String starts again after every variable [Bash]

Question

I'm trying to use multiple variables in a string but after each variable the string starts again and overrides the beginning:

#!/bin/bash
var1="ABCDEFG"
var2="hi"
echo "${var1} ${var2}"
echo "$var1 $var2"

It should output

ABCDEFG hi

But both echos output

 hiDEFG

Also if I only use one variable and put text after the variable it still overrides...

bla=hello
laber=kthx
echo "${bla}ohai${laber}bye"

This should output:

helloohaikthxbye

But it outputs:

byeikthx

I'm starting the .sh with

sudo bash path/bash.sh

But with sudo sh it does not work as well...

I use Ubuntu 16.04 (as a virtual machine)

Thanks Dennis

This is strange. Can you past output the same command only from shell(not from script). #> echo "${var1} ${var2}" — KlapenHz, Aug 02 '17 at 08:35
Just tried it, in the shell it works just normal, outputs "ABCDEFG hi". — Dennis Ebner, Aug 02 '17 at 08:38

score 2 · Accepted Answer · answered Aug 02 '17 at 09:04

2

It might be a carriage return character in your input string(s).

You can pipe your echo to a tr command to remove it from output string:

 echo "${var1} ${var2}" | tr -d '\r'

answered Aug 02 '17 at 09:04

SLePort

This works, thanks!! So how do I remove the carriage return character, or do I always have to pipe with tr :S ? – Dennis Ebner Aug 02 '17 at 09:28
You can clean you script with sed(GNU sed): `sed -i 's/\x0D//g' script.sh`. – SLePort Aug 02 '17 at 09:41

1 Answers1