删除注释而不影响配置文件中的值

Question

我有一个配置文件，我需要删除以 # 开头到行尾的 cmets。 但它不应该影响双引号/单引号中的值。

我的输入文件：

# comment1
# comment2
#hbase_table_name=mytable # hbase table.
hbase_table_name=newtable # hbase table.
hbase_txn_family=txn
app_name= "cust#100"  # Name of the application
app_user= 'all#50,all2#100'  # users
hbase.zookeeper.quorum=localhost
zookeeper.znode.parent=/hbase-secure
hbase.zookeeper.property.clientPort=2181

我正在尝试的 perl 命令

perl -0777 -pe ' s/^\s*$//gms ; s/#.*?$//gm; s/^\s*$//gms;s/^$//gm' config.txt

我得到的输出是

hbase_table_name=newtable
hbase_txn_family=txn
app_name= "cust
app_user= 'all
hbase.zookeeper.quorum=localhost
zookeeper.znode.parent=/hbase-secure
hbase.zookeeper.property.clientPort=2181

但是需要的输出是

hbase_table_name=newtable
hbase_txn_family=txn
app_name= "cust#100"
app_user= 'all#50,all2#100'
hbase.zookeeper.quorum=localhost
zookeeper.znode.parent=/hbase-secure
hbase.zookeeper.property.clientPort=2181

我正在寻找使用任何工具的 bash 解决方案 - awk 或 perl 可以解决这个问题。

一种罕见的情况可能是像这样的配置条目

app_user= 'all#50,all2#100'  # users - "all" of them

结果应该是app_user= 'all#50,all2#100'

Answer 1

这是一个 perl 脚本：

#!/usr/bin/perl

use strict;

while (<DATA>){
    if (m/^\h*#/) {next;};
    if (m/((['"])[^\2]*\2)/) {print substr $_, 0, @+[0]; print "\n"; next; }
    s/#.*$//; print ;
}

__DATA__
# comment1
# comment2
#hbase_table_name=mytable # hbase table.
hbase_table_name=newtable # hbase table.
hbase_txn_family=txn
app_name= "cust#100"  # Name of the application
#app_name= "cust#100"  # Name of the application
app_user= 'all#50,all2#100'  # users
hbase.zookeeper.quorum=localhost
zookeeper.znode.parent=/hbase-secure
hbase.zookeeper.property.clientPort=2181
# from comments, other lines
hbase_table_name=newtable ## hbase table.
app_user= 'all#50,all2#100'  # users - "all" of them

输出：

hbase_table_name=newtable 
hbase_txn_family=txn
app_name= "cust#100"
app_user= 'all#50,all2#100'
hbase.zookeeper.quorum=localhost
zookeeper.znode.parent=/hbase-secure
hbase.zookeeper.property.clientPort=2181
hbase_table_name=newtable 
app_user= 'all#50,all2#100'

将<DATA> 更改为<> 并用于文件...

Answer 2

您能否尝试以下操作（使用所示示例编写和测试）。

awk '
/^#/{
  next
}
/".*"|\047.*\047/{
  match($0,/.*#/)
  print substr($0,RSTART,RLENGTH-1)
  next
}
{
  sub(/#.*/,"")
}
1
'  Input_file

说明：为上述代码添加详细说明。

awk '                                   ##Starting awk program from here.
/^#/{                                   ##Checking condition if a line starts from #  then do following.
  next                                  ##next will skip all further statements from here.
}
/".*"|\047.*\047/{                      ##Checking condition if a line matching regex from " to * OR single quote to single quote in current line.
  match($0,/.*#/)                       ##If above TRUE then come inside block; using match to match everything till # here.
  print substr($0,RSTART,RLENGTH-1)     ##Printing substring which prints from starting to length of matched regex with -1 to remove # in it.
  next                                  ##next willskip all further statements from here.
}
{
  sub(/#.*/,"")                         ##This statement will executewhen either a line is NOT starting from # OR  does not have single/double quote in it.
}
1                                       ##1 will print edited/non-edited lines here.