bash script to read my files and execute the command - bash

Question

I have a script which one can run as ./analyze file.txt file.root log.txt
file.txt is input file which contains executable root files with their paths, others are output. My problem is I have almost 30 input files and i do not want to write down the command each time to run the code. Bash script would be nice but I did not manage. It gives an error. see an example of the code below, I try to run:
#!/bin/bash
do
echo
./analyze runlist1.txt runlist1.root log1.txt
./analyze runlist2.txt runlist2.root log.txt
./analyze runlist3.txt runlist3.root log.txt
./analyze runlist4.txt runlist4.root res4.txt
done
I get the error "syntax error near unexpected token `do' ", Any help would be appreciated.

Answer 1

When all files are called Brunsplit.txt, the following will help
for file in Brunsplit*.txt; do
tmp=${file%.txt}
nr=${tmp#Brunsplit}
./analyze ${file} runlist${nr}.root res${nr}.txt
done
The tmp and nr vars are filled with special syntax, it is something like cutting off .txt at the end and Brunsplit from the start.

Answer 2

Like others indicated in comments, the do out of context is a syntax error (it is a valid keyword right after a for, while, and until control statement).
Apparently, there is no systematic mapping between input file names and the corresponding log file names, so you need a script which maintains this mapping. Something like this, then:
while read suffix logfile; do
./analyze "runlist$suffix.txt" "runlist$suffix.root" "$logfile"
done <<'____HERE'
1 log1.txt
2 log.txt
3 log.txt
4 res4.txt
____HERE
The here document (the stuff between << delimiter and the delimiter alone on a line) is just like a text file, except it is embedded as part of the script.

Answer 3

Check it out the primitive but the modified version.
!/bin/bash
for file in Brunsplit1.txt Brunsplit2.txt
do
echo $file
./analyze Brunsplit1.txt runlist1.root res1.txt
./analyze Brunsplit2.txt runlist2.root res2.txt
done
Thanks.