bash run command for every file in directory one at a time

Question

I want to execute the following command:

$SCHRODINGER/utilities/prepwizard [options] inputfile outputfile

$SCHRODINGER is the export of a software and in [options] I must control some variables of the calculation within the program that are not a problem:

for example:

$SCHRODINGER/utilities/prepwizard -watdist 0 fillsidechains -fillloops inputfile outputfile

I want to run it for multiple files with extension .pdb, but the issue is that I must run it one by one, not all at the same time. That is, as soon as one of the following executes, the output can also be in .pdb

I have written this code but it executes all the files to me at the time and it is not what I want:

for i in `ls *pdb`
do
$ SCHRODINGER / utilities / prepwizard [options] $i prep_$i 
done

Answer 1

First, never parse the output of ls. Bad things can happen. Characters like \r and \n can actually be in filenames. To combat this, bash globbing is at your service:

for file in *.pdb
do
    ${SCHRODINGER}/utilities/prepwizard [options] "$file" "prep_${file}"
done

Or if you want a one-liner:

for file in *.pdb; do ${SCHRODINGER}/utilities/prepwizard [options] "$file" "prep_${file}"; done

This will execute each file IN TURN and will wait until the prepwizard application completes before continuing to the next file.

Answer 2

To execute several commands sequentially (one being executed after the previous one has finished), you can separate them with a semicolon.

Thus, you could prepare all your commands in a string, and eval them after your loop:

allcommands=""
for i in `ls *pdb`
do
allcommands="$allcommands;SCHRODINGER options"
done
eval $allcommands