load bash comands from file one per line and execute them for each file in a directory

Question

I'm trying to write a bash queue which will be able to load command sets from a file and then execute each over a set of files. The prepared file should serve as a template for execution in manual bash queues

prepared file sed.sh: _{(I have also comments in the file but i don't know how to add them here)}

sed -e 's,something_complex_long,something_even_longer,'
sed -e 's,other_complex_long_thing,something_so_much_longer,'
sed -e 's,another_long_stuff,something_hell_long,'

and there is directory with set of quite huge files with generic names such as aa ab ac etc. _{(sliced down even bigger file using split)}

So I tried:

sedie=\`grep -v -e '^#' ../../sed.sh\`
for i in *
do
  for b in $sedie
  do
    cat $i | $b > ${i}.new
    mv ${i}.new $i
  done
done

Which of course totally failed, since the second for breaks up statements in $sedie by spaces.

Then I tried to go along using IFS as I have found somewhere but still not much progress. Then I thought I should load the commands in some kind of array to help differentiate what I want ... maybe change sed.sh like arr=("sed " "sed " "sed ") and then source sed.sh and then for b in ${arr[*]}?

Anybody please know how to deal with this in an elegant way (e.g., reusable) manually ?

My IFS try was rather cumbersome and didn't work anyway... $b became a single statement including the spaces and all and there were no such command

SAVEIFS=$IFS
IFS=$(echo -en "\n\b")
sedie=\`grep -v -e '^#' ../../sed.sh\`
for i in *
do
  for b in $sedie
  do
    echo cat $i \| $($b) \> ${i}.new
    echo mv ${i}.new $i
  done
done
IFS=$SAVEIFS

which was not elegant at all.

UPDATE #1

Using this Q: extract file contents into array using bash scripting and this one: How to execute command stored in a variable?, now I'm able to ..

$ cat > myfile 
aaaaaaaa
aaaaaaa
aaaaaa
aaaa
aa
aaaa

$ cat sed.sh 
sed -e 's/a/b/g'
sed -e 's/b/c/g'
sed -e 's/c/f/g'

$ declare -a pole
$ readarray pole < sed.sh
$ for ((i=0;i<${#pole[@]};i++));do \
    eval ${pole[i]} xfile > xfile.b; mv xfile.b xfile; done; 

$ cat xfile
ffffffff
fffffff
ffffff
ffff
ff
ffff

Which is almost what I needed. So my question is basically two:

how to read file into array?

and

how to execute command in variable?

Rather than invoking sed repeatedly in a loop, why not just put the sed expressions in a file and apply them to each file using sed -f script-file <yourfiles>? — steeldriver
– steeldriver, Commented Jul 17, 2015 at 1:19
@sudo - please do not use the answer section for follow up information - use the edit button below your Q to refine it as new information becomes available. — slm
– slm ♦, Commented Jul 17, 2015 at 7:30

Kep Brown · Accepted Answer · 2015-07-17 00:26:15Z

This may not be exactly what you are after but since it looks like you are using sed the you could create a file sed.cmd with just the sed commands you want to execute.

s,something_complex_long,something_even_longer,
s,other_complex_long_thing,something_so_much_longer,
s,another_long_stuff,something_hell_long,

I used the following in my test sed.cmd

s/dog/cat

Then create a script that reads those commands into an array and call sed passing in those commands

array=()

# Read the file in parameter and fill the array named "array"
getArray() {
    i=0
    while read line # Read a line
    do
        array[i]=$line # Put it into the array
        i=$(($i + 1))
    done < $1
}

getArray "sed.cmd"

for cmd in "${array[@]}"
do
    ls -1 | while read line 
do 
    cat $line | sed -e $cmd > ${line}.new
    mv ${line}.new $line
done
done

You will probably want to change the ls command to order the files but the above seems to meet your requirements.

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load bash comands from file one per line and execute them for each file in a directory

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