In a bash script, replace character in string with multiple characters

Question

I want to replace "." in this result: "172.16.0.25" with " dot ".

Here is my code:

#!/bin/bash    
connection=`netstat -tn | grep :1337 | awk '{print $5}' | cut -d: -f1`
#this returns "172.16.0.25"
replace=" dot "
final=${connection/./$replace}
echo "$final"

Which returns: test.sh: 4: test.sh: Bad substitution

I tried using tr '.' ' dot ' but that only replaced the '.' with a space (' ')

I know this is a really dumb question, but I'm new to Shell Script.

Also, if it changes anything, I'm on a Raspberry Pi 2 running Raspbian.

Arjun Mathew Dan · Accepted Answer · 2015-06-03 05:59:01Z

2

You can do the same with awk alone :

netstat -tn | awk '/:1337/{sub(/:.*/,"",$5);gsub(/\./," dot ",$5);print $5}'

If pattern :1337 is matched, take the 5th field. Now remove the :number part. Also replace . with dot and print the field.

answered Jun 3, 2015 at 5:59

Arjun Mathew Dan

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4 Comments

Patrick Cook Over a year ago

I do have a question though, is there anyway I could replace an enter/return in the string?

Arjun Mathew Dan Over a year ago

Can u provide a sample input and output? Now you r getting the different addresses in one line... is that the concern?

Patrick Cook Over a year ago

Input: "172.16.0.25 (ENTER)172.16.0.25" Output: "172.16.0.25 and 172.16.0.26"

Arjun Mathew Dan Over a year ago

You can use sed ':a;N;$!ba;s/\n/ and /g' .Refer stackoverflow.com/questions/1251999/…

Amadan · Accepted Answer · 2015-06-03 05:58:24Z

2

connection=`netstat -tn | grep :1337 | awk '{print $5}' | cut -d: -f1 | sed 's/\./ dot /g'`

You can even simplify by staying in awk:

connection=`netstat -tn | awk '/1337:/ && !x { split($5, a, /:/); x = a[1]; gsub(/[.]/, " dot ", x); print x }'`

(I added && !x to make sure only one row is fetched, just in case.)

edited Jun 3, 2015 at 5:58

answered Jun 3, 2015 at 5:48

Amadan

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2 Comments

Patrick Cook Over a year ago

and the second one returns " "

Amadan Over a year ago

Hmm... could be something with Raspian... I believe it works on my end :(

Jahid · Accepted Answer · 2015-06-03 10:27:10Z

2

That line looks fine to me (although it will only replace the first dot; use ${connection//./$replace} to replace all of them), so the most likely thing is that you're not actually using bash.

The bash error message has a lower case b in bad substitution and puts the word line before the line number. The error message shown looks like it is coming from /bin/sh.

If you are running the script with the command

sh test.sh

then the system will use /bin/sh instead of /bin/bash.

edited Jun 3, 2015 at 10:27

Jahid

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answered Jun 3, 2015 at 6:03

rici

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Comments

ring bearer · Accepted Answer · 2015-06-03 07:45:51Z

1

While @Amadan's answer nails it, I am posting a variation(just for enthusiasts), risking negative votes :)

connection=$(netstat -tn | grep :1337 | awk '{gsub("\.","dot", $5); print $5}' | cut -d: -f1)

edited Jun 3, 2015 at 7:45

answered Jun 3, 2015 at 6:06

ring bearer

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Comments

Jahid · Accepted Answer · 2015-06-03 10:14:06Z

0

You can simply use:

final="${connection//./$replace}"

Example:

#!/bin/bash
connection="172.16.0.25"
replace=" dot "
final="${connection//./$replace}"
echo "$final"

Output:

172 dot 16 dot 0 dot 25

answered Jun 3, 2015 at 10:14

Jahid

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