awk在可变空间分隔符之间打印单词作为数组

Supposing Ive a row like this:

假设我有这样的一行：

LOCUS       NG_052676              31180 bp    DNA     linear   PRI 08-AUG-2017

Which is being selected by match($0, /LOCUS\s*([^\n]*)/, o)

哪个是匹配选择的（$ 0，/ LOCUS \ s *（[^ \ n] *）/，o）

And printed by print o[1]

并通过打印o [1]打印

But this selects/prints the entire row because of the white space:

但由于白色空间，这会选择/打印整行：

NG_052676              31180 bp    DNA     linear   PRI 08-AUG-2017

How can I capture the first two strings as an array, o, such that: o[1] = NG_052676 and o[2] = 31180 ?

如何将前两个字符串捕获为数组o，这样：o [1] = NG_052676和o [2] = 31180？

NB I don't want to change the FS variable as that is being used for something else

NB我不想更改FS变量，因为它被用于其他东西

NB2 This is the entire awk function I am using:

NB2这是我正在使用的整个awk功能：

BEGIN{RS="//";FS=OFS="|"}
{
  match($0, /LOCUS\s*([^\n]*)/, o)
  match($0, /\(([^)]+)\)/, a)
  match($0, /\/gene="([^"]+)"/, b)
  match($0, /\/product="([^"]+)"/, c)
  match($0, /\/chromosome="([^"]+)"/, d)
  match($0, /\/map="([^"]+)"/, e)
  match($0, /Summary:\s([^\[]+)/, f)

  print o[1] " ", a[1] " ",b[1] " ", gensub(/\s\s+/, " ", "g1", c[1]) " ", 
  d[1] " ", e[1] " ",
    gensub(/\s\s+/, " ", "g2", f[1])
}

2 个解决方案

#1

With GNU awk (which you're already using) for the 3rd arg to match():

使用GNU awk（你已经在使用）为第3个arg匹配（）：

$ awk 'match($0, /LOCUS\s+(\S+)\s+(\S+)/, o) { print o[1], o[2] }' file
NG_052676 31180

#2

Since by default awk is using white space as FS , why not to consider the simpliest awk format?

因为默认情况下awk使用空格作为FS，为什么不考虑最简单的awk格式呢？

$ f1="LOCUS       NG_052676              31180 bp    DNA     linear   PRI 08-AUG-2017"
$ awk '{o[1]=$2;o[2]=$3}{print o[1],o[2]}' <(echo "$f1")
NG_052676 31180

You can still combine it with your regex:

您仍然可以将它与正则表达式结合使用：

$ awk '/LOCUS/{o[1]=$2;o[2]=$3;print o[1],o[2]}' <(echo "$f1")

#1