replacing ls command in a bash script

Question

Everybody says it is bad, bad, very bad to use ls to parse. This is a script that will pick out a log file. I don't want to glob the whole directory, $1 is the name of the file I want to look at, and of those files, I only want to see the last two. How do I do this without using ls ?

for i in $(ls -ltr ~/logs/autosys/*${1}* | tail -2) ; do
    if [[ ${i} =~ ".out" ]] ; then
    tailpath=$(echo ${i%.*} | awk '{print $9}')
    fi
done

I think it's "take the two newest files in ~/logs/autosys with names containing $1, then do something with them". — Benjamin W.
– Benjamin W., Commented Aug 7, 2018 at 21:18
If you want to compare by last modification time and can't use the filename: BashFAQ/003 — Benjamin W.
– Benjamin W., Commented Aug 7, 2018 at 21:19
@Benjamin - you are correct - that is what the snippet does - it is part of something larger — capser
– capser, Commented Aug 7, 2018 at 21:26
ls -t | head -n2 will give you just the filename, you don't need to parse the ls output. — karakfa
– karakfa, Commented Aug 7, 2018 at 21:59

glenn jackman · Accepted Answer · 2018-08-08 02:29:07Z

1

I suspect you want

logfile=$(stat -c "%Y:%n" ~/logs/autosys/*"$1"*.out | sort -t: -k1,1n | tail -1 | cut -d: -f2-)

answered Aug 8, 2018 at 2:29

glenn jackman

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4 Comments

Benjamin W. Over a year ago

Shouldn't it be tail -2 (the two latest ones, as per question)?

glenn jackman Over a year ago

The logic of that loop is odd. The variable is set each tie through, so only the last iteration counts

Benjamin W. Over a year ago

I think it's incorrect in the code in the question, just pointing out that the text says "last two".

glenn jackman Over a year ago

It appears that the OP wants the ".out" file that is in the last two. I'm short-circuiting that

chepner · Accepted Answer · 2018-08-07 22:08:42Z

Without using ls, you would iterate over the files unconditionally, calling stat on each one, and keeping track of the two newest files yourself.

t0=0
t1=0

for i in ~/logs/autosys/*"${1}"*; do
    # Assuming GNU stat; syntax for other implementations may vary
    t=$(stat -c "%Y" "$i")
    if (( t > t0 )); then
        f1=$f0
        f0=$i
        t1=$t0
        t0=$t
    elif (( t > t1 )); then
        t1=$t
        f1=$i
    fi
done

Now $f0 and $f1 contain the names of the two newest files.

The fact is, bash isn't really suited for doing this task easily and efficiently. (The bash distribution does contain a built-in stat command, but it doesn't seem to be commonly installed.)

zsh, on the other hand, makes this task rather easy:

% files=(~/logs/autosys/*${1}*(Om[-2,-1]))

The glob qualifiers Om and [-2,-1] sort by modification date and limit the match to the last two elements, respectively. ((om[1,2]) also works.)

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