Find string while knowing part of it and return string

Question

I have a string, for example

"Icecream123 AirplaneBCD CompanyTL1 ComputerYU1"

Let's say I know that my string will contain for sure the substring IceCream but I don't know what follows it.

It might be 123 as in my example or it might be something different.

While I can use grep to detect if "Icecream" substring exists in my string with the following command

echo $string | grep -oF 'Icecream';

Which will print

Icecream

I want with a command to get it to print the whole substring, which in my example is

Icecream123

Of course what follows Icecream is random and not known beforehand so I can't just do

$SUBSTRING=$(echo $string | grep -oF 'Icecream')
$SUBSTRINGTRAIL=123
echo $SUBSTRING$SUBSTRINGTRAIL

Answer 1

If your grep supports perl compatible regular expressions, you could match non-greedily up to the next word boundary:

echo "$string" | grep -oP 'Icecream.*?\b'

Otherwise, match the longest sequence of non-blank characters:

echo "$string" | grep -o 'Icecream[^[:blank:]]*'

Or keep everything in the shell and remove the longest trailing sequence of characters starting with a space:

echo "${string%% *}"

Answer 2

Since you tagged bash:

[[ $string =~ (Icecream[^ ]*) ]] && result=${BASH_REMATCH[1]}

More generally, for a search term in $search:

[[ $string =~ ($search[^ ]*) ]] && result=${BASH_REMATCH[1]}

... or with parameter expansion:

# remove any leading text up to -and through- the search text:
x=${string##*$search}

# remove any trailing space onwards
result=$search${x%% *}

Answer 3

Using a grep that knows about -o:

$ printf '%s\n' "$string" | grep -o '\<Icecream[^[:blank:]]*'
Icecream123

The pattern \<Icecream[^[:blank:]]* matches the string Icecream (where the I is preceded by a non-word character, or the start of the line) followed by zero or more non-blanks (not spaces or tabs).

---

Using awk:

$ printf '%s\n' "$string" | awk -v RS=' ' '/^Icecream/'       
Icecream123

The awk program divides the string into space-separated records, and tests each one. It will print the ones that start with the string Icecream.

Using mawk or GNU awk, you may also use

printf '%s\n' "$string" | awk -v RS='[[:blank:]]' '/^Icecream/'

since they interpet RS as a regular expression if it contains more than one character.

---

With sed, in a similar fashion as with grep:

$ printf '%s\n' "$string" | sed 's/.*\(\<Icecream[^[:blank:]]*\).*/\1/'
Icecream123

---

Using /bin/sh:

set -- Icecream123 AirplaneBCD CompanyTL1 ComputerYU1
for string; do
    case $string in
        Icecream*)
            printf '%s\n' "$string"
            break
    esac
done

---

Perl (with a little help from tr):

$ printf '%s\n' "$string" | tr ' ' '\n' | perl -ne '/Icecream\S*/ && print'
Icecream123

or just

$ printf '%s\n' "$string" | perl -ne '/(Icecream\S*)/ && print $1, "\n"'
Icecream123

Answer 4

For example, if you use GNU grep:

$ echo "Icecream123 AirplaneBCD CompanyTL1 ComputerYU1" | grep -oP '\bIcecream.*?(\s|$)' --color

It uses PCRE.

Answer 5

A little bit simpler perhaps, especially since you say that your version of grep does not support perl regex:

$ echo $string | tr ' ' '\n' | grep 'Icecream' Icecream123

The tr splits the string into lines by replacing all the spaces with newlines. Then you can use grep easily.

You can also write the following to obtain only what follows the word you are looking for:

$ echo $string | tr ' ' '\n' | sed -n 's/Icecream//p' 123