2.com
domain3.com
#domain4.com
domain5.com
Output should be:
输出应该是：
address=/domain1.com/127.0.0.1
address=/domain2.com/127.0.0.1
address=/domain3.com/127.0.0.1
address=/domain5.com/127.0.0.1
I will drop the script in /etc/dnsmasq.d/directory so that dnsmaq.confcan process it when dnsmasqis started.
我会将脚本放在/etc/dnsmasq.d/目录中，以便dnsmaq.conf在dnsmasq启动时可以处理它。
采纳答案by ДМИТРИЙ МАЛИКОВ
It's safer to use [[ "$line" = "\#*" ]]
使用起来更安全 [[ "$line" = "\#*" ]]
Btw, address="\\${line}\\127.0.0.1"
顺便提一句， address="\\${line}\\127.0.0.1"
UPD:
更新：
If I've understand you right you need to change every uncommented domains to address=\domain\127.0.0.1. It could be done fast and easy with sed, there is no need in bash-program.
如果我理解正确，您需要将每个未注释的域更改为address=\domain\127.0.0.1. 使用 可以快速轻松地完成sed，不需要 bash 程序。
$> cat ./text
domain1.com
domain2.com
domain3.com
#domain4.com
domain5.com
$> sed -r -e 's/(^[^#]*$)/address=\/\/127.0.0.1/g' ./text2
address=/domain1.com/127.0.0.1
address=/domain2.com/127.0.0.1
address=/domain3.com/127.0.0.1
#domain4.com
address=/domain5.com/127.0.0.1
If you need to remove commented lines, sed can do it too with /matched_line/d
如果您需要删除注释行， sed 也可以使用 /matched_line/d
$> sed -r -e 's/(^[^#]*$)/address=\/\/127.0.0.1/g; /^#.*$/d' ./text2
address=/domain1.com/127.0.0.1
address=/domain2.com/127.0.0.1
address=/domain3.com/127.0.0.1
address=/domain5.com/127.0.0.1
UPD2: if you want to do all that stuff inside the bash script, here is your code modification:
UPD2：如果你想在 bash 脚本中做所有这些事情，这里是你的代码修改：
file="./text2"
while read -r line; do
[[ "$line" =~ ^#.*$ ]] && continue
echo "address=/${line}/127.0.0.1"
done < "$file"
And it's output:
它的输出：
address=/domain1.com/127.0.0.1
address=/domain2.com/127.0.0.1
address=/domain3.com/127.0.0.1
address=/domain5.com/127.0.0.1
回答by Keith Thompson
To skip lines startingwith #:
要跳过线开始有#：
grep -v '^#' myfile | while read -r file ; do
...
done
Modify the grepcommand as needed to, for example, skip lines starting with whitespace and a #character.
grep根据需要修改命令，例如，跳过以空格和#字符开头的行。
回答by briceburg
Comment lines can and often do begin with whitespace. Here's a bash native regex solution that handles any preceeding whitespace;
注释行可以并且经常以空格开头。这是一个 bash 原生正则表达式解决方案，可以处理任何前面的空格；
while read line; do
[[ "$line" =~ ^[[:space:]]*# ]] && continue
...work with valid line...
done
回答by Pipo
Only one working for me was:
只有一个对我来说有效的是：
while IFS=$'\n' read line
do
if [[ "$line" =~ \#.* ]];then
logDebug "comment line:$line"
else
logDebug "normal line:$line"
fi
done < myFile
回答by JoJo Brandon Coffman
[ "${line:0:1}" = "#" ] && continue
This takes the string, gets the substring at offset 0, length 1:
"${line:0:1}"
and checks if it is equal to #
并检查它是否等于 #
= "#"
and continues looping if so
并继续循环如果是这样
&& continue
回答by Claes Wikner
You can filter with awk:
你可以过滤awk：
awk '!/^#/{print"address=/"file="myfile"
sed -i".backup" 's/^#.*$//' $file
"/127.0.0.1"}' file
回答by chown
This could also be accomplished with 1 sedcommand:
这也可以通过 1 个sed命令来完成：
**awk -F'#' '{print }' t.txt | awk 'NF > 0' | awk '{print "address=/"address=/domain1.com/127.0.0.1
address=/domain2.com/127.0.0.1
address=/domain3.com/127.0.0.1
address=/domain5.com/127.0.0.1
"/127.0.0.1"}'**
This will modify the file in-place (creating a backup copy first), removing all lines starting with a #.
这将就地修改文件（首先创建备份副本），删除所有以#.
回答by Abhi km
It has 3 parts. Please read each to understand clearly
它有 3 个部分。请阅读每一个以清楚地理解
- To remove # line -----
awk -F'#' '{print $1}' t.txt - To remove a blank line created by # ----
awk 'NF > 0' - To print in required format. ------
awk '{print "address=/"$0"/127.0.0.1"}'
- 删除 # 行 -----
awk -F'#' '{print $1}' t.txt - 删除由 # ---- 创建的空行
awk 'NF > 0' - 以所需格式打印。------
awk '{print "address=/"$0"/127.0.0.1"}'
So Total Script Needed is,
所以需要的总脚本是，
awk '{ if ([[ "$line"~="#.*" ]] && continue
!~ /^#/){printf "address=/%s/127.0.0.1 \n",##代码##}}' <your_input_file>
Output :
输出 ：
##代码##回答by louie_45
回答by uzsolt
Maybe you can try
也许你可以试试
##代码##Check the ~in operand!
检查~in 操作数！
