UNIX - Setting string variable with other variables inside

Question

I am having trouble setting my variables to the correct string in my UNIX shell script. Here is my code ($1 is a filename):

nameIndex=$(echo "$lineCount+3" |bc);
name=$(sed -n -e "$nameIndexp" -e "$nameIndexq" $1);
echo "$name";

When $lineCount is 270, the output is just a new line. That is, $name is not printed or not evaluated.

If $lineCount is 270, I want the output to be sed -n -e 273p -e 273q filename.txt which essentially prints the 273rd line of filename.txt

I appreciate your time!

`lineCount` is unlikely to be a floating-point value, so you can use `nameIndex=$(( lineCount + 3 ))` instead of calling the external `bc` program. — chepner, Nov 19 '13 at 13:18

kfsone · Accepted Answer · 2013-11-19T18:08:01.253

4

$nameIndexp and $nameIndexq look like unique variable names, so you need to use ${name} syntax, i.e.

${nameIndex}p

and

${nameIndex}q

to separate the variable name from the plain text.

nameIndex=$(echo "$lineCount+3" |bc)
name=$(sed -n -e "${nameIndex}p" -e "${nameIndex}q" $1)
echo "$name"

edited Nov 19 '13 at 18:08

answered Nov 19 '13 at 06:37

kfsone

3

Its better to not use back tics. Not sure why you have changed the original parentheses one one line to back ticks and not the other. Use: `nameIndex=$(echo "$lineCount+3" |bc)` – Jotne Nov 19 '13 at 06:56

Jotne · Answer 2 · 2013-11-19T07:30:09.440

1

You can do math in bash without using bc like this:

nameIndex=$(($lineCount+3))

edited Nov 19 '13 at 07:30

answered Nov 19 '13 at 06:59

Jotne

score 0 · Answer 3 · answered Nov 19 '13 at 06:38

0

try this:

nameIndex=`echo "$lineCount+3" |bc`
par=$1
name=$(sed -n -e "$nameIndexp" -e "$nameIndexq" $par)
echo "$name"

answered Nov 19 '13 at 06:38

its same as question `kfsone` is right `"${nameIndex}p"` is better approach – Ashish Nov 19 '13 at 06:43

3 Answers3