bash 使用 sed 命令从文本文件中提取字符串
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Strings extraction from text file with sed command
提问by user1492786
I have a text file which contains some lines as the following:
我有一个文本文件，其中包含以下几行：
ASDASD2W 3ASGDD12 SDADFDFDDFDD W11 ACC=PNO23 DFSAEFEA EAEDEWRESAD ASSDRE
AERREEW2 3122312 SDADDSADADAD W12 ACC=HH34 23SAEFEA EAEDEWRESAD ASEEWEE
A15ECCCW 3XCXXF12 SDSGTRERRECC W43 ACC=P11 XXFSAEFEA EAEDEWRESAD ASWWWW
ASDASD2W 3122312 SDAFFFDEEEEE SD3 ACC=PNI22 ABCEFEA EAEDEWRESAD ASWEDSSAD
...
I have to extract the substring between the '=' character and the following blank space for each line , i.e.
我必须为每一行提取 '=' 字符和以下空格之间的子字符串，即
PNO23
HH34
P11
PNI22
I've been using the sedcommand but cannot figure out how to ignore all characters following the blank space.
我一直在使用sed命令，但无法弄清楚如何忽略空格后面的所有字符。
Any help?
有什么帮助吗？
采纳答案by Ignacio Vazquez-Abrams
Use the right tool for the job.
为工作使用正确的工具。
$ awk -F '[= ]+' '{ print }' input.txt
PNO23
HH34
P11
PNI22
回答by Jo So
Sorry, but have to add another one because I feel the existing answers are just to complicated
对不起，但必须添加另一个，因为我觉得现有的答案太复杂了
sed 's/.*=//; s/ .*//;' inputfile
回答by potong
This might work for you:
这可能对你有用：
sed -n 's/.*=\([^ ]*\).*//p' file
or, if you prefer:
或者，如果您愿意：
sed 's/.*=\([^ ]*\).*//p;d' file
回答by Sisay Chala
A chain of grep can do the trick.
一连串的 grep 可以解决问题。
grep -o '[=][a-zA-Z0-9]*' file | grep -o '[a-zA-Z0-9]*'
回答by tripleee
Put the string you want to capture in a backreference:
将要捕获的字符串放在反向引用中：
sed 's/.*=\([^ =]*\) .*//'
or do the substitution piecemeal;
或者逐步替换；
sed -e 's/.*=//' -e 's/ .*//'
回答by Paused until further notice.
sed 's/[^=]*=\([^ ]*\) .*//' inputfile
Match all the non-equal-sign characters and an equal sign. Capture a sequence of non-space characters. Match a space and the rest of the line. Substitute the captured string.
匹配所有非等号字符和一个等号。捕获一系列非空格字符。匹配一个空格和该行的其余部分。替换捕获的字符串。
