Cut the string in half with the last specific character shows up in the string

Question

For example, with the string a_link_list_java, use the _ character to separate the string in half would result in: a_link_list and java.

list_java would result in: list and java.

I want to take these two parts and store them into two variables.

I tried the cut command but it only works for only one character shown in the string

How can I accomplish this in a bash shell script?

Answer 1

The shell's parameter expansions can do that:

$ v="a_link_list_java"
$ printf "%s\n" "${v%_*}" "${v##*_}"
a_link_list
java

Though note that if the patterns in the expansion don't match, nothing is removed. So a value without an underscore would be unchanged by both transformations:

$ v=foo
$ printf "%s\n" "${v%_*}" "${v##*_}"
foo
foo

Answer 2

With sed:

var1=$(sed 's/_[^_]*$//' file)
var2=$(sed 's/.*_//' file)

Notice that sed is very greedy, so in the second case glob * swallows all characters, including previous underscores _.

Output:

$ echo "$var1"
a_link_list
$ echo "$var2"
java

Answer 3

Using awk you can have all in separate defining delimiter as _

VAR1=$(awk -F'_' '{print $1}' infile.txt) 

You can print which fields you want based on its index above instead of $1 that points to first field.

Or you can use bash and read as an array.

IFS='_' read -r -a array <<< "$myString"

then simply read with individual index number:

echo "${array[2]}"

Or even simply you can use bash parameter expansion:

VAR1="${myString%_*}"   # will gives you 'a_link_list' 
VAR2="${myString##*_}"  # will gives you 'java' 
VAR3="${VAR1%_*}"       # will gives you 'a_link'