如果前面有空格，则 BASH 将行附加到前面的行答案

【问题标题】：BASH append line to earlier line if preceded by spaces如果前面有空格，则 BASH 将行附加到前面的行
【发布时间】：2017-04-08 01:05:32
【问题描述】：

我有一个看起来像这样的文件：

t2_this_is_some_output_80_pool
            address 12.34.56.78
            state down
            address 13.34.56.78
            state down
t2_this_is_a_different_output_80_pool
            address 14.34.56.78
t2_this_is_another_output_80_pool
            address 15.34.56.78
            state up

我想输出它，所以它看起来像：

t2_this_is_some_output_80_pool address 12.34.56.78 state down
t2_this_is_some_output_80_pool address 13.34.56.78 state down
t2_this_is_a_different_output_80_pool address 14.34.56.78
t2_this_is_another_output_80_pool 15.34.56.78 state up

我一直在尝试使用 BASH、awk 和 sed，但我所做的一切都无法为我提供所需的输出。

我尝试过的事情之一：

用 = 替换开头的 12 个空格，如果行以 = 开头，则追加到上一行

cat file.txt | sed 's/            /=/' | sed -e :a -e '$!N;s/\n=/ /;ta' -e 'P;D'

但这不起作用...

任何帮助都会得到很大的帮助:-)

【问题讨论】：

我很确定 sed 可以解决问题。
请显示 bash、awk 或 sed 尝试，以便我们在您遇到问题的地方提供帮助。
已更新。我粘贴在帖子的第一个 daft 中，而不是第二个 face palm

标签： bash awk sed

【解决方案1】：

大概是这样的：

$ cat f1
t2_this_is_some_output_80_pool
            address 12.34.56.78
            state down
            address 13.34.56.78
            state down
t2_this_is_a_different_output_80_pool
            address 14.34.56.78
t2_this_is_another_output_80_pool
            address 15.34.56.78
            state up    

$ echo $(cat f1) | sed 's/t2/\nt2/g'

t2_this_is_some_output_80_pool address 12.34.56.78 state down address 13.34.56.78 state down 
t2_this_is_a_different_output_80_pool address 14.34.56.78 
t2_this_is_another_output_80_pool address 15.34.56.78 state up

【讨论】：

【解决方案2】：

sed ':a; $!N;s/\n[ ]\+/ /;ta;P;D' yourfile.txt

$!N 如果不是最后一行，则追加下一行。

s\n[ ]\+/ / 用一个空格替换新行之后的所有空格。

ta如果有匹配，转到标签:a

Pprint 修改的行

D删除旧版行

【讨论】：

【解决方案3】：

下面是一个简单的选项，将 sed 反向引用和您的 = 标识符结合起来：

sed -r 's/^([^ ])/=\1/g' file.txt |tr '\n' ' ' |tr '=' '\n' |sed -r 's/ +/ /g'

如果我正确地回答了您的问题，这将给出以下输出，这可能不是您正在寻找的内容（因为它不会在每个地址的开头使用相应的t2_ 开始一个新行):

t2_this_is_some_output_80_pool address 12.34.56.78 state down address 13.34.56.78 state down 
t2_this_is_a_different_output_80_pool address 14.34.56.78 
t2_this_is_another_output_80_pool address 15.34.56.78 state up

说明：

sed -r 's/^([^ ])/=\1/g'：在每行的开头放置一个 = 字符，而不是以空格开头。

输出：

=t2_this_is_some_output_80_pool
            address 12.34.56.78
            state down
            address 13.34.56.78
            state down
=t2_this_is_a_different_output_80_pool
            address 14.34.56.78
=t2_this_is_another_output_80_pool
            address 15.34.56.78
            state up

tr '\n' ' '：将每个换行符替换为空格字符。

输出：

=t2_this_is_some_output_80_pool             address 12.34.56.78             state down             address 13.34.56.78             state down =t2_this_is_a_different_output_80_pool             address 14.34.56.78 =t2_this_is_another_output_80_pool             address 15.34.56.78             state up

tr '=' '\n'：将每个 = 字符替换为换行符。

输出：

t2_this_is_some_output_80_pool             address 12.34.56.78             state down             address 13.34.56.78             state down 
t2_this_is_a_different_output_80_pool             address 14.34.56.78 
t2_this_is_another_output_80_pool             address 15.34.56.78             state up

sed -r 's/ +/ /g'：将所有连续的空格字符替换为单个空格字符。

输出：

t2_this_is_some_output_80_pool address 12.34.56.78 state down address 13.34.56.78 state down 
t2_this_is_a_different_output_80_pool address 14.34.56.78 
t2_this_is_another_output_80_pool address 15.34.56.78 state up

【讨论】：

【解决方案4】：

awk '
/^[^[:blank:]+]/{                     # search for record/line does not start with blank
            if(s){print s; s=""}      # This is whenever state not found then print variable s
            p=$0                      # store record in varibale p   
            next                      # stop processing go to next line
}
{                             
      gsub(/^[ \t]+/,"")              # suppress starting space/tab char 
      s = s ? s OFS $0: p OFS $0      # if variable s has something then concatenate variable s with current record else variable p and current record
}
/state/{                              # if record has word state then 
      print s; s=""                   # print variable s and reset variable
}
END{                                  # end block
     if(s)print s                     # if s has something print s
}' file

