Get something between separators with regular expressions

Question

I am trying to extract the middle value from a register that is something like this:

NAME=PATH=USER=DATE

I need to get the path part and all I came up with is this:

=[^=]*=

the problem is that it includes the = at the beginning and at the end. How could i not include them?

Also, which unix command should I use to extract it? I was thinking sed, but I usually use it to replace strings not to get them. grep maybe?

I am new in bash programming...

Is there any place where I can learn some regex? because I think I will need to really know how to work with them.

Answer 1

Use captured groups:

if [[ $str =~ =([^=]+)= ]]
then
    echo "Part between = and = is ${BASH_REMATCH[1]}."
fi

Answer 2

In bash:

IFS="="
a="NAME=PATH=USER=DATE"
read -a b <<< "$a"
echo "${b[1]}"

UPDATE as suggested by mklement0

a="NAME=PATH=USER=DATE"
IFS="=" read -a b <<< "$a"
echo "${b[1]}"

Answer 3

So as to have several possibilities, you can also use read and make an array with all your fields:

var="NAME=PATH=USER=DATE"
IFS== read -r -a var_ary <<< "$var"
echo "field1: ${var_ary[0]}"
echo "field2: ${var_ary[1]}"
echo "field3: ${var_ary[2]}"
echo "field4: ${var_ary[3]}"

will output:

field1: NAME
field2: PATH
field3: USER
field4: DATE

this will also enable you to check that you have the correct number of fields:

if ((${#var_ary[@]}==4)); then
    echo "Cool I have 4 fields"
else
    echo "Oh no, I don't have 4 fields (I have ${#var_ary[@]} fields)"
fi

Answer 4

No need for complicated regex, a simple awk can do it:

echo "NAME=PATH=USER=DATE" | awk -F= '{print $2}'
PATH

Answer 5

You can use bash parameter to remove the leading and trailing pieces.

$ s='NAME=PATH=USER=DATE'
$ s=${s#*=} && echo "${s%%=*}"
PATH

%% removes the longest match from behind and # removes the shortest match from front. Using them together allows you to remove pieces you don't need. You can read more about bash parameter expansion here.

Answer 6

NAME=PATH=USER=DATE

Multiple ways to extract this data. The easiest may be pattern filtering. Pattern filtering has four forms:

• ${VAR#PATTERN} - Remove the smallest left most part of the string that matches the pattern.
• ${VAR##PATTERN} - Remove the largest left most part of the string that matches.
• ${VAR%PATTERN} - Remove the **smallest right most part of the string that matches.
• ${VAR%%PATTERN} - Remove the largest right most part of the string that matches.

You can remember that # is to the left of % on the keyboard, so # is left and % is right.

STRING="NAME=PATH=USER=DATE"
PATH=${STRING#*=}  # Removes NAME=
PATH=${PATH%%=*}   # Removes =USER=DATE
echo $PATH         # Echoes "PATH"

You might be able to use the read to get all four at once. I am on an iPad, so I can't test this right now.

OLD_IFS="$IFS"
IFS="="
read NAME PATH USER DATE <<<"$STRING"
IFS="$OLDIFS"

$IFS is thee Input File Separator and is set to space/tab/NL by default. I save the value of $IFS before I change it. I set it to = which separates your various values in your input string..

The read will read in the values using $IFS to separate each one. The <<< is a way to get the shell variable in as input.

Once I finish getting the values, I reset IFS. Otherwise, I would have problems later on.

Answer 7

Excellent place to start for regex on Stackoverflow is : Reference - What does this regex mean?

For your actual question - you were looking for a regex and a way to use it in bash so:

josephs-mbp$more temp.txt 
NAME=PATH=USER=DATE
josephs-mbp$sed 's/^.*=\([^=]*\)=.*=.*$/\1/' temp.txt 
PATH
josephs-mbp$

Let's break down the important bit. Occasionally you want to refer to bits of a regular expression separately, it turns out that the easy way of doing this is by putting things in parenthesis '(' ')' - but in regex land we have to escape them so it's '(' and ')'. These make no difference at all to the match but they make a difference to what happens now.

In general the bit of a regex between a '(' and ')' is stored in a location that can be accessed later with \1, \2,... with the first pair stored in \1 and the second in \2 and so on. Here I just put the bit of the regex you wanted into the brackets and then replaced (thats the 's/fu/bar' bit) in sed. It's ugly and there are probably much more effective ways of doing it, but I think you are starting from about the same point I am and I think that this is the next step for you.