The end of git status looks like this:
# Untracked files:
# (use "git add <file>..." to include in what will be committed)
#
# Classes/Default.png
# Classes/Default#2x.png
...
Since you might have any number of untracked files, I'm trying to tail from the end of the file to "Untracked files" and save it to a temp file, strip out the first three lines and convert the filenames to git add Classes/...
I can't seem to find a good way (other than maybe a different language) to tail up to a searchable expression. Thanks!
Use sed to print everything from "Untracked files" to the end:
git status | sed -n '/Untracked files:$/,$p'
Then you just have to parse the filenames by removing the # character.
You can also use git status -s to get a shorter, more easily parsed output:
~$ git status -s
?? Classes/Default.png
?? Classes/Default#2x.png
This is a good application of awk, which lets you grep and extract at the same time:
~$ git status -s | awk '/\?\?/{print $2}'
Classes/Default.png
Classes/Default#2x.png
Alternatively: awk '{if ($1 == "??") print $2}'
You can also, of course, use git add to list (and add) untracked files.
Use the tail command:
tail -$(($(wc -l < file.txt) - $(grep -n "Untracked files" file.txt|cut -d: -f1) - 2)) file.txt
How it works:
total number of lines in file = wc -l < file.txt
line number of "Untracked files" = grep -n "Untracked files" file.txt|cut -d: -f1
The -2 is to remove the top lines
Complete command with git add:
tail -$(($(wc -l < file.txt) - $(grep -n "Untracked files" file.txt|cut -d: -f1) - 2)) file.txt | tr -d '#'| while read line; do echo git add $line; done
Pipe it to:
perl -e 'my ($o, $p) = (0, shift); while(<>){print if $o or $o = /$p/}' "$MY_REGEX"
Where $MY_REGEX is your pattern. In your case, probably '^\s{7}'.
Solution using shell scripting.
First start reading the file in a while loop, keep a count of the number of lines read, break the loop when the required line is found.
Using the count of lines tail the file and then extract file names using awk.
i=0;
l=`wc -l filename | awk '{print $1}'`;
while read line;
do i=`echo $i + 1 | bc`;
if [[ $line == "# Untracked files:" ]];
then break;
fi;
done < "filename";
tail -`echo $l -$i -2 | bc` filename | awk -F "/" '{print $NF}'
Here "filename" is the file you want to process
Related
I'm still new to the shell and need some help.
I have a file stapel_old.
Also I have in the same directory files like english_old_sync, math_old_sync and vocabulary_old_sync.
The content of stapel_old is:
english
math
vocabulary
The content of e.g. english is:
basic_grammar.md
spelling.md
orthography.md
I want to manipulate all files which are given in stapel_old like in this example:
take the first line of stapel_old 'english', (after that math, and so on)
convert in this case english to english_old_sync, (or after that what is given in second line, e.g. math to math_old_sync)
search in english_old_sync line by line for the pattern '.md'
And append to each line after .md :::#a1
The result should be e.g. of english_old_sync:
basic_grammar.md:::#a1
spelling.md:::#a1
orthography.md:::#a1
of math_old_sync:
geometry.md:::#a1
fractions.md:::#a1
and so on. stapel_old should stay unchanged.
How can I realize that?
I tried with sed -n, while loop (while read -r line), and I'm feeling it's somehow the right way - but I still get errors and not the expected result after 4 hours inspecting and reading.
Thank you!
EDIT
Here is the working code (The files are stored in folder 'olddata'):
clear
echo -e "$(tput setaf 1)$(tput setab 7)Learning directories:$(tput sgr 0)\n"
# put here directories which should not become flashcards, command: | grep -v 'name_of_directory_which_not_to_learn1' | grep -v 'directory2'
ls ../ | grep -v 00_gliederungsverweise | grep -v 0_weiter | grep -v bibliothek | grep -v notizen | grep -v Obsidian | grep -v z_nicht_uni | tee olddata/stapel_old
# count folders
echo -ne "\nHow much different folders: " && wc -l olddata/stapel_old | cut -d' ' -f1 | tee -a olddata/stapel_old
echo -e "Are this learning directories correct? [j ODER y]--> yes; [Other]-->no\n"
read lernvz_korrekt
if [ "$lernvz_korrekt" = j ] || [ "$lernvz_korrekt" = y ];
then
read -n 1 -s -r -p "Learning directories correct. Press any key to continue..."
