bash Variables  with hex 0a on the end

Posted on 2014-11-24
Last Modified: 2014-12-08
I am getting data in bash and storing it in Variables.

Half the time I get the string, the other half of the time I get the string terminated with hex 0a on the end if I
pipe it to hexedit -C

is there an easy way to strip this from the variable  with a $(varname. ) or something like it ?

I am having this issue a lot

I am using the later version of redhat :)
Question by:TIMFOX123
  • 3
LVL 12

Expert Comment

ID: 40463209
Hi Tim,

Hex 0A is a newline (AKA line-feed) character, which is the standard UNIX/Linux line terminator.

Q1. Are you able to post your code so we can see how that's happening?

> "Half the time I get the string, the other half..."
Q2. When you say "half the time", do you mean in half of the variables, or do you get different results with different data, or what?

> "I am using the later version of redhat :)"
Q3. Later than what?
LVL 68

Assisted Solution

woolmilkporc earned 500 total points
ID: 40463288
Which way do you "pipe" the variable to the hex editor? If you use something like

echo $A | od -tcx1

you will always see something like this (x"0a" at the end):

0000000    s   t   r   i   n   g  \n
          73  74  72  69  6e  67  0a

That's because "echo" adds a linefeed automatically.

Instead, if you use the "-n" flag of "echo ("do not output the trailing newline")

echo -n $A | od -tcx1

then you'll see this:

0000000    s   t   r   i   n   g
          73  74  72  69  6e  67

i. e. no trailing x"0a".

I used "od" because my hexedit doesnt support a "-C" flag.

If your variable actually contains a linefeed at the end:


echo -n "$A"

will preserve this linefeed, while

echo -n $A

will not.

As tel2 suggested above, please give us more detail!

Author Comment

ID: 40467200
thank you both

example, If I:

grep root /etc/passwd | hexdump -C
you will see the line ends with a hex 0a

if you want to get rid of the hex 0a  you can:
 grep root /etc/passwd | tr -d '\n' | hexdump -C

There is a part of bash that I have used but to not understand where you can modify stuff with ${}

what I would like to do is to take a varable that has a hex 0a an the end and when I use it I can "chomp" it off

echo ${varname xxxxxx }
where the xxxx is code that I do not know.

also when I was saying about "later version of Linux" I was not being clear.  I should have said "currently supported" Linux like 5,6,7.  

thank you so much
LVL 68

Assisted Solution

woolmilkporc earned 500 total points
ID: 40467273
You see the linefeed just because "grep" adds it.
 The variable itself does not contain x'0a'.

varname=$(grep root /etc/passwd)
echo -n ${varname} | hexdump -C

Try this:

echo ${#A}

#varname is the length of the variable varname.

The above will give 4, not 5

Or, if you want "grep"

echo 1234 > testfile
A=$(grep 1234 testfile)
echo ${#A}

With "${varname xxxxxx }" you probably mean variable editing (aka "parameter expansion"), using e.g. ${varname%string} (or %%), ${varname#string} (or ##) or ${varname/old/new}.

I'm not aware of a way specifying hex values with the above.
LVL 68

Accepted Solution

woolmilkporc earned 500 total points
ID: 40467363
I could tell you how to strip the last character which might be a linefeed (or might not):

echo ${#A}
-> 5
echo ${#A}
-> 4

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