bash 使用 sed 提取大括号中的子字符串
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Using sed to extract a substring in curly brackets
提问by TheMightyLlama
I've currently got a string as below:
我目前有一个字符串如下：
integration@{Wed Nov 19 14:17:32 2014} branch: thebranch
This is contained in a file, and I parse the string. However I want the value between the brackets {Wed Nov 19 14:17:32 2014}
这包含在一个文件中，我解析了该字符串。但是我想要括号之间的值{Wed Nov 19 14:17:32 2014}
I have zero experience with Sed, and to be honest I find it a little cryptic.
我对 Sed 的经验为零，老实说，我觉得它有点神秘。
So far I've managed to use the following command, however the output is still the entire string.
到目前为止，我已经设法使用以下命令，但是输出仍然是整个字符串。
What am I doing wrong?
我究竟做错了什么？
sed -e 's/[^/{]*"\([^/}]*\).*//'
采纳答案by Avinash Raj
To get the values which was between {, }
要获得介于 之间的值{，}
$ sed 's/^[^{]*{\([^{}]*\)}.*//' file
Wed Nov 19 14:17:32 2014
回答by Jotne
This is very simple to do with awk, not complicate regex.
这很简单awk，而不是复杂的正则表达式。
awk -F"{|}" '{print }' file
Wed Nov 19 14:17:32 2014
It sets the field separator to {or }, then your data will be in the second field.
它将字段分隔符设置为{或}，那么您的数据将位于第二个字段中。
FScould be set like this to:
FS可以这样设置：
awk -F"[{}]" '{print }' file
To see all field:
查看所有字段：
awk -F"{|}" '{print "field#1=""\nfield#2=""\nfield#3="}' file
field#1=integration@
field#2=Wed Nov 19 14:17:32 2014
field#3= branch: thebranch
回答by nu11p01n73R
This might work
这可能有效
sed -e 's/[^{]*\({[^}]*}\).*//g'
Test
测试
$ echo "integration@{Wed Nov 19 14:17:32 2014} branch: thebranch" | sed -e 's/[^{]*{\([^}]*\)}.*//g'
Wed Nov 19 14:17:32 2014
2014 年 11 月 19 日星期三 14:17:32
Regex
正则表达式
[^{]*Matches anything other than the{, That isintegration@([^}]*)Capture group 1\{Matches{[^}]*matches anything other than}, That isWed Nov 19 14:17:32 2014\}matches a}
.*matches the rest
[^{]*匹配除 之外的任何内容{，即integration@([^}]*)捕获组 1\{火柴{[^}]*匹配除 之外的任何内容}，即Wed Nov 19 14:17:32 2014\}匹配一个}
.*匹配其余的
回答by user6093493
Simply, below command also get the data...
简单地说，下面的命令也获取数据......
echo "integration@{Wed Nov 19 14:17:32 2014} branch: thebranch" | sed 's/.*{\(.*\)}.*//g'