else
read -n 1 -s -r -p "Learning directories not correct, please change in line 4. Press any key to continue..."
exit
fi
echo -e "\n_____________________________\n$(tput setaf 6)$(tput setab 5)Found cards:$(tput sgr 0)$(tput setaf 6)\n"
#GET && WRITE FOLDER NAMES into olddata/stapel_old
anzahl_zeilen=$(cat olddata/stapel_old |& tail -1)
#GET NAMES of .md files of every stapel and write All to 'stapelname'_old_sync
i=0
name="var_$i"
for (( num=1; num <= $anzahl_zeilen; num++ ))
do
i="$((i + 1))"
name="var_$i"
name=$(cat olddata/stapel_old | sed -n "$num"p)
find ../$name/ -name '*.md' | grep -v trash | grep -v Obsidian | rev | cut -d'/' -f1 | rev | tee olddata/$name"_old_sync"
done
(tput sgr 0)
I tried to add:
input="olddata/stapel_old"
while IFS= read -r line
do
sed -n "$line"p olddata/stapel_old
done < "$input"
The code to change only the english_old_sync is:
lines=$(wc -l olddata/english_old_sync | cut -d' ' -f1)
for ((num=1; num <= $lines; num++))
do
content=$(sed -n "$num"p olddata/english_old_sync)
sed -i "s/"$content"/""$content":::#a1/g"" olddata/english_old_sync
done
So now, this need to be a inner for-loop, of a outer for-loop which holds the variable for english, right?
stapel_old should stay unchanged.
You could try a while + read loop and embed sed inside the loop.
#!/usr/bin/env bash
while IFS= read -r files; do
echo cp -v "$files" "${files}_old_sync" &&
echo sed '/^.*\.md$/s/$/:::#a1/' "${files}_old_sync"
done < olddata/staple_old
convert in this case english to english_old_sync, (or after that what is given in second line, e.g. math to math_old_sync)
cp copies the file with a new name, if the goal is renaming the original file name from the content of the file staple_old then change cp to mv
The -n and -i flag from sed was ommited , include it, if needed.
The script also assumes that there are no empty/blank lines in the content of staple_old file. If in case there are/is add an addition test after the line where the do is.
[[ -n $files ]] || continue
It also assumes that the content of staple_old are existing files. Just in case add an additional test.
[[ -e $files ]] || { printf >&2 '%s no such file or directory.\n' "$files"; continue; }
Or an if statement.
if [[ ! -e $files ]]; then
printf >&2 '%s no such file or directory\n' "$files"
continue
fi
See also help test
See also help continue
Combining them all together should be something like:
#!/usr/bin/env bash
while IFS= read -r files; do
[[ -n $files ]] || continue
[[ -e $files ]] || {
printf >&2 '%s no such file or directory.\n' "$files"
continue
}
echo cp -v "$files" "${files}_old_sync" &&
echo sed '/^.*\.md$/s/$/:::#a1/' "${files}_old_sync"
done < olddata/staple_old
Remove the echo's If you're satisfied with the output so the script could copy/rename and edit the files.
I have a textfile called log.txt, and it logs the file name and the path it was gotten from. so something like this
2.txt
/home/test/etc/2.txt
basically the file name and its previous location. I want to use grep to grab the file directory save it as a variable and move the file back to its original location.
for var in "$#"
do
if grep "$var" log.txt
then
# code if found
else
# code if not found
fi
this just prints out to the console the 2.txt and its directory since the directory has 2.txt in it.
thanks.
Maybe flip the logic to make it more efficient?
f=''
while read prev
do case "$prev" in
*/*) f="${prev##*/}"; continue;; # remember the name
*) [[ -e "$f" ]] && mv "$f" "$prev";;
done < log.txt
That walks through all the files in the log and if they exist locally, move them back. Should be functionally the same without a grep per file.
If the name is always the same then why save it in the log at all?
