Bash loop

Johng · 05-09-2023, 02:50 AM

I have 15 jpg files which want to reduce in size and rename xxx-01.jpg to xxx-15.jpg
As a starter I have tried:

Code:

#!/bin/bash

for i in $(ls *.jpg) ; do
convert $i $(basename $i) -resize 600x400 $i R.jpg
done

but it's not looping. What do I need to do to complete the loop?

Turbocapitalist · 05-09-2023, 02:54 AM

I would try:

Code:

for i in *.jpg ; do

as per why you shouldn't parse the output of ls(1).

pan64 · 05-09-2023, 02:57 AM

Please use shellcheck to check your script, it will definitely give you some ideas to make it better.
Would be nice to tell us what's happening, what is printed. Also you can use set -xv to debug your script and insert an echo to see what's going on.

Code:

# set -xv
for i in *.jpg ; do  # do not use ls here
    echo converting $i
    convert ...
done

Johng · 05-09-2023, 03:44 AM

When I run:

Code:

#!/bin/bash

for i in $(ls *.jpg) ; do
convert $i $(basename $i) -resize 600x400   $i *R.jpg
done

It gives me three files: *R-0.jpg *R-1.jpg and *R-2.jpg
All are the same image (last on the list), the first two reduced, the third full size

When I run:

Code:

# set -xv
for i in *.jpg ; do  # do not use ls here
    echo converting $i
    convert -resize 600x400
done

I get:

Code:

converting JPEG_20230509_131747_191157587250982617.jpg
convert-im6.q16: missing an image filename `600x400' @ error/convert.c/ConvertImageCommand/3226.

fifteen times

Johng · 05-09-2023, 03:48 AM

Note error copying second script, should have been:

Code:

#!/bin/bash

# set -xv
for i in *.jpg ; do  # do not use ls here
    echo converting $i
    convert -resize 600x400
done

pan64 · 05-09-2023, 03:56 AM

convert -resize 600x400 <here missing operands>

Turbocapitalist · 05-09-2023, 03:57 AM

Quote:

Originally Posted by Johng

fifteen times

Ok. The loop part is solved then.

What do the file names look like? Do they already contain the incremented number 01 - 15 or should your script add that too?

rizitis · 05-09-2023, 04:15 AM

Code:

#!/bin/bash

counter=1
for file in *.jpg; do
    # Rename file to Rxxx.jpg
    mv "$file" "R$(printf '%03d' $counter).jpg"
    
    # Convert to 600x400
    convert "R$(printf '%03d' $counter).jpg" -resize 600x400\! "R$(printf '%03d' $counter).jpg"
    
    counter=$((counter+1))

take this script as an example and modify it for your needs...

Johng · 05-09-2023, 04:25 AM

Thank you for your help(s)

pan64:

Quote:

convert -resize 600x400 <here missing operands>

I don't understand

Turbocapitalist:

Quote:

The loop part is solved then

No, only 3 loops out of needed 15
and as per my reply the three were numbered: *R-0.jpg *R-1.jpg and *R-2.jpg note the unwanted "*" and missing leading zero

I want them to be the series: R-1.jpg R-2.jpg R-3.jpg . . . . up to R-15.jpg

Racho · 05-09-2023, 01:53 PM

Never used convert before, but man convert says:

Code:

 convert-im6.q16 [input-option] input-file [output-option] output-file

So you need to tell convert the input-file and the output-file names.
Try this:

Code:

for i in * ; do 
   echo converting $i to R${i%.*}
   convert $i -resize 600x400 R${i%.*}.jpg; 
done

This code will convert MyPic1.jpg to RMyPic1.jpg.
But if RMyPic1.jpg already exists it will be overwritten!

So it may be better to create an empty subdirectory first and store there the new files:

Code:

mkdir out_dir
for i in * ; do 
   echo converting $i to out_dir/R${i%.*}
   convert $i -resize 600x400 out_dir/R${i%.*}.jpg; 
done

Johng · 05-09-2023, 04:24 PM

Thank you all for your suggestions.

rizitis: I ran your script but got a syntax error: unexpected end of file

Code:

#!/bin/bash

counter=1
for file in *.jpg; do
    # Rename file to Rxxx.jpg
    mv "$file" "R$(printf '%03d' $counter).jpg"

    # Convert to 600x400
   convert "R$(printf '%03d' $counter).jpg" -resize 600x400\! "R$(printf '%03d' $counter).jpg"

Racho: your script made a directory with all 15 files converted files with original file name preceded with letter R

Would it be possible to have the name of the output files starting with R01.jpg to R15.jpg??

Racho · 05-09-2023, 04:55 PM

Ok, try this:

Code:

mkdir out_dir
let num=1
for i in *.jpg ; do
   outfilename=$(printf "out_dir/R%02i.jpg" $num)
   echo converting $i to $outfilename
   convert $i -resize 600x400 $outfilename
   let num++
done

Before running it try to understand every step so you can adapt it to your needs whenever you want.
Understanding code before running it has the double purpose of learn from it and keeping you safe from potential dangerous code.

Johng · 05-09-2023, 05:42 PM

Thank you Racho, that worked perfectly

Johng · 05-23-2023, 11:11 PM

A new need:
to convert the image files to out_dir with original name

Johng · 05-23-2023, 11:24 PM

Cancel that, sorted:

Code:

mkdir out_dir
let num=1
for i in *.jpg ; do
   outfilename=$(printf "out_dir/$i.jpg")
   echo converting $i to $outfilename
   convert $i -resize 600x400 $outfilename
   let num++
done