variable error in bash

Question

I am writing my first code in bash. I am stuck from 2 hours. This is my code:

#!/bin/bash declare -i l=0 declare -i d=0 declare -i s1=0 declare -i s2=0 declare -i t=0 declare -i o=0 declare -i p=0 for i in demo_room/subject_1/?* do if [ "${i:0:1}" = "l" ]; then ((l++)); echo "l=$l" python motempl_n.py $i $l elif [ "${i:0:1}" = "d" ]; then ((d++)); echo "d=$d" python motempl_n.py $i $d elif [ "${i:0:1}" = "o" ]; then o=$((o+1)); echo "o=$o" python motempl_n.py $i $o elif [ "${i:0:1}" = "p" ]; then p=$((p+1)); python motempl_n.py $i $p elif [ "${i:0:1}" = "t" ]; then t=$((t+1)); python motempl_n.py $i $t elif [ "${i:0:7}" = "slide_1" ]; then s1=$((s1+1)); python motempl_n.py $i $s1 #elif [ "${i:0:7}" == 'slide_2' ] else s2=$((s2+1)); python motempl_n.py $i $s2 fi done 

So I am having a folder demo_room/subject_1. In this folder I have 140 avi videos their names are:

20 videos have name: dislike_01 to dislike_20

20 videos have name: like_01 to like_20

20 videos have name: ok_01 to ok_20

20 videos have name: point_01 to point_20

20 videos have name: slide_1_01 to slide_1_20

20 videos have name: slide_2_01 to slide_2_20

20 videos have name: take_a_picture_01 to take_a_picture_1_20

What I want to do is first find the class of the input video then give its occurrence as input to python file. First 20 videos of subject_1 folder are dislike one so this code works fine but the 21st video is like_01 but the parameter it passes to the python code is 21.

But it should be 1 because this is first video of like class in the for loop. And each time it prints value of $d. It means each time it goes in 2nd if condition. In the python code I can verify that the name of the video is like_01 but second value passed is 21. Why? This happens for all the 140 videos.

Answer 1

Your $i will contain the full path (demo_room/subject_1/file), so will always start with d.

You could do something like:

case "${i##*/}" in (d*) ...;; (l*) ...;; (slide1_*) ...;; ... esac 

Where ${i##*/} is $i with the leading part matching */ removed.

Also beware that parameter expansions need to be quoted in sh/bash:

python motempl_n.py "$i" "$s2" 

Answer 2

While your code is well written, there's a lot of duplication. Here's a more condensed version:

#!/bin/bash declare -A count for file in demo_room/subject_1/*; do base=${file##*/} case $base in slide_[12]*) key=${base:0:7};; [ldopt]*) key=${base:0:1};; *) ;; # other types ignored esac (( count[$key]++ )) echo "$key=${count[$key]}" python motempl_n.py "$file" "${count[$key]}" done 

I'm using an associative array instead of distinct integer variables, but fortunately bash arithmetic handles (( an_unset_variable++ )) gracefully and assigns "1" to the variable.

Note that the ? in your ?* was redundant as file names always contain at least one character, so there's no file that * (0 or more characters) would match and ?* (1 or more characters) wouldn't.

Answer 3

Your $i variable always start with a d: demo_room … …
You need to extract the file name. And since we are extracting, we may as well extract both the name and the number of the file.

Thus, a shorter version of your script could be:

#!/bin/bash declare -Ai count for i in ./demo_room/subject_1/?*; do j=${i##*/} ; name=${j%_*} ; number=${j##*_} ((count[$name]++)) echo \ python motempl_n.py "$i" "${count[$name]}" done echo "List of keys: " "${!count[@]}" 

This script use only one array variable count, the list of (automatically) keys used (from $name) is printed at the end of the script.

Comment out the echo \ line once you like how the script works to actually execute the commands.