Getting substring by index position - grep

Question

i need to extract a substring from a line using the initial and final position. I think it should be easily done with grep, but still haven't figured out how.
An example, a line of n chars, i want to extract the substring starting at k position and ending in the l position of that line.
With obviously l, k < n and l > k

Answer 1

Cut is a good choice for this. You can select ranges and individual fields:
$ echo "123456" | cut -c2-4
234
$ echo "123456" | cut -c1,3,6
136
$ echo "123456" | cut -c1-3,6
1236

Answer 2

Why not use awk?
echo "12345678" | awk '{print substr($0, 3, 2);}'
# prints '34'

Answer 3

If you're using bash you could just do:
LINE="strings"
K=3 ## 4th character starting from index 0
L=5 ## 6th character starting from index 0
echo "${LINE:K:L - K + 1}"
On a loop for a file:
while -r read LINE; do echo "${LINE:K - 1:L - K}"; done < file
As for awk basing that L means the position and not the length where 0 is the starting index:
awk -v k="3" -v l="5" '{ n = length; print substr($0, k, l - k + 1);}' < file

Answer 4

Using sed:
k=3
echo '123456789' | sed 's/^.\{'$k'\}//'
Output:
456789

Answer 5

Is this what your looking for?
String test = "ateststring";
String testa = test.substring(0,4);
String test2 = testa.substring(0,3);
int testc = test.indexOf("test");
String test3 = test.substring(testc,5);
System.out.println(test + ":" + testa + ":" + test2 + ":" +test3);