shell script to replace space in 2nd column after delimiter in same file

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i have a file with below content in linux:

servername: tesing1001

os: Unix

bu:

aid: 132

location: anywhere

environment: dev-stg

application: ABCD testing space

city: kk

I want to replace content in same file with below data:

servername: tesing1001

 os: Unix

**bu: BLANK**  **>>>>here since value is empty i will set it as BLANK hardcoded**

aid: 132

location: anywhere

environment: dev-stg

**application: ABCD_testing_space**  **>>>>>here we will replace string space with "_"**

city: kk

all this thing we will do in same file . So, far i am trying to acheive above output with below logic

    #!/bin/bash
    cp -p /opsunix/dyfacter.txt /tmp/customized.txt.tmp
    awk -F ":" '{
                 if ($2 == "")
                  {
                    print $0  " blank"
                  } else {
                      print $0
                        } 
                }' /opsunix/dyfacter.txt > /tmp/customized.txt.tmp && mv /tmp/customized.txt.tmp /opsunix/dyfacter.txt

with the help of above code, am able to identify null value and replace it with "blank" string.

servername: tesing1001

 os: Unix

**bu: blank**  **>>>>done**

aid: 132

location: anywhere

environment: dev-stg

**application: ABCD testing space**  **>>>>>still need to correct**

city: kk

however 2nd last line i.e: application: ABCD testing space am not able to convert it as application: ABCD_testing_space.

appliying sed -i 's/ /_/g' /opsunix/dyfacter.txt is replacing all spaces in file after : .

purpose is to replace space in string .

Please help!

sed awk shell linux
4 Answers

Using sed, this will change all null values to BLANK and strings with consecutive spaces between the values to change the space to _ after the delimiter :

$ sed -E ':a;s/^([^ ]*: +)([^ ]*) +/\1\2_/;ta;s/: +?$/: BLANK/' input_file
servername: tesing1001

os: Unix

bu: BLANK

aid: 132

location: anywhere

environment: dev-stg

application: ABCD_testing_space

city: kk
user_163721090

Maybe this will help you. I replace all spaces with an _ After that i replace all ":_" with ": "

foo=${str// /_}
echo $foo
echo ${foo/:_/: }
user_164541760

Using awk (gawk in particular):

awk -F ': ' '/^application/ { gsub(" ","_",$2) } /^bu/ { gsub("","BLANK",$2)}1' file

Set the field delimiter to ": " and then where the line starts with "application", use awks, gsub function to replace all instances of " " with "_" in the second field. With lines starting with bu, replace "" with "BLANK" in the second field

This will output the amendments to the screen, to write the data back to the file, use a temporary file and so:

awk -F ': ' '/^application/ { gsub(" ","_",$2) } /^bu/ { gsub("","BLANK",$2)}1' file > file.tmp && cp -f file.tmp file
user_78404400

Using awk you could set the FS (Field Separator) to a colon followed by optional spaces, and the OFS (Ouput Field Separator) to :

Using a pattern, you could match if the line starts with non whitespace chars followed by : and then optional spaces only.

If that is the case, print out the first field with the OFS and "BLANK" and move to the next line as there is nothing to substitute on this line.

For all the other lines, replace " " with _

awk '
BEGIN{
  FS=":[[:space:]]*";
  OFS=": "
}
/^[^[:space:]]*:[[:space:]]*$/{
  print $1 OFS "BLANK"
  next
}
{
  gsub(" ","_", $2)
}1
' file

Output

servername: tesing1001

os: Unix

bu: BLANK

aid: 132

location: anywhere

environment: dev-stg

application: ABCD_testing_space

city: kk

With gnu awk you might also match the whole line using a pattern and 2 capture groups, and do the replacement based on the value of capture group 2.

awk '
match($0, /^([^[:blank:]]+:[[:blank:]]*)(.+)?$/, a) {
  if (a[2] != "") {
    print a[1] gensub(" ", "_", "g", a[2])
    next
  }
  print a[1] "BLANK"
}
' file
user_54249880
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