Reputation: 23
I'm hoping this is a simple question, since I've never done shell scripting before. I'm trying to filter certain files out of a list of results. While the script executes and prints out a list of files, it's not filtering out the ones I don't want. Thanks for any help you can provide!
#!/bin/bash
# Purpose: Identify all *md files in H2 repo where there is no audit date
#
#
#
# Example call: no_audits.sh
#
# If that call doesn't work, try ./no_audits.sh
#
# NOTE: Script assumes you are executing from within the scripts directory of
# your local H2 git repo.
#
# Process:
# 1) Go to H2 repo content directory (assumption is you are in the scripts dir)
# 2) Use for loop to go through all *md files in each content sub dir
# and list all file names and directories where audit date is null
#
#set counter
count=0
# Go to content directory and loop through all 'md' files in sub dirs
cd ../content
FILES=`find . -type f -name '*md' -print`
for f in $FILES
do
if [[ $f == "*all*" ]] || [[ $f == "*index*" ]] ;
then
# code to skip
echo " Skipping file: " $f
continue
else
# find audit_date in file metadata
adate=`grep audit_date $f`
# separate actual dates from rest of the grepped line
aadate=`echo $adate | awk -F\' '{print $2}'`
# if create date is null - proceed
if [[ -z "$aadate" ]] ;
then
# print a list of all files without audit dates
echo "Audit date: " $aadate " " $f;
count=$((count+1));
fi
fi
done
echo $count " files without audit dates "
Upvotes: 0
Views: 147
Reputation: 295736
First, to address the immediate issue:
[[ $f == "*all*" ]]
is only true if the exact contents of f is the string *all*
-- with the wildcards as literal characters. If you want to check for a substring, then the asterisks shouldn't be quoted:
[[ $f = *all* ]]
...is a better-practice solution. (Note the use of =
rather than ==
-- this isn't essential, but is a good habit to be in, as the POSIX test
command is only specified to permit =
as a string comparison operator; if one writes [ "$f" == foo ]
by habit, one can get unexpected failures on platforms with a strictly compliant /bin/sh
).
That said, a ground-up implementation of this script intended to follow best practices might look more like the following:
#!/usr/bin/env bash
count=0
while IFS= read -r -d '' filename; do
aadate=$(awk -F"'" '/audit_date/ { print $2; exit; }' <"$filename")
if [[ -z $aadate ]]; then
(( ++count ))
printf 'File %q has no audit date\n' "$filename"
else
printf 'File %q has audit date %s\n' "$filename" "$aadate"
fi
done < <(find . -not '(' -name '*all*' -o -name '*index*' ')' -type f -name '*md' -print0)
echo "Found $count files without audit dates" >&2
Note:
find -print0
, read with IFS= read -r -d ''
; this is discussed in [BashFAQ #1].find
.awk
using grep
, as awk
is capable of searching through input files itself.< <(...)
is used to avoid the behavior in BashFAQ #24, wherein content piped to a while
loop causes variables set or modified within that loop to become unavailable after its exit.printf '...%q...\n' "$name"
is safer than echo "...$name..."
when handling unknown filenames, as printf
will emit printable content that accurately represents those names even if they contain unprintable characters or characters which, when emitted directly to a terminal, act to modify that terminal's configuration.Upvotes: 2
Reputation: 23
Nevermind, I found the answer here:
bash script to check file name begins with expected string
I tried various versions of the wildcard/filename and ended up with:
if [[ "$f" == *all.md ]] || [[ "$f" == *index.md ]] ;
The link above said not to put those in quotes, and removing the quotes did the trick!
Upvotes: 0