Bash Shell - 在第二次出现某个字符后返回子字符串
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Bash Shell - Return substring after second occurrence of certain character
提问by Leo
I need to return everything after a delimeter I decide but still don't fully know how to use sed. What I need to do is:
我需要在我决定的分隔符之后返回所有内容，但仍然不完全知道如何使用 sed。我需要做的是：
$ echo "ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,," \
| sed <some regexp>
For this example the return should be (substring)everything after the second comma:
对于这个例子，返回应该是（子字符串）第二个逗号之后的所有内容：
123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,
I can do this with cut like this:
echo "ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,," | cut -d',' -f 2
我可以像这样用 cut 做到这一点：
echo "ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,," | cut -d',' -f 2
but I've been told cut is slower than sed...
但有人告诉我 cut 比 sed 慢...
Can some guru who has them (and wants to... :) ) give me a few minutes of his time and advice me please? Thanks! Leo
一些拥有它们（并想要...... :)）的大师可以给我几分钟的时间并给我建议吗？谢谢！狮子座
采纳答案by Thor
In my experience cutis always faster than sed.
以我的经验cut总是比sed.
To do what you want with sedyou could use a non-matching group:
要执行您想要的操作，sed您可以使用不匹配的组：
echo 'ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,' |
sed -r 's/([^,]*,){2}//'
This removes the first two fields (if the fields do not contain commas themselves) by removing non-comma characters [^,]followed by a comma twice {2}.
这通过删除非逗号字符[^,]后跟逗号两次来删除前两个字段（如果字段本身不包含逗号）{2}。
Output:
输出：
123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,
回答by chepner
You could also try doing the extraction in bashwithout spawning an external process at all:
您也可以尝试在bash不产生外部进程的情况下进行提取：
$ [[ 'ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,' =~ [^,]*,[^,]*,(.*) ]]
$ echo "${BASH_REMATCH[@]}"
123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,
or
或者
$ FOO='ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,'
$ echo ${FOO/+([^,]),+([^,]),}
or
或者
$ IFS=, read -a FOO <<< 'ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,'
$ echo ${FOO[@]:2}
(Assuming this is for a one-off match, not iterating over the contents of a file.)
（假设这是一次性匹配，而不是迭代文件的内容。）
回答by Vijay Nirmal
This method is by find the index of second occurrence of a character and using bash substring to get the required result
这种方法是通过查找字符第二次出现的索引并使用bash子字符串来获得所需的结果
input="ABC DE,FG_HI J,123.XYZ-A1,DD/MM/YYYY HH24:MI:SS,,,"
index=$(($(echo $input| grep -aob '/' | grep -oE '[0-9]+' | awk 'NR==2') + 1))
result=${input:$index}
