问题

I want to print lmn.

~$ echo abcde^10=fgi^11=jkl^12=lmn|grep -o '^12=[^)]*\^'

~$ echo abcde^11=jkl^12=lmn^10=fgi|grep -o '^12=[^)]*\^'

~$ echo 12=lmn^11=jkl^10=fgi^abcde|grep -o '12=lmn^'

abcde^10=fgi^11=jkl^12=lmn 

abcde^11=jkl^12=lmn^10=fgi

12=lmn^11=jkl^10=fgi^abcde 

~$ echo abcde^10=fgi^11=jkl^12=lmn|grep -o '\^12=[^)]*\^' -> this doesnt print anything 

~$ echo abcde^11=jkl^12=lmn^10=fgi|grep -o '\^12=[^)]*\^' -> this prints ^12=lmn^

答案1

得分: 2

With GNU awk for multi-char RS:

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='[\n^]' -F'=' '$1==12{print $2}'
lmn

$ echo '12=lmn^11=jkl^10=fgi^abcde' | awk -v RS='[\n^]' -F'=' '$1==12{print $2}'
lmn

or using any awk:

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='^' -F'[=\n]' '$1==12{print $2}'
lmn

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='^' -F'[=\n]' '$1==12{print $2}'
lmn

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='[\n^]' -F'=' '$1==12{print $2}'
lmn

$ echo '12=lmn^11=jkl^10=fgi^abcde' | awk -v RS='[\n^]' -F'=' '$1==12{print $2}'
lmn

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='^' -F'[=\n]' '$1==12{print $2}'
lmn

$ echo 'abcde^10=fgi^11=jkl^12=lmn' | awk -v RS='^' -F'[=\n]' '$1==12{print $2}'
lmn

答案2

得分: 1

使用在GNU awk中显示的示例，您可以在awk的while循环中使用正则表达式(^|\^)12=([^^]*)的match函数。

{
  while(match($0,/(^|\^)12=([^^]*)/,arr)){
     print arr[2]
     $0=substr($0,RSTART+RLENGTH)
  }
}

awk '
{
  while(match($0,/(^|\^)12=([^^]*)/,arr)){
     print arr[2]
     $0=substr($0,RSTART+RLENGTH)
  }
}
' Input_file

答案3

得分: 0

如果您想提取第一个`12=`和下一个`^`或字符串末尾（不确定）之间的部分，无需外部工具。 `bash` 替换就足够了：

s='abcde^10=fgi^11=jkl^12=lmn'
s="${s#12=}"；echo "${s%%^}"

或者：

[[ "abcde^10=fgi^11=jkl^12=lmn" =~ .12=([^^]).* ]] && echo "${BASH_REMATCH[1]}"

但如果您喜欢`sed`：

sed 's/.12=([^^]).*/\1/' <<< 'abcde^10=fgi^11=jkl^12=lmn'

<details>
<summary>英文:</summary>

If you want to extract the part between the first `12=` and the next `^` or the end of the string (not sure of that) you don&#39;t need external utilities. `bash` substitutions are enough:

s='abcde^10=fgi^11=jkl^12=lmn'
s="${s#12=}"; echo "${s%%^}"

Or:

[[ "abcde^10=fgi^11=jkl^12=lmn" =~ .12=([^^]).* ]] && echo "${BASH_REMATCH[1]}"

But if you prefer `sed`:

sed 's/.12=([^^]).*/\1/' <<< 'abcde^10=fgi^11=jkl^12=lmn'

</details>



# 答案4
**得分**: 0

使用正向后视断言的grep（仅打印匹配项）

grep -Po '(?<=12=)[^^]+'

使用sed（完全打印没有匹配的行）

sed -E 's/(^|.^)12=([^^]+)./\2/'

<details>
<summary>英文:</summary>

With grep using positive lookbehinds (prints only matches)

    grep -Po &#39;(?&lt;=12=)[^^]+&#39;

With sed (prints lines wihtout matches entirely)

    sed -E &#39;s/(^|.*\^)12=([^^]+).*//&#39;

</details>



# 答案5
**得分**: 0

echo &#39;
abcde^10=fgi^11=jkl^12=lmn 
abcde^11=jkl^12=lmn^10=fgi
12=lmn^11=jkl^10=fgi^abcde&#39; | 

nawk &#39;gsub(/.*12=|\^.+/,_)^_&#39;
&gt;

gawk &#39;$_=$2&#39; FS=&#39;.*12=|\\^.+&#39;
&gt;

mawk ++NF OFS= FS=&#39;.*12=|\^.+&#39;

lmn 
lmn
lmn

<details>
<summary>英文:</summary>

you mean like this ?

    echo &#39;
    abcde^10=fgi^11=jkl^12=lmn 
    abcde^11=jkl^12=lmn^10=fgi
    12=lmn^11=jkl^10=fgi^abcde&#39; | 
---
    nawk &#39;gsub(/.*12=|\^.+/,_)^_&#39;
&gt;
    gawk &#39;$_=$2&#39; FS=&#39;.*12=|\\^.+&#39;
&gt;
    mawk ++NF OFS= FS=&#39;.*12=|\^.+&#39;
---
    lmn 
    lmn
    lmn



</details>



# 答案6
**得分**: 0

lmn
lmn
lmn

<details>
<summary>英文:</summary>

    $ echo &#39;abcde^10=fgi^11=jkl^12=lmn
    abcde^11=jkl^12=lmn^10=fgi
    12=lmn^11=jkl^10=fgi^abcde
    &#39; | awk -F&#39;(^|\\^)12=&#39; &#39;NF&gt;1{sub(/\^.*/,&quot;&quot;,$NF); print $NF}&#39;
    lmn
    lmn
    lmn

</details>