Bash – How to Split a String by the Third Dot Delimiter

bashshellsplitstring

I did a few research, I can split the string by '.' but the only thing I want to replace is the third '.', and the version number is actually variable so how can I do the replace/split?

version: 1.8.0.110

but What I want is the output like this:

version: 1.8.0-110

Best Answer

Used sed for example:

$ echo 'version: 1.8.0.110' | sed 's/\./-/3'
version: 1.8.0-110

Explanation:

sed s/search/replace/x searches for a string and replaces the second. x determines which occurence to replace - here the 3rd. Often g is used for x to mean all occurances.

Here we wish to replace the dot . but this is a special character in the regular expression sed expects in the search term. Therefore we backslashify the . to \. to specify a literal ..

Since we use special characters in the argument to sed (here, the backslash \) we need to put the whole argument in single quotes ''. Many people always use quotes here so as not to run into problems when using characters that might be special to the shell (like space ).

Related Solutions

Bash – Splitting String by First Occurrence of a Delimiter

cut sounds like a suitable tool for this:

bash-4.2$ s='id;some text here with possible ; inside'

bash-4.2$ id="$( cut -d ';' -f 1 <<< "$s" )"; echo "$id"
id

bash-4.2$ string="$( cut -d ';' -f 2- <<< "$s" )"; echo "$string"
some text here with possible ; inside

But read is even more suitable:

bash-4.2$ IFS=';' read -r id string <<< "$s"

bash-4.2$ echo "$id"
id

bash-4.2$ echo "$string"
some text here with possible ; inside

Bash String Split – How to Split a String into an Array in Bash

Pure bash solution, no external tools used to process the strings, just parameter expansion:

#! /bin/bash
str='battery.charge: 90 battery.charge.low: 30 battery.runtime: 3690 battery.voltage: 230.0 device.mfr: MGE UPS SYSTEMS device.model: Pulsar Evolution 500'

IFS=: read -a fields <<< "$str"

for (( i=0 ; i < ${#fields[@]} ; i++ )) ; do
    f=${fields[i]}

    notfirst=$(( i>0 ))
    last=$(( i+1 == ${#fields[@]} ))

    (( notfirst )) && echo -n ${f% *}

    start=('' $'\n' ' ')
    colon=('' ': ')
    echo -n "${start[notfirst + last]}${f##* }${colon[!last]}"
done
echo

Explanation: $notfirst and $last are booleans. The part before the last space ${f% *} isn't printed for the first field, as there is no such thing. $start and $colon hold various strings that separate the fields: at the first item, notfirst + last is 0, so nothing is prepended, for the rest of the lines, $notfirst is 1, so a newline is printed, and for the last line, the addition gives 2, so a space is printed. Then, the part after the last space is printed ${f##* }. Colon is printed for all lines except the last one.

Best Answer

Related Solutions

Bash – Splitting String by First Occurrence of a Delimiter

Bash String Split – How to Split a String into an Array in Bash

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