Remove carriage return from the variable

Shell Programming and Scripting
1

Hi,

I try to handle very large numbers with a bash script. I run ssh command in a remote server and store the output in a local variable. But this output contains a return carriage at the end. So I try to remove it by tr But I can't figure out the right notation with printf. So my problem is:

 
var1=$( some command )

echo $var1
2.80985e+09
 
var2=`expr $var1 + 10000000`
expr: non-integer argument

printf "%f\n" $var1
2809850000.000000

printf "%s\n" $var1
2.80985e+09

If I use tr -dc '[:digit:]' with this printf output, it will remove all non character not only return carriage at the end.

So for first notation it will give me a number like 2809850000000000 and for second 28098509 and both incorrect. I need a display like 2809850000 so tr could only remove return carriage. So which parameter of printf would produce this notation? (w/o exponential nor decimal notation)

Thanks

2

You could use awk also

$ a="2.80985e+09\r\r\n"
$ echo $a
2.80985e+09\r\r\n

$ b=$(awk -v var=$a 'BEGIN{print var+10000000}')
$ echo $b
2819850000
3

Actually I have a lot of variables, 5,6 sometimes even 10. So I should perform arithmetic operations with all of them.

So I need something like (after your example)

a=20000000000000000000^m
b=300000000000000000000^m
c=40000000000000000000^m
d=250000000000000^m

e=$( awk -v var1=$a var2=$b var3=$c var4=$d 'BEGIN{print var1 * var2 / var3 + var4}' )

But this example above didn't work. Do you have any suggestion for this?

4

I didn't see calculation part -v is missing

e=$( awk -v var1=$a -v var2=$b -v var3=$c  -v var4=$d 'BEGIN{print var1 * var2 / var3 + var4}' )
1 Like
5

Oh I see. I need a separate -v for each variable. It worked like that. Thanks buddy. This saved my life :slight_smile: :b::b:

6

Yes, it's for defining variable

from man page of awk

-v var=val
       --assign var=val
              Assign the value val to the variable var, before execution of the program begins.  Such variable values are  available  to  the
              BEGIN block of an AWK program.
7

Please be careful - that trick with awk works because when converting a field to a numerical value, awk will start at the beginning of the field and scan until an unconvertible char is encountered - which is "\" "r" in post#2. Yes - it's not <CR> but two normal chars.

 a="2.80985e+09\r\r\n"
echo $a xxx
2.80985e+09\r\r\n xxx
a="2.80985e+09"$'\r'
echo $a xxx
 xxx985e+09

The second assignment really has a <CR> char in it, making echo overprint the first chars.
To remove the <CR>, you can try this bashism (pattern substitution expansion):

echo ${a/$'\r'} xxx
2.80985e+09 xxx

or e.g.

echo $a  |hd
00000000  32 2e 38 30 39 38 35 65  2b 30 39 0d 0a
echo $a |tr -d '\r' |hd
00000000  32 2e 38 30 39 38 35 65  2b 30 39 0a