Print last character in the given path

Question

I am trying to concat "comnp.dat_CY*" files into one file. File name, source path and destination path are the parameters. I am trying to find the last character in the path to check if it is "/" or not. But when I try to do that I am getting error. Below is my script

if [ "$#" -ne 3 ]; then
    echo "Script requires 3 inputs"
    echo "1. File pattern (comnp.dat_CY)"
    echo "2. Source path"
    echo "3. Destination path"
    exit
fi

last_char1 = `echo $2|awk -F  '{if (NF>1) {print $NF}}`
last_char2 = `echo $3|awk -F  '{if (NF>1) {print $NF}}`

if [ $last_char1 != "/" ];then
    ap_src_path = "$2/"
fi
if [ $last_char2 != "/" ];then
    ap_dest_path = "$3/"
fi

cat $ap_src_path$1* > $ap_dest_pathcomnp.dat

Below are my errors:

comnp.sh: Broken pipe
comnp.sh[14]: last_char1:  not found
comnp.sh[15]: last_char2:  not found
comnp.sh[17]: test: argument expected
comnp.sh[20]: test: argument expected

What is wrong in my script?

I am using ksh AIX

Answer 1

Variable assignment in shell syntax is not allowed to have spaces. last_char = ... is a command called last_char with the arguments = and ..... last_char=... is an assignment that stores ... in the shell variable last_char.

---

This is a much more efficient and direct way to do what you want:

case $2 in
  */) ap_src_path=$2;; 
  *) ap_src_path=$2/;;
esac
case $3 in
  */) ap_dest_path=$3;;
  *) ap_dest_path=$3/;;
esac

You don't need to extract the last character to do a simple comparison. Also, command substitution is inefficient and clunky for simple shell variable manipulation like this.

---

More importantly, you don't need to do any of this at all. For the most part, Unix path resolution does not care about duplicated slashes, so path///to//a/file is just as valid as path/to/a/file. In scripts, you can take advantage of this by simply not worrying about duplicate /. In other words you can replace the entire script with

if [ $# -ne 3 ]; then
   # ... same as above ...
fi
pattern=$1
ap_src_path=$2
ap_dst_path=$3

cat "$ap_src_path/$pattern"* > "$ap_dest_path/compn.dat"

Answer 2

The problem here is that you cannot use space around =. The shell is very picky about that. And awk commands are totally buggy.

Moreover, in 2014 better use $( ) than backticks. So awk lines should be :

last_char1=$(echo "$2" | awk '{print $NF}' FS='')
last_char2=$(echo "$3" | awk '{print $NF}' FS='')

check http://mywiki.wooledge.org/BashFAQ/082

Answer 3

last_char1=$(echo ${2:${#2}-1}); 
last_char2=$(echo ${3:${#3}-1}); 

if [[ "$last_char1" != "/" ]]; then 

Tips:

• Use $(command), instead of old `command`: Reference
• Use [[ ]], instead of [ ]: Reference

Answer 4

There are multiple problems with your script. Spaces around = should not be present, single quotes ' are not paired, and additionally awk -F requires a parameter - a field separator, so you should rather write awk -F '' '{if (NF>1) {print $NF}}'.

But in fact if you use any reasonable shell like bash or zsh then you can simply write

cat "${ap_src_path}/${1}*" > "${ap_dest_path}/comnp.dat"

without a need of last characters and all those if conditions. Possible additional / are just ignored, so for example a path /var//log///mysql////mysql.log is perfectly fine syntax.