[英]How can i print specific fields, and get last word from last field that contain delimiter in each line
This maybe simple, but with lack of my programming skill, i can't figure it how to accomplish this.这可能很简单,但是由于缺乏我的编程技能,我无法弄清楚如何实现这一点。
I have content of file.txt like this:我有这样的file.txt内容:
1 22 July 2003 Path /Documents/Photo
2 23 July 2003 Path /Documents/Photo
3 24 July 2003 Path /Documents/Photo
and i only want to print field 2,3,4 and last word from last field that contain separator '/' each line.我只想打印字段 2,3,4 和最后一个字段的最后一个单词,每行包含分隔符“/”。
like this:像这样:
22 July 2003 Photo
23 July 2003 Photo
24 July 2003 Photo
i've tried this with awk command, but its delete all word from last field.我已经用 awk 命令尝试了这个,但是它从最后一个字段中删除了所有单词。
awk '{ print $2,$3,$4,$(NF=$NF) }'
Lynx:~ root# cat file.txt | awk '{ print $2,$3,$4,$(NF=$NF) }'
22 July 2003
23 July 2003
24 July 2003
how can i accomplish this with awk,sed or grep?我如何使用 awk、sed 或 grep 来完成此操作?
Thanks in Advance!提前致谢!
You can split the last field on /
and get the last item from the array您可以在
/
上拆分最后一个字段并从数组中获取最后一项
awk '{
i = split($NF,a,"/")
print $2,$3,$4,a[i]
}' file
Output Output
22 July 2003 Photo
23 July 2003 Photo
24 July 2003 Photo
You could also remove all until the last occurence of /
in the last field.您也可以在最后一个字段中最后一次出现
/
之前删除所有内容。
awk '{
sub(/.*\//, "", $NF)
print $2,$3,$4,$NF
}' file
With your shown samples, please try following.使用您显示的示例,请尝试以下操作。 Simply setting field separator to space OR / here and printing 2nd, 3rd, 4th and last field's values for each lines.
只需在此处将字段分隔符设置为空格或 / 并为每行打印第二、第三、第四和最后一个字段的值。
awk -F'[ /]' '{print $2,$3,$4,$NF}' Input_file
another one另一个
$ awk '{print $2,$3,$4,a[split($NF,a,"/")]}' file
or或者
$ awk '{sub(/.*\//,"",$NF); print $2,$3,$4,$NF}'
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