If it is, then
while read prev
do f="${prev##*/}" # strip the path info
[[ -e "$f" ]] && mv "$f" "$prev"
done < <( grep / log.txt )
Having the file names on the same line would significantly simplify your script. But maybe try something like
# Convert from command-line arguments to lines
printf '%s\n' "$#" |
# Pair up with entries in file
awk 'NR==FNR { f[$0]; next }
FNR%2 { if ($0 in f) p=$0; else p=""; next }
p { print "mv \"" p "\" \"" $0 "\"" }' - log.txt |
sh
Test it by replacing sh with cat and see what you get. If it looks correct, switch back.
Briefly, something similar could perhaps be pulled off with printf '%s\n' "$#" | grep -A 1 -Fxf - log.txt but you end up having to parse the output to pair up the output lines anyway.
Another solution:
for f in `grep -v "/" log.txt`; do
grep "/$f" log.txt | xargs -I{} cp $f {}
done
grep -q (for "quiet") stops the output
I have many files on a server which contains many lines:
201701010530.contentState.csv.gz
201701020530.contentState.csv.gz
201701030530.contentState.csv.gz
201701040530.contentState.csv.gz
I would like with one line command this result:
170033|20170101
169865|20170102
170010|20170103
170715|20170104
The goal is to have the number of lines of each file, just by keeping the date which is already in the filename of the file.
I tried this but the result is not in one line but two...
for f in $(ls -1 2017*gz);do zcat $f | wc -l;echo $f | awk '{print substr($0,1,8)}';done
Thanks in advance guys.
Just use zcat file | wc -l to get the number of lines.
For the name, I understand it is enough to extract the first 8 characters:
$ t="201701030530.contentState.csv.gz"
$ echo "${t:0:8}"
20170103
All together:
for file in 2017*gz;
do
lines=$(zcat "$file" | wc -l)
printf "%s|%s\n" "$lines" "${file:0:8}"
done > myresult.csv
Note the usage of for file in 2017*gz; to go through the files matching the 2017*gz pattern: this suffices, no need to parse ls!
Use zgrep -c ^ file to count the lines, here encapsulated in awk:
$ awk 'FNR==1{ "zgrep -c ^ " FILENAME | getline s; print s "|" substr(FILENAME,1,8) }' *.gz
12|20170101
The whole "zgrep -c ^ " FILENAME should probably be in a var (s) and then s | getline s.
I have a problem creating a script that reads specific value from all the files of an entire folder
I have a number of email files in a directory and I need to extract from each file, 2 specific values.
After that I have to put them into a new file that looks like that:
--------------
To: value1
value2
--------------
This is what I want to do, but I don't know how to create the script:
# I am putting the name of the files into a temp file
`ls -l | awk '{print $9 }' >tmpfile`
# use for the name of a file
`date=`date +"%T"
# The first specific value from file (phone number)
var1=`cat tmpfile | grep "To: 0" | awk '{print $2 }' | cut -b -10 `
# The second specific value from file(subject)
var2=cat file | grep Subject | awk '{print $2$3$4$5$6$7$8$9$10 }'
# Put the first value in a new file on the first row
echo "To: 4"$var1"" > sms-$date
# Put the second value in the same file on the second row
echo ""$var2"" >>sms-$date
.......
and do the same for every file in the directory
I tried using while and for functions but I couldn't finalize the script
Thank You
I've made a few changes to your script, hopefully they will be useful to you:
#!/bin/bash
for file in *; do
var1=$(awk '/To: 0/ {print substr($2,0,10)}' "$file")
var2=$(awk '/Subject/ {for (i=2; i<=10; ++i) s=s$i; print s}' "$file")
outfile="sms-"$(date +"%T")
i=0
while [ -f "$outfile" ]; do outfile="sms-$date-"$((i++)); done
echo "To: 4$var1" > "$outfile"
echo "$var2" >> "$outfile"
done
The for loop just goes through every file in the folder that you run the script from.
I have added added an additional suffix $i to the end of the file name. If no file with the same date already exists, then the file will be created without the suffix. Otherwise the value of $i will keep increasing until there is no file with the same name.
I'm using $( ) rather than backticks, this is just a personal preference but it can be clearer in my opinion, especially when there are other quotes about.
There's not usually any need to pipe the output of grep to awk. You can do the search in awk using the / / syntax.
I have removed the cut -b -10 and replaced it with substr($2, 0, 10), which prints the first 10 characters from column 2.
