Using echo command with -n option doesn't print the variable

Question

I am extracting a string variable from a file and then print it to the screen. The problem is that using just echo works, but if I used echo -n, then nothing is printed. Here is an example:

$ Var=$(grep -a MrEdbType FileName)
$ echo $Var
Prot
$ echo -n $Var
$

Similarly, if I use printf, it works only if I add \n:

$ Var=$(grep -a MrEdbType FileName)
$ printf "%s" $Var
$
$ printf "%s\n" $Var
Prot

What is wrong with the string I am extracting? The file is a DICOM file containing a mixture of text and binary data, hence the use of -a option.

As suggested in the comments, using hexdump -C option generated this result:

$ Var=$(grep -a MrEdbType FileName | hexdump -C)
$ echo $Var
00000000 50 72 6f 74 0d 0a |Prot..| 00000006

Add output of `grep -a MrEdbType FileName | hexdump -C` to your question. — Cyrus, Sep 19 '19 at 16:19
You are suffering from CRLF endings. To demonstrate try `$ echo "foo\r"` and `$ echo -n "foo\r"` — kvantour, Sep 19 '19 at 16:20
Also related: [Are shell scripts sensitive to encoding and line endings?](https://stackoverflow.com/questions/45772525) — kvantour, Sep 19 '19 at 16:23
If you look at http://www.asciitable.com/, then you notice that `0D` is the hexadecimal value for CR (`\r`) — kvantour, Sep 19 '19 at 16:25

score 0 · Answer 1 · answered Sep 19 '19 at 16:29

0

A suggested link led to an answer - using sed 's/\r$//' at the end of the grep search fixed the problem!

answered Sep 19 '19 at 16:29

Renat

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Using echo command with -n option doesn't print the variable

1 Answers1