Finding files in linux based on regex, with multiple alternatives

Question

Let's say I have a list of strings. I want to find files with filenames starting with these string.

Example,

Strings: filename.could.start.with.this.restoffilename a.file.could.have.this.at.beginning.restoffilename

This list of strings can be an arbitrary number (obviously much more than just two as above).

I want to use a bash script for this and only use what I consider basic linux applications (ls, grep, sed, find etc.), so no python, perl, ruby or other such languages.

Answer 1

prefixes=("filename.could.start.with.this" "a.file.could.have.this.at.beginning")

# Turn the prefix array into a find expression (in array form)
matchlist=()
for prefix in "${prefixes[@]}"; do
    matchlist+=("-o" "-name" "$prefix*")
done
matchlist=("${matchlist[@]:1}") # remove the extra "-o" from the beginning

# Search the current directory for plain files with names starting with one of the prefixes
find -x . "(" "${matchlist[@]}" ")" -type f

If your list of filename prefixes wasn't in array form, you'd have to modify this appropriately, e.g.

prefixes="filename.could.start.with.this a.file.could.have.this.at.beginning"

...
for prefix in $prefixes; do
...

Also, this will not behave well if given an empty list of filename prefixes; if that's a possibility, check for this condition first.

Answer 2

BASH FAQ entry #50: "I'm trying to put a command in a variable, but the complex cases always fail!"

Just keep adding find predicates to the array until you have your command.