CODEWITHSUNDEEP
bash
Oded Sayar
Oded Sayar

Reputation: 429

List files whose content begin with a letter in Bash

I need to write a bash script which lists all files in a directory (and its subdirectories) that start with a certain letter, say A/a. How can it be done if the content of the file should begin with the letter, rather than its name?

Upvotes: 1

Views: 1185

Answers (2)

Benjamin W.
Benjamin W.

Reputation: 52291

With just Bash and the globstar option (shopt -s globstar) to enable **/*:

for file in **/*; do [[ $(< "$file") == [Aa]* ]] && echo "$file"; done

This loops through all files (and directories) recursively and checks if their content starts with A or a. If so, the filename gets printed.

It actually also tries to compare directories to [Aa]*, but just returns a non-successful exit status for $(< "$file") when $file is a directory.

Alternatively, without Bashsims, using find:

find . -type f -exec sh -c '[ "$(head -c 1 "$1" | tr A a)" = a ]' _ {} \; -print

This filters for files (-type f), then spawns a subshell, in which the test "$(head -c 1 "$1" | tr A a)" = a is run (get first character of file, transpose potential uppercase A to a, compare to a, and if the exit status of that is successful, print the file name.

Upvotes: 2

Arjun Mathew Dan
Arjun Mathew Dan

Reputation: 5298

grep -rn "^[A|a]" * | awk -F: '$2==1{print $1}'

grep recursively(-r) for the pattern ^[A|a] in the current folder. If found (need not be on the first line), grep output would look like filname:linenumber:match. Use awk with : as field seperator and for cases where linenumber is 1(match found on the first line, i.e file beginning), print the 1st field which is the filename. grep output will contain the linenumber since we r giving the -n option.

Upvotes: 1

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