英文:
sed/grep to extract word between 2 characters
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^'
英文:
My sentence is
abcde^10=fgi^11=jkl^12=lmn
The sequence can be jumbled also
for eg.
abcde^11=jkl^12=lmn^10=fgi
The sequence can also be :
12=lmn^11=jkl^10=fgi^abcde
I want to print lmn.
I have gotten till thus far:
~$ 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
英文:
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
<p>
$ 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)}}
英文:
With your shown samples in GNU awk, you can use match function of awk with regex (^|\^)12=([^^]*) in while loop.
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'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 "${BASH_REMATCH[1]}"
But if you prefer `sed`:
sed 's/.12=([^^]).*/\1/' <<< 'abcde^10=fgi^11=jkl^12=lmn'
</details># 答案4**得分**: 0使用正向后视断言的grep（仅打印匹配项）
grep -Po '(?<=12=)[^^]+'
使用sed（完全打印没有匹配的行）
sed -E 's/(^|.^)12=([^^]+)./\2/'
<details><summary>英文:</summary>With grep using positive lookbehinds (prints only matches)grep -Po '(?<=12=)[^^]+'With sed (prints lines wihtout matches entirely)sed -E 's/(^|.*\^)12=([^^]+).*//'</details># 答案5**得分**: 0echo 'abcde^10=fgi^11=jkl^12=lmnabcde^11=jkl^12=lmn^10=fgi12=lmn^11=jkl^10=fgi^abcde' |nawk 'gsub(/.*12=|\^.+/,_)^_'>gawk '$_=$2' FS='.*12=|\\^.+'>mawk ++NF OFS= FS='.*12=|\^.+'lmnlmnlmn<details><summary>英文:</summary>you mean like this ?echo 'abcde^10=fgi^11=jkl^12=lmnabcde^11=jkl^12=lmn^10=fgi12=lmn^11=jkl^10=fgi^abcde' |---nawk 'gsub(/.*12=|\^.+/,_)^_'>gawk '$_=$2' FS='.*12=|\\^.+'>mawk ++NF OFS= FS='.*12=|\^.+'---lmnlmnlmn</details># 答案6**得分**: 0lmnlmnlmn<details><summary>英文:</summary>$ echo 'abcde^10=fgi^11=jkl^12=lmnabcde^11=jkl^12=lmn^10=fgi12=lmn^11=jkl^10=fgi^abcde' | awk -F'(^|\\^)12=' 'NF>1{sub(/\^.*/,"",$NF); print $NF}'lmnlmnlmn</details>
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