Oneliner

$ awk '/^[^[:blank:]+]/{if(s){print s; s=""}p=$0; next}{gsub(/^[ \t]+/,"");s = s ? s OFS $0: p OFS $0; }/state/{print s; s=""}END{if(s)print s}' file

输入

$ cat f
t2_this_is_some_output_80_pool
            address 12.34.56.78
            state down
            address 13.34.56.78
            state down
t2_this_is_a_different_output_80_pool
            address 14.34.56.78
t2_this_is_another_output_80_pool
            address 15.34.56.78
            state up

输出

$ awk '
/^[^[:blank:]+]/{
            if(s){print s; s=""}
            p=$0; 
            next
}
{
      gsub(/^[ \t]+/,"")
      s = s ? s OFS $0: p OFS $0 
}
/state/{
      print s; s=""
}
END{
     if(s)print s
}' f
t2_this_is_some_output_80_pool address 12.34.56.78 state down
t2_this_is_some_output_80_pool address 13.34.56.78 state down
t2_this_is_a_different_output_80_pool address 14.34.56.78
t2_this_is_another_output_80_pool address 15.34.56.78 state up

【讨论】：

Akshay，太棒了！我认为我可以看到它是如何工作的 :-) 非常感谢您的帮助！非常感谢所有给我答案的人！猜猜我需要再次挖掘我的书！

【解决方案5】：

sed '
# for every new section (starting with non space)
   /^[^[:blank:]]/ {
# copy it to the hold buffer
      h
# delete the line (and go to next line)
      d
      }
# other line(s)
# add next line
   N
# add holding buffer to current line (with new line between)
   G
# reformat, removing new line followed by spaces and putitnging last line in front
   s/[[:blank:]]*\([^[:blank:]].*\)\^J[[:blank:]]*\([^[:blank:]].*\)\^J\(.*\)/\3 \1 \2/
# print the resulting line
   ' YourFile

在线POSIX中如此

sed -e '/^[^[:blank:]]/{h;d' -e '};N;G;s/[[:blank:]]*\([^[:blank:]].*\)\^J[[:blank:]]*\([^[:blank:]].*\)\^J\(.*\)/\3 \1 \2/' YourFile

注意：\^J 是真正的换行符 CTRL+V+J。 \n avalaible for gawk 和 \ with new line for POSIX 是等效的

【讨论】：

【解决方案6】：

我相信使用简单的 awk 脚本会更容易理解，例如：

awk '
BEGIN {
    separator_address = "address" # Put what you want here
    line=""
    addr=""
    args=""
}
{
    if (substr($0, 1, 1) != " ") { # Start by a space
        if (addr != "") {
            print line" "addr" "args
        }
        line=$0
        addr = ""
        args = ""
        next # Nothing more to do here
    }
    gsub(/^[ \t]+/,"",$0) # Remove spaces
    if (substr($0, 1, length(separator_address)) == separator_address) { 
        if (addr == "") { 
            addr = $0
        } else { 
            print line" "addr" "args
            addr = $0
            args = ""
        } 
    } else { 
        args = $0
    } 
}
END {
    if (addr != "") {
        print line" "addr" "args
    }
}' file.txt

Sed 确实非常强大，但在多行处理方面并不是很方便。但是因为我真的不希望如何分割行，所以我使用了分隔符（“地址”）。如果它是两行，则可以使用标志而不是分隔符（例如，初始化为 0，如果读取以空格开头的某个值，则 +1，如果达到 2，则打印，例如，在每次打印后重新初始化）。

结果：

t2_this_is_some_output_80_pool address 12.34.56.78 state down
t2_this_is_some_output_80_pool address 13.34.56.78 state down
t2_this_is_a_different_output_80_pool address 14.34.56.78
t2_this_is_another_output_80_pool 15.34.56.78 state up

【讨论】：

【解决方案7】：

@try：还有一种方法。

awk '{ORS=$0 ~ /^ +/?"":RS} {$1=$1} 1; END{print RS}'  Input_file

编辑：现在在这里添加代码解释也很成功。

awk '{
ORS=           ##### Setting ORS(Output record separator) here, where output field separator is awk's default keyword whose default value is a new line.
$0 ~ /^ +/     ##### Checking if any record/line is starting from space here.
?              ##### Conditional operator here, which will execute it's following statements if above mentioned condition is TRUE.
""             ##### If above mentioned conditions is TRUE then setting ORS's value to NULL.
:              ##### : is a conditional operator here, which will execute the statements following it when conditions is FALSE.
RS}            ##### Setting ORS's value to RS(Record separator) whose default value will be a new line.
{$1=$1}        ##### Resetting the $1(first field) here, so that value of ORS could be reflected here.
1;             ##### Mentioning 1 here. awk works on condition then action method, so I am making condition TRUE here and not mentioning any action here so by default print action will happen here.
END{print RS}  ##### in this block printing the RS(record separator)'s value here.
'  Input_file  ##### mentioning the Input_file here.

【讨论】：

【解决方案8】：

如果您不介意开头有一个额外的换行符，而结尾处没有，那么这个简短的脚本可以：

awk '/^[^ ]/ { printf "\n" } $1=$1' ORS=' '

您可以通过分别添加NR>1 和END { print "\n" } 来解决这两个问题。

【讨论】：