I want to monitor a directories + sub dir and get the first 7 characters of file that created in the folder or sub folder. I'm using these codes for my problem :
inotifywait -mqr /path/ -e create |
while read path file ; do
line=$(head -c 7 $file) ;
echo $line ;
done;
but when I run the code it won't print only succeeded for some text file (first 7 character = "setruk"), and not print anythings for some text file (first 7 characters = "kukuryu"). Can anyone help me ? pleasee ?
For me, this is the output of inotifywait:
$ inotifywait -mqr . -e create
./test/ CREATE five
./test/ CREATE six
./test/ CREATE seven
I don't know if you get the CREATE part, but you're probably missing the path to the directory. You likely want:
inotifywait -mqr /path/ -e create |
while read path action file ; do
line=$(head -c 7 "$path$file") ;
echo $line ;
done;
If you don't have the CREATE part in the line, just change the head command to include the $path part of the filename:
line=$(head -c 7 "$path$file") ;
What's probably happening is that files in sub-directories are not getting their characters printed but files in the main directory are.
Related
I have a lot of daily files that are sort by hours which comes from a data-logger (waveform). I downloaded inside a USB stick, now I need to save them inside folders named with the first 8 characters of waveform.
Those files have the following pattern:
Year-Month-Day-hourMinute-##.Code_Station_location_Channel
for example, inside the USB I have:
2020-10-01-0000-03.AM_REDDE_00_EHE; 2020-10-01-0100-03.AM_REDDE_00_EHE; 2020-10-02-0300-03.AM_REDDE_00_EHE; 2020-10-20-0000-03.AM_REDDE_00_EHE; 2020-10-20-0100-03.AM_REDDE_00_EHE; 2020-11-15-2000-03.AM_REDDE_00_EHE; 2020-11-15-2100-03.AM_REDDE_00_EHE; 2020-11-19-0400-03.AM_REDDE_00_EHE; 2020-11-19-0900-03.AM_REDDE_00_EHE;
I modified a little a code from #user3360767 (shell script to create folder daily with time-stamp and push time-stamp generated logs) to speed up the procedure of creating a folder and moving the files to them
for filename in 2020-10-01*EHE; do
foldername=$(echo "$filename" | awk '{print (201001)}');
mkdir -p "$foldername"
mv "$filename" "$foldername"
echo "$filename $foldername" ;
done
2020-10-01*EHE
Here I list all hours from 2020-10-01-0000-03.AM_REDDE_00_EHE
foldername=$(echo "$filename" | awk '{print (201001)}');
Here I create the folder that belongs to 2020-10-01 and with the following lines create the folder and then move all files to created folder.
mkdir -p "$foldername"
mv "$filename" "$foldername"
echo "$filename $foldername" ;
As you may notice, I will always need to modify the line for filename in 2020-10-01*EHE each time the file changes the date.
Is there a way to try to create folders with the first 8 number of the file?
Tonino
Use date
And since the foldername doesn't change, you don't need to keep creating one inside the loop.
files="$(date +%Y-%m-%d)*EHE"
foldername=$(date +%Y%m%d)
mkdir -p "$foldername"
for filename in $files; do
mv "$filename" "$foldername"
echo "$filename $foldername"
done
Edit:
If you want to specify the folder each time, you can pass it as an argument and use sed to get the filename pattern
foldername=$1
files=$(echo $1 | sed 's/\(....\)\(..\)\(..\)/\1-\2-\3/')
filepattern="$files*EHE"
mkdir -p "$foldername"
for filename in $filepattern; do
mv "$filename" "$foldername"
echo "$filename $foldername"
done
You call it with
./<yourscriptname>.sh 20101001
I think you want to move all files whose names end in *EHE into subdirectories. The subdirectories will be created as necessary and will be named according to the date at the start of each filename without the dashes/hyphens.
Please test the following on a copy of your files in a temporary directory somewhere.
#!/bin/bash
for filename in *EHE ; do
# Derive folder by deleting all dashes from filename, then taking first 8 characters
folder=${filename//-/}
folder=${folder:0:8}
echo "Would move $filename to $folder"
# Uncomment next 2 lines to actually move file
# mkdir -p "$folder"
# mv "$filename" "$folder"
done
Sample Output
Would move 2020-10-01-0000-03.AM_REDDE_00_EHE to 20201001
Would move 2020-10-01-0100-03.AM_REDDE_00_EHE to 20201001
Note that the 2 lines:
folder=${filename//-/}
folder=${folder:0:8}
use "bash parameter substitution", which is described here if you want to learn about it, and obviate the need to create whole new processes to run awk, sed or cut to extract the fields.
