Question

我有一个包含多行的文件

http://example.com/part-1   this    number 1 one 
http://example.com/part--2  this is number 21 two
http://example.com/part10   this is an number 12 ten
http://example.com/part-num-11  this is an axample  number 212 eleven

如何删除第一个columd和“number x”之间“number x”+之后的所有字符...我想要这样的输出

http://example.com/part-1    1
http://example.com/part--2   21 
http://example.com/part10    12
http://example.com/part-num-11   212

另一个案例：输入：

http://server1.example.com/00/part-1    this    number 1 one 
http://server2.example.com/1a/part--2   this is section 21 two two
http://server3.example.com/2014/5/part10    this is an Part 12 ten  ten ten
http://server5.example.com/2014/7/part-num-11   this is an PARt number 212 eleven

我想要相同的输出....而且数字总是在最后一个数字字段

Answer 1

这是一种方式：

awk -F"number" '{split($1,a," ");split($2,b," ");print a[1],b[1]}' file
http://example.com/part-1 1
http://example.com/part--2 21
http://example.com/part10 12
http://example.com/part-num-11 212

如果您想拥有的号码始终位于倒数第二个字段，那么也应该这样做：

awk '{print $1,$(NF-1)}' file
http://example.com/part-1 1
http://example.com/part--2 21
http://example.com/part10 12
http://example.com/part-num-11 212

Answer 2

sed -r 's/^([^0-9]*[0-9]+)[^0-9]*([0-9]+).*/\1 \2/' file

输出：

http://example.com/part-1 1
http://example.com/part--2 21
http://example.com/part10 12
http://example.com/part-num-11 212

Answer 3

试试这个：

sed 's/ .*number \([0-9]+\).*/ \1/' myfile.txt

Answer 4

感谢大家......从您的评论中，我有自己的解决方案：

sed -re 's/([0-9]*[0-9]+)/#\1#/g' | sed -re 's/(^.*#).*/\1/g' | sed 's/#//g' | awk '{print $1"  "$NF}'

我的想法：用＃[numbers]＃替换所有数字组，然后选择从行首到“＃”的所有字符（sed将选择最后一个＃）并删除所有其余字符。接下来是awk

谢谢大家（y）

删除匹配字符后的所有字符

4 个答案: