I'm looking for a programmatic way to add a file's date to the filename. I'm on a Mac, Yosemite (10.10).
Using Bash, I have put a fair amount of effort into this, but just too new to get there so far. Here's what I have so far:
#!/bin/bash
#!/bin/bash
(IFS='
'
for x in `ls -l | awk '{print$9" "$7"-"$6"-"$9}'`
do
currentfilename=$(expr "$x" : '\($substring\)')
filenamewithdate=$(expr "$x" : '.*\($substring\)')
echo "$currentfilename"
echo "$filenamewithdate"
done)
The idea here is to capture detailed ls output, use awk to capture the strings for the columns with the filename ($9), and also date fields ($7 and $6), then loop that output to capture the previous filename and new filename with the date to mv the file from old filename to new. The awk statement adds a space to separate current filename from new. The echo statement is there now to test if I am able to parse the awk ouput. I just don't know what to add for $substring to get the parts of the string that are needed.
I have much more to learn about Bash scripting, so I hope you'll bear with me as I learn. Any guidance?
Thanks.
Looking at the stat man page, you'd want:
for file in *; do
filedate=$(stat -t '%Y-%m-%dT%H:%M:%S' -f '%m' "$file")
newfile="$file-$filedate"
echo "current: $file -> new: $newfile"
done
Adjust your preferred datetime format to your taste.
You could save a line with
for file in *; do
newfile=$(stat -t '%Y-%m-%dT%H:%M:%S' -f '%N-%m' "$file")
Related
I have a very large selection of files eg.
foo_de.vtt, foo_en.vtt, foo_es.vtt, foo_fr.vtt, foo_pt.vtt, baa_de.vtt, baa_en.vtt, baa_es.vtt, baa_fr.vtt, baa_pt.vtt... etc.
I have created a tab separated file, filenames.txt containing the current string and replacement string eg.
foo 1000
baa 1016
...etc
I want to rename all of the files to get the following:
1000_de.vtt, 1000_en.vtt, 1000_es.vtt, 1000_fr.vtt, 1000_pt.vtt, 1016_de.vtt, 1016_en.vtt, 1016_es.vtt, 1016_fr.vtt, 1016_pt.vtt
I know I can use a utility like rename to do it manually term by term eg:
rename 's/foo/1000/g' *.vtt
could i chain this into an awk command so that it could run through the filenames.txt?
or is there an easier way to do it just in awk? I know I can rename with awk such as:
find . -type f | awk -v mvCmd='mv "%s" "%s"\n' \
'{ old=$0;
gsub(/foo/,"1000");
printf mvCmd,old,$0;
}' | sh
How can I get awk to process filenames.txt and do all of this in one go?
This question is similar but uses sed. I feel that being tab separated this should be quite easy in awk?
First ever post so please be gentle!
Solution
Thanks for all your help. Ultimately I was able to solve by adapting your answers to the following:
while read new old; do
rename "s/$old/$new/g" *.vtt;
done < filenames.txt
I'm assuming that the strings in the TSV file are literals (not regexes nor globs) and that the part to be replaced can be located anywhere in the filenames.
With that said, you can use mv with shell globs and bash parameter expansion:
#!/bin/bash
while IFS=$'\t' read -r old new
do
for f in *"$old"*.vtt
do
mv "$f" "${f/"$old"/$new}"
done
done < file.tsv
Or with GNU rename (more performant):
while IFS=$'\t' read -r old new
do
rename "$old" "$new" *"$old"*.vtt
done < file.tsv
This might work for you (GNU sed and rename):
sed -E 's#(.*)\t(.*)#rename -n '\''s/\1/\2/'\'' \1*#e' ../file
This builds a script which renames the files in the current directory using file to match and replace parts of the filenames.
Once you are happy with the results, remove the -n and the renaming will be enacted.
I have a long file name that I'm converting and a creating a new file from it. Im having a bad brain lapse and cant figure out how to take the time stamp from the original file and append it to the end of my new file.
heres an example of the original file name, the 20171005084554 is the part i want to append to my new file.
527448423osel5108_k8lx2og.527448423_20171005-095125.20171005084554
I run my script
#/bin/sh -x
grep "COLUMN:WM\|COLUMN:Vendor" ${1} | sed 's/^COLUMN://' | sed 's/\.//g' > cust_$(date +%Y%m%d)_converted.txt
grep "DATA:" ${1} | sed 's/^DATA://' >> cust_$(date +%Y%m%d)_converted.txt
# mv ${1} processed/
Where i have $(date +%Y%m%d) is where i want to the other time stamp.
Anyone have any unix / shell magic to help me out?
Using just parameter expansion:
#!/usr/bin/env bash
ext=${1##*.} # ${1: -14} also possible
file=cust_${ext}_converted.txt
echo "$file"
Output:
cust_20171005084554_converted.txt
Due to a variety of complex photo library migrations that had to be done using a combination of manual copying and importing tools that renamed the files, it seems I wound up with a ton of files with a similar structure. Here's an example:
2009-05-05 - 2009-05-05 - IMG_0486 - 2009-05-05 at 10-13-43 - 4209 - 2009-05-05.JPG
What it should be:
2009-05-05 - IMG_0486.jpg
The other files have the same structure, but obviously the individual dates and IMG numbers are different.
Is there any way I can do some command line magic in Terminal to automatically rename these files to the shortened/correct version?
I assume you may have sub-directories and want to find all files inside this directory tree.
This first code block (which you could put in a script) is "safe" (does nothing), but will help you see what would be done.
datep="[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]"
dir="PUT_THE_FULL_PATH_OF_YOUR_MAIN_DIRECTORY"
while IFS= read -r file
do
name="$(basename "$file")"
[[ "$name" =~ ^($datep)\ -\ $datep\ -\ ([^[:space:]]+)\ -\ $datep.*[.](.+)$ ]] || continue
date="${BASH_REMATCH[1]}"
imgname="${BASH_REMATCH[2]}"
ext="${BASH_REMATCH[3],,}"
dir_of_file="$(dirname "$file")"
target="$dir_of_file/$date - $imgname.$ext"
echo "$file"
echo " would me moved to..."
echo " $target"
done < <(find "$dir" -type f)
Make sure the output is what you want and are expecting. I cannot test on your actual files, and if this script does not produce results that are entirely satisfactory, I do not take any responsibility for hair being pulled out. Do not blindly let anyone (including me) mess with your precious data by copy and pasting code from the internet if you have no reliable, checked backup.
Once you are sure, decide if you want to take a chance on some guy's code written without any opportunity for testing and replace the three consecutive lines beginning with echo with this :
mv "$file" "$target"
Note that file names have to match to a pretty strict pattern to be considered for processing, so if you notice that some files are not being processed, then the pattern may need to be modified.
Assuming they are all the exact same structure, spaces and everything, you can use awk to split the names up using the spaces as break points. Here's a quick and dirty example:
#!/bin/bash
output=""
for file in /path/to/files/*; do
unset output #clear variable from previous loop
output="$(echo $file | awk '{print $1}')" #Assign the first field to the output variable
output="$output"" - " #Append with [space][dash][space]
output="$output""$(echo $file | awk '{print $5}')" #Append with IMG_* field
output="$output""." #Append with period
#Use -F '.' to split by period, and $NF to grab the last field (to get the extension)
output="$output""$(echo $file | awk -F '.' '{print $NF}')"
done
From there, something like mv /path/to/files/$file /path/to/files/$output as a final line in the file loop will rename the file. I'd copy a few files into another folder to test with first, since we're dealing with file manipulation.
All the output assigning lines can be consolidated into a single line, as well, but it's less easy to read.
output="$(echo $file | awk '{print $1 " - " $5 "."}')""$(echo $file | awk -F '.' '{print $NF}')"
You'll still want a file loop, though.
Assuming that you want to convert the filename with the first date and the IMG* name, you can run the following on the folder:
IFS=$'\n'
for file in *
do
printf "mv '$file' '"
printf '%s' $(cut -d" " -f1,4,5 <<< "$file")
printf "'.jpg"
done | sh
I have a folder where there are books and I have a file with the real name of each file. I renamed them in a way that I can easily see if they are ordered, say "00.pdf", "01.pdf" and so on.
I want to know if there is a way, using the shell, to match each of the lines of the file, say "names", with each file. Actually, match the line i of the file with the book in the positión i in sort order.
<name-of-the-book-in-the-1-line> -> <book-in-the-1-position>
<name-of-the-book-in-the-2-line> -> <book-in-the-2-position>
.
.
.
<name-of-the-book-in-the-i-line> -> <book-in-the-i-position>
.
.
.
I'm doing this in Windows, using Total Commander, but I want to do it in Ubuntu, so I don't have to reboot.
I know about mv and rename, but I'm not as good as I want with regular expressions...
renamer.sh:
#!/bin/bash
for i in `ls -v |grep -Ev '(renamer.sh|names.txt)'`; do
read name
mv "$i" "$name.pdf"
echo "$i" renamed to "$name.pdf"
done < names.txt
names.txt: (line count must be the exact equal to numbered files count)
name of first book
second-great-book
...
explanation:
ls -v returns naturally sorted file list
grep excludes this script name and input file to not be renamed
we cycle through found file names, read value from file and rename the target files by this value
For testing purposes, you can comment out the mv command:
#mv "$i" "$name"
And now, simply run the script:
bash renamer.sh
This loops through names.txt, creates a filename based on a counter (padding to two digits with printf, assigning to a variable using -v), then renames using mv. ((++i)) increases the counter for the next filename.
#!/bin/bash
i=0
while IFS= read -r line; do
printf -v fname "%02d.pdf" "$i"
mv "$fname" "$line"
((++i))
done < names.txt
I have a few files that I want to copy and rename with the new file names generated by adding a fixed string to each of them.
E.g:
ls -ltr | tail -3
games.txt
files.sh
system.pl
Output should be:
games_my.txt
files_my.sh
system_my.pl
I am able to append at the end of file names but not before *.txt.
for i in `ls -ltr | tail -10`; do cp $i `echo $i\_my`;done
I am thinking if I am able to save the extension of each file by a simple cut as follows,
ext=cut -d'.' -f2
then I can append the same in the above for loop.
do cp $i `echo $i$ext\_my`;done
How do I achieve this?
You can use the following:
for file in *
do
name="${file%.*}"
extension="${file##*.}"
cp $file ${name}_my${extension}
done
Note that ${file%.*} returns the file name without extension, so that from hello.txt you get hello. By doing ${file%.*}_my.txt you then get from hello.txt -> hello_my.txt.
Regarding the extension, extension="${file##*.}" gets it. It is based on the question Extract filename and extension in bash.
If the shell variable expansion mechanisms provided by fedorqui's answer look too unreadable to you, you also can use the unix tool basename with a second argument to strip off the suffix:
for file in *.txt
do
cp -i "$file" "$(basename "$file" .txt)_my.txt"
done
Btw, in such cases I always propose to apply the -i option for cp to prevent any unwanted overwrites due to typing errors or similar.
It's also possible to use a direct replacement with shell methods:
cp -i "$file" "${file/.txt/_my.txt}"
The ways are numerous :)