It's not much shorter but I used a loop rather than the $2$3..., I think it looks a bit neater.
There's no need for all the extra " in the two output lines.
I sugest to try the following:
#!/bin/sh
RESULT_FILE=sms-`date +"%T"`
DIR=.
fgrep -l 'To: 0' "$DIR" | while read FILE; do
var1=`fgrep 'To: 0' "$FILE" | awk '{print $2 }' | cut -b -10`
var2=`fgrep 'Subject' "$FILE" | awk '{print $2$3$4$5$6$7$8$9$10 }'`
echo "To: 4$var1" >>"$RESULT_FIL"
echo "$var2" >>"$RESULT_FIL"
done
I have number of files which have similar names like
DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out
DWH_Export_AUSTA_20120701_20120731_v1_2.csv.397.dat.2012-10-02 04-03-12.out
DWH_Export_AUSTA_20120801_20120831_v1_1.csv.397.dat.2012-10-02 04-04-16.out
etc.
I need to get number before .csv(1 or 2) from the file name and put it into end of every line in file with TAB separator.
I have written this code, it finds number that I need, but i do not know how to put this number into file. There is space in the filename, my script breaks because of it.
Also I am not sure, how to send to script list of files. Now I am working only with one file.
My code:
#!/bin/sh
string="DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out"
out=$(echo $string | awk 'BEGIN {FS="_"};{print substr ($7,0,1)}')
awk ' { print $0"\t$out" } ' $string
for file in *
do
sfx=$(echo "$file" | sed 's/.*_\(.*\).csv.*/\1/')
sed -i "s/$/\t$sfx/" "$file"
done
Using sed:
$ sed 's/.*_\(.*\).csv.*/&\t\1/' file
DWH_Export_AUSTA_20120701_20120731_v1_1.csv.397.dat.2012-10-02 04-01-46.out 1
DWH_Export_AUSTA_20120701_20120731_v1_2.csv.397.dat.2012-10-02 04-03-12.out 2
DWH_Export_AUSTA_20120801_20120831_v1_1.csv.397.dat.2012-10-02 04-04-16.out 1
To make this for many files:
sed 's/.*_\(.*\).csv.*/&\t\1/' file1 file2 file3
OR
sed 's/.*_\(.*\).csv.*/&\t\1/' file*
To make this changed get saved in the same file(If you have GNU sed):
sed -i 's/.*\(.\).csv.*/&\t\1/' file
Untested, but this should do what you want (extract the number before .csv and append that number to the end of every line in the .out file)
awk 'FNR==1 { split(FILENAME, field, /[_.]/) }
{ print $0"\t"field[7] > FILENAME"_aaaa" }' *.out
for file in *_aaaa; do mv "$file" "${file/_aaaa}"; done
If I understood correctly, you want to append the number from the filename to every line in that file - this should do it:
#!/bin/bash
while [[ 0 < $# ]]; do
num=$(echo "$1" | sed -r 's/.*_([0-9]+).csv.*/\t\1/' )
#awk -e "{ print \$0\"\t${num}\"; }" < "$1" > "$1.new"
#sed -r "s/$/\t$num/" < "$1" > "$1.mew"
#sed -ri "s/$/\t$num/" "$1"
shift
done
Run the script and give it names of the files you want to process. $# is the number of command line arguments for the script which is decremented at the end of the loop by shift, which drops the first argument, and shifts the other ones. Extract the number from the filename and pick one of the three commented lines to do the appending: awk gives you more flexibility, first sed creates new files, second sed processes them in-place (in case you are running GNU sed, that is).
Instead of awk, you may want to go with sed or coreutils.
Grab number from filename, with grep for variety:
num=$(<<<filename grep -Eo '[^_]+\.csv' | cut -d. -f1)
<<<filename is equivalent to echo filename.
With sed
Append num to each line with GNU sed:
sed "s/\$/\t$num" filename
Use the -i switch to modify filename in-place.
With paste
You also need to know the length of the file for this method:
len=$(<filename wc -l)
Combine filename and num with paste:
paste filename <(seq $len | while read; do echo $num; done)
Complete example
for filename in DWH_Export*; do
num=$(echo $filename | grep -Eo '[^_]+\.csv' | cut -d. -f1)
sed -i "s/\$/\t$num" $filename
done