grep on a variable

Question

Let's say I have a variable

line="This is where we select from a table."

now I want to grep how many times does select occur in the sentence.

grep -ci "select" $line

I tried that, but it did not work. I also tried

grep -ci "select" "$line"

It still doesn't work. I get the following error.

grep: This is where we select from a table.: No such file or directory

Answer 1

Have grep read on its standard input. There you go, using a pipe...

$ echo "$line" | grep select

... or a here string...

$ grep select <<< "$line"

Also, you might want to replace spaces by newlines before grepping :

$ echo "$line" | tr ' ' '\n' | grep select

... or you could ask grep to print the match only:

$ echo "$line" | grep -o select

This will allow you to get rid of the rest of the line when there's a match.

Edit: Oops, read a little too fast, thanks Marco. In order to count the occurences, just pipe any of these to wc(1) ;)

Another edit made after lzkata's comment, quoting $line when using echo.

Answer 2

test=$line i=0
while case "$test" in (*select*)
test=${test#*select};;(*) ! :;;
esac; do i=$(($i+1)); done

You don't need to call grep for such a simple thing.

Or as a function:

occur() while case "$1" in (*"$2"*) set -- \
        "${1#*"$2"}" "$2" "${3:-0}" "$((${4:-0}+1))";;
        (*) return "$((${4:-0}<${3:-1}))";;esac
        do : "${_occur:+$((_occur=$4))}";done

It takes 2 or 3 args. Providing any more than that will skew its results. You can use it like:

_occur=0; occur ... . 2 && echo "count: $_occur"

...which prints the occurrence count of . in ... if it occurs at least 2 times. Like this:

count: 3

If $_occur is either empty or unset when it is invoked then it will affect no shell variables at all and return 1 if "$2" occurs in "$1" fewer than "$3" times. Or, if called with only two args, it will return 1 only if "$2" is not in "$1". Else it returns 0.

And so, in its simplest form, you can do:

occur '' . && echo yay || echo shite

...which prints...

shite

...but...

occur . . && echo yay || echo shite

...will print...

yay

You might also write it a little differently and omit the quotes around $2 in both the (*"$2"*) and "${1#*"$2"}" statement. If you do that then you can use shell globs for matches like sh[io]te for the match test.

Answer 3

I would just use sed to break up the sentence for me and then compare each line in a loop because this seems like the easiest way to me (no offence to anyone here, and also I am not an expert so there may be a reason people don't do it this way, not sure)

string="a a a a a asd gsam en"
count=$(echo $string | sed 's/ /\n/g' | grep -w 'a' | wc -l)
echo $count

when I run it

mark@gamerblock:~$ string="a a a a a asd gsam en"
mark@gamerblock:~$ count=$(echo $string | sed 's/ /\n/g' | grep -w 'a' | wc -l)
mark@gamerblock:~$ echo $count
5

As a script with input argument

#!/bin/bash
echo "Enter your string"
read string
echo "Enter what you want substring counted"
read substring
count=$(echo $string | sed 's/ /\n/g' | grep $substring | wc -l)
echo $count

when ran

mark@gamerblock:~$ ./substring.sh
Enter your string
peter piper picked a pack of pickled peppers. a pack of pickled peppers peter piper did pick.
Enter what you want substring counted
peter
2
mark@gamerblock:~$ ./substring.sh
Enter your string
peter piper picked a pack of pickled peppers. a pack of pickled peppers peter piper did pick.
Enter what you want substring counted
p
0

If you don't want exact matches, remove the -w argument from grep

mark@gamerblock:~$ ./substring.sh
Enter your string
peter piper picked a pack of pickled peppers. a pack of pickled peppers peter piper did pick.
Enter what you want substring counted
p
12
mark@gamerblock:~$