shell

zhxjfl

有个文件如下：
2015.03.24 00:00:00|81845|81148|2769588|2764883
2015.03.24 00:01:00|81846|81149|2769614|2764908
2015.03.24 00:02:00|81848|81151|2769647|2764941
2015.03.24 00:03:00|81850|81153|2769676|2764970
2015.03.24 00:04:00|81851|81154|2769704|2764998
2015.03.24 00:05:00|81852|81155|2769727|2765021
2015.03.24 00:06:00|81854|81157|2769755|2765049
2015.03.24 00:07:00|81855|81158|2769781|2765075
2015.03.24 00:08:00|81855|81158|2769817|2765111
2015.03.24 00:09:00|81855|81158|2769847|2765141


怎么处理成如下结果呢（拿第一行举例）？就是把日期变成YYYYMMDD HH:MM，第一个|变成@，其它|变成:
'20150324 00:00'@81845:81148:2769588:2764883

谢谢！

我是一隻羊

先来个傻瓜版..

1. sed -r "s/^/'/;s/\.//g;s/:[0-9]{2}\|/'@/;s/\|/:/g" inputfile

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klainogn

1. sed 's/^/\'/;s/:..|/\'@/;s/\.//g;s/|/:/g' file

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zhxjfl

klainogn 发表于 2015-03-24 12:49

这个可以，不过要把前后两个'换成"

zsszss0000

1. sed "s/\.//g;s/:[0-9]\{2\}//2;s/|/@/1;s/|/:/g;s/\(.*\)\(@.*\)/'\1'\2/" file

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感觉这个写的太麻烦了

reyleon

原来还要去掉点号

1. awk -F\| '{gsub(/\./,"",$1);for(i=2;i<=NF;i++)s=s?s":"$i:$i;printf("\047%s\047@%s\n",substr($1,1,length($1)-3),s);s=""}' file

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fenghedong_007

[root@promote bash]# cat 2.txt |sed -nr 's/[.]//g;s/[|]/:/g;s/[:]/@/2p'
20150324 00:00@00:81845:81148:2769588:2764883
20150324 00:01@00:81846:81149:2769614:2764908
20150324 00:02@00:81848:81151:2769647:2764941
20150324 00:03@00:81850:81153:2769676:2764970
20150324 00:04@00:81851:81154:2769704:2764998
20150324 00:05@00:81852:81155:2769727:2765021
20150324 00:06@00:81854:81157:2769755:2765049
20150324 00:07@00:81855:81158:2769781:2765075
20150324 00:08@00:81855:81158:2769817:2765111
20150324 00:09@00:81855:81158:2769847:2765141

zxy877298415

1. awk '{sub(/^/,"\047",$1);gsub(/\./,"",$1);sub(/:00\|/,"\047@",$2);gsub(/\|/,":",$2)}1' file
   
2. '20150324 00:00'@81845:81148:2769588:2764883
   
3. '20150324 00:01'@81846:81149:2769614:2764908
   
4. '20150324 00:02'@81848:81151:2769647:2764941
   
5. '20150324 00:03'@81850:81153:2769676:2764970
   
6. '20150324 00:04'@81851:81154:2769704:2764998
   
7. '20150324 00:05'@81852:81155:2769727:2765021
   
8. '20150324 00:06'@81854:81157:2769755:2765049
   
9. '20150324 00:07'@81855:81158:2769781:2765075
   
10. '20150324 00:08'@81855:81158:2769817:2765111
    
11. '20150324 00:09'@81855:81158:2769847:2765141

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回复 1# zhxjfl


   

tuyajie

1. 
   
2. sed 's/[.]//g;s/:[0-9][0-9]|/@/;s/|/:/g'
   

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jcdiy0601

1. sed -e 's/|/@/' -e 's/|/:/g' test

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