concatenation of two command in one single command

Question

while read pos; do
string1= echo $pos | cut -c 20-38
string2="=$pos"
string3="$string1 $string2"
echo "$string3"
done < file

This is the code of my script. I want to display the output in one single line, but the output comes on different lines.

pos = abcdefghi1234567890QWERTYUXY.tar.gz
string1 = 1234567890QWERTYUXY
string2 = abcdefghi1234567890QWERTYUXY.tar.gz

Output wanted:

1234567890QWERTYUXY abcdefghi1234567890QWERTYUXY.tar.gz

But the two strings are displayed in two different lines. I want to display the output in 1 line.

Answer 1

A couple changes and it should work:
Generally, when you use a variable like you did in the echo, the original variable remains as it was, so your use of string 2 is totally unnecessary. Also, I am in the habit of using curly braces to {enclose} variables so they can be referenced right inside text, the curly braces work to delimit the variable name, just put the dollar sign in front (example to follow).

1- When you first set string1, do not put spaces on either side of the equal sign.
2- when assigning the results of a command to a variable, you have to force the command to be executed, single back ticks (the character to the left of the #1 key on US keyboards) surrounding the command do this.
3- watch for the number of characters you cut, your code and example do not agree. So that line should look like:

string1=`echo $pos | cut -c 20-38`

The other thing is when you set one value from another do not include the extra equal sign. So the code to set string2 should look like:

string2="$pos"

I believe the following will do as you request, I adjusted the character counts to give the output of your example:

The input "file" for my test contained:
abcdefghi1234567890QWERTYUXY.tar.gz

And my script is:

 #!/bin/sh
while read pos; do
string1=`echo ${pos} | cut -c 10-28`
echo "${string1} ${pos}"
done < file

The output looks like:
1234567890QWERTYUXY abcdefghi1234567890QWERTYUXY.tar.gz

Which is exactly what I believe you are asking for.

Answer 2

You can pipe the output through sed, notably:

echo "$string3" | sed ':a;N;$!ba;s/\n/ /g'

All credit to Zsolt Botykai on StackOverflow for his answer

Answer 3

Bash has powerful variable substitution that will save you from having to create extra variables and pipe to another command. Maybe try something like this

while read pos; do
  echo "${pos} ${pos%.tar.gz}"
done

General form is ${var%Pattern} -- Remove from $var the shortest part of $Pattern that matches the back end of $var.

http://www.tldp.org/LDP/abs/html/parameter-substitution.html