在值之间找到多个字符串并在bash中用换行符替换

Question

I need to write a bash script to list values from an sql database.

I've got so far but now I need to get the rest of the way.

The string so far is

10.255.200.0/24";i:1;s:15:"10.255.207.0/24";i:2;s:14:"192.168.0.0/21

I now need to delete everything between the speech marks and send it to a new line.

desired output:

10.255.200.0/24
10.255.207.0/24
192.168.0.0/21

any help would be greatly appreciated.

Answer 1

$ tr '"' '\n' <<< $string | awk 'NR%2'

10.255.200.0/24
10.255.207.0/24
192.168.0.0/21

Answer 2

You could use :

echo 'INPUT STRING HERE' | sed $'s/"[^"]*"/\\\n/g'

Explanation :

• sed 's/<PATTERN1>/<PATTERN2/g' : we substitute every occurrence of PATTERN1 by PATTERN2
• [^"]* : any character that is not a " , any number of time
• \\\\\\n : syntax for newline in sed ( reference here )

Answer 3

Considering that your Input_file is same as shown sample then could you please try following.

awk '
{
  while(match($0,/[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+\/[0-9]+/)){
    print substr($0,RSTART,RLENGTH)
    $0=substr($0,RSTART+RLENGTH)
  }
}'   Input_file

Answer 4

This might work for you (GNU sed):

sed 's/"[^"]*"/\n/g' file

Or using along side Bash:

sed $'/"[^"]*"/\\n/g' file

Or using most other sed's:

sed ':a;/"[^"]*"\(.*\)\(.\|$\)/{G;s//\2\1/;ba}' file

This uses the feature that an unadulterated hold space contains a newline.