用 bash/shell 在 echo 中显示 2 个变量的疯狂问题

Question

I have a stupid problem completely crazy with shell/bash script. I have a configuration file with 2 variables defined...See below my file called test.conf :

WEB_DIR="www"
TPL_DIR="tpl"

Now I have my shell/bash script called test.sh with this content :

#!/bin/bash
source test.conf
echo $WEB_DIR
echo $TPL_DIR
echo -e "$WEB_DIR - $TPL_DIR"
exit 0

As you can seen this is very simple because this kind of script should return this :

www
tpl
www - tpl

But, I don't know why, when I launch this script, he returns this :

www
tpl
- tpl

We can see that the first variable $WEB_DIR is missing and I don't know why because she is defined.

Do you have an idea why?

Thanks L.

Answer 1

test.conf has DOS line endings, so the value of each variable ends with a carriage return. Compare:

$ WEB_DIR=$'www\r'
$ TPL_DIR=$'tpl\r'
$ echo -e "$WEB_DIR - $TPL_DIR"
 - tpl

The carriage return at the end of WEB_DIR is "displayed" by moving the cursor to the beginning of the line, so that - tpl overwrites www .

Remove the DOS line endings with dos2unix , or use your text editor to save the file using Unix line endings.