提取方括号之间的两个文本之间的字符串

Question

I have strings similar to below ones

1. the quick brown `[fox].[jumps]` [over] the lazy dog
 2. the quick brown fox [jumps] [over] the lazy dog
 3. `[the].[quick]` brown `[fox].[jumps]` [over] the lazy dog

I would need to extract below values

 1. fox.jumps
 2. <Nothing>
 3. the.quick, fox.jumps

Please could you help me with the regular expressions in shell scripts?

Answer 1

With GNU awk for multi-char RS, RT, and gensub():

$ awk -v RS='[[][^]]*][.][[][^]]*]' 'RT{print gensub(/[][]/,"","g",RT)}' file
fox.jumps
the.quick
fox.jumps

Answer 2

With your shown samples, could you please try following. Written and tested in GNU awk .

awk '
{
  val=""
  for(i=1;i<=NF;i++){
    if($i~/^\[.*]\.\[.*]$/){
      gsub(/[][]/,"",$i)
      val=(val?val ", ":"")$i
    }
  }
  print (val==""?"<Nothing>":val)
}'  Input_file

Sample output will be as follows as per shown samples.

fox.jumps
<Nothing>
the.quick, fox.jumps

Answer 3

This is another one-liner... from shell point of view (you can remove \ and newline to make it one line).

Make sure \ is always last character of the line and no space after that.

gawk '{\
  for(i=1;i<NF;i++)\
  {\
     if(match($i,/\]\.\[/)>0)\
     {\
         for(k=1;k<length($i);k++)\
         {\
            c=substr($i,k,1);\
            if(c!="[" && c!="]")\
            printf("%s",c);\
         }\
         printf(" ");\
     }\
  }\
  printf("\n");\
}' example.txt

Anyway, it would be useful to put the gawk-code in between ' and ' into a file (file.awk, in file.awk remove all \ ) and then call, gawk like so, meaning test.awk starts with { and ends with } . It might not be an elegant solution, but you can add a lot more to this, like many variables, a whole program, subroutines, ...

gawk -f test.awk example.txt

Output:
fox.jumps 

the.quick fox.jumps

Answer 4

With sed (that supports \n in s/ commands).

sed '
    s/$/\n/
    : again
    /\([^\n]*\)\[\([^]]*\)\]\.\[\([^]]*\)\]\([^\n]*\)\n/{
        s//\1\4\n\2.\3\n/
        b again
    }
    s/[^\n]*\n//
    s/\n$//
    s/\n/, /g
'

• s/$/\n/ add a newline on the end of read line. Important in case of lines without any regex.
• : again define label again that you can go to
• /../ - match a regex
  • \([^\n]*\) match any non-newline and remember it in \1
  • \[\([^]]*\)\]\.\[\([^]]*\)\] Match [somethign].[something] and remember parts in \2 and \3
  • \([^\n]*\) - match any non-newline
• /.../{ - when the regex is matched - s// - reuse last regex, ie. the one above - /\1\4\n\2.\3\n/ - shuffle input so that place the non-interesting parts before the newline, and extracted interesting part after the newline - b again - go to again, to match another pattern
• s/[^\n]*\n// remove the non-matched part of line
• s/\n$// - remove trailing newline
• s/\n/, /g separate parts with comma and a space.

Example:

$ sed 's/$/\n/; : again; /\([^\n]*\)\[\([^]]*\)\]\.\[\([^]]*\)\]\([^\n]*\)\n/{ s//\1\4\n\2.\3\n/; b again; }; s/[^\n]*\n//; s/\n$//; s/\n/, /g' <<EOF
the quick brown [fox].[jumps] [over] the lazy dog
the quick brown fox [jumps] [over] the lazy dog
[the].[quick] brown [fox].[jumps] [over] the lazy dog
EOF

outputs:

fox.jumps

the.quick, fox.jumps

If you do not want the empty line in between, then do not output anything from sed in such case where no patterns where found. Add sed -n and on the end of script do not output if empty - /^$/!p , like so:

sed -n 's/$/\n/; : again; /\([^\n]*\)\[\([^]]*\)\]\.\[\([^]]*\)\]\([^\n]*\)\n/{ s//\1\4\n\2.\3\n/; b again; }; s/[^\n]*\n//; s/\n$//; s/\n/, /g; /^$/!p'

Answer 5

With GNU awk for FPAT (using regexp and gensub function from Ed Morton's code):

awk -v OFS=', ' -v FPAT='[[][^]]*][.][[][^]]*]' '{for (i=1; i<=NF; i++) printf "%s%s", gensub(/[][]/,"","g",$i), (i<NF?OFS:ORS)}' file
fox.jumps
the.quick, fox.jumps