bash script: "echo" being missed when script is executed

L_Carver · 11-06-2017, 09:15 AM

My installed shellcheck passes it as having no errors.
Is it my choice of variables? I've used a few, and here is the latest script.

Code:

#!/bin/bash
c="$1"
mapfile -t honest<"$c"
fg=${honest[0]}
if [[ ! -f $fg ]]; then
 echo -e "File $fg is not in this directory.\nCan't continue."
 exit 0
fi
cap1=${honest[1]}
caplen=${#cap1}
comm1=${honest[2]}
comlen=${#comm1}
k=${honest[3]}
kk=$(echo "$k" | tr ',' ' ' | wc -w)
echo "$kk"
echo "All the lines look right. Here goes."
echo "READING information FROM $c"
echo "Caption is $caplen characters long (spaces included)."
echo "Comment is $comlen characters long (likewise)"
echo -e "$fg: There are $kk keywords."

Line 20, (echo -e "$fg: There are $kk keywords."), works when executed from stdout, but not during execution. What could be wrong?

Carver

MensaWater · 11-06-2017, 09:35 AM

bash shell has a built in "echo" command.

Separately there is a binary command called "echo" typically /bin/echo.

The "-e" is valid for the binary but not for the built in.

Change your command line to the path of the echo command e.g.

Code:

/bin/echo -e "File $fg is not in this directory.\nCan't continue."

and

Code:

echo -e "$fg: There are $kk keywords."

OR you could set a variable at start of script:

Code:

ECHO=/bin/echo

Then use that:

Code:

$ECHO -e "File $fg is not in this directory.\nCan't continue."

and

Code:

$ECHO -e "$fg: There are $kk keywords."

michaelk · 11-06-2017, 09:36 AM

I would start with debugging your script. You can add set -x at the beginning of your script as well as adding extra echo commands at other places in your script to verify variables are being assigned as desired.

http://tldp.org/LDP/Bash-Beginners-G...ect_02_03.html

MadeInGermany · 11-06-2017, 12:49 PM

Replace echo (non-portable) with printf (portable and a shell built-in).
Especially if you use variables with unknown contents; e.g. a leading -n or -e in the variable contents can mess it up. In contrast, in printf all formatting is done by the first argument, and the variables are safely put into the further arguments.

Code:

kk=$(printf "%s\n" "$k" | tr ',' ' ' | wc -w)

Code:

printf "%s: There are %s keywords.\n" "$fg" "$kk"

Code:

 printf "File %s is not in this directory.\nCan't continue.\n" "$fg"