I have two folders which have same file names, but different contents. So, I am trying to generate a script to get the difference and to see what is being changed. I wrote a script below :
folder1="/opt/dir1"
folder2=`ls/opt/dir2`
find "$folder1/" /opt/dir2/ -printf '%P\n' | sort | uniq -d
for item in `ls $folder1`
do
if [[ $item == $folder2 ]]; then
diff -r $item $folder2 >> output.txt
fi
done
I believe this script has to work, but it is not giving any output on output folder.
So the desired output should be in one file . Ex:
cat output.txt
diff -r /opt/folder1/file1 /opt/folder2/file1
1387c1387
< ALL X'25' BY SPACE
---
> ALL X'0A' BY SPACE
diff -r /opt/folder1/file2 /opt/folder2/file2
2591c2591
< ALL X'25' BY SPACE
---
> ALL X'0A' BY SPACE
Any help is appreciated!
Ok. So twofold:
First get the files in one folder. Never use ls. Forget it exists. ls is for nice printing in our console. In scripts, use find.
Then for each file do some command. A simple while read loop.
So:
{
# make find print relative to `/opr/dir1` director
cd /opt/dir1 &&
# Use `%P` so that print without leading `./`
find . -mindepth 1 -type f -print "%P\n"
} |
while IFS= read -r file; do
diff /opt/dir1/"$file" /opt/dir2/"$file" >> output/"$file"
done
Notes:
always quote your variable
Why you shouldn't parse the output of ls(1)
I'm trying to write a bash script that if I pass a text file containing some extension and a folder returns me an output file with the list of all files that match the desired extension, searching recursively in all sub-directories
the folder is my second parameter the extension list file my first parameter
I have tried:
for i in $1 ; do
find . -name $2\*.$i -print>>result.txt
done
but doesn't work
As noted from in comment:
It is not a good idea to write to a hard coded file name.
The given example fixes only the given code from the OP question.
Yes of course, it is even better to call with
x.sh y . > blabla
and remove the filename from the script itself. But my intention is not to fix the question...
The following bash script, named as x.sh
#!/bin/bash
echo -n >result.txt # delete old content
while read i; do # read a line from file
find $2 -name \*.$i -print>>result.txt # for every item do a find
done <$1 # read from file named with first arg from cmdline
with an text file named y with following content
txt
sh
and called with:
./x.sh y .
results in a file result.txt which contents is:
a.txt
b.txt
x.sh
OK, lets give some additional hints as got from comments:
If the results fiel should not collect any other conntent from other results of the script it can be simplified to:
#!/bin/bash
while read i; do # read a line from file
find $2 -name \*.$i -print # for every item do a find
done <$1 >result.txt # read from file named with first arg from cmdline
And as already mentioned:
The hard coded result.txt could be removed and the call can be something like
./x.sh y . > result.txt
Give this one-liner command a try.
Replace /mydir with the folder to search.
Change the list of extensions passed as argument to the egrep command:
find /mydir -type f | egrep "[.]txt|[.]xml" >> result.txt
After the egrep, each extension should be separated with |.
. char must be escaped with [.]
I have folder which contains more then 100 files. I want to insert line number into each file.
nl command gives output to standard output on terminal. But I want to add line number in all files of folder.
Can you suggest me how to do this?
Following on #Gianluca's answer, and using bash instead:
for i in *.c *.h ; do ( nl $i > $i.numbered ) && mv $i.numbered $i ; done
This replaces all files ending with .c or .h in the current directory with line-numbered versions.
Using tcsh you can do something like
foreach f (`ls *`)
nl $f >> $f.out
mv $f.out $f
end
You can delete them mv command if you don't want to rename the files
(try the script on a copy ;-) )
I am in need of some command line fu.
I have a wack of files, starting with 4 numbers then a dash then various letters then extentsion, eg.
0851_blahblah_p.dbf
0754_asdf_l.dbf
What I want is to move the four numbers to the end of the filename (keeping the extension in tact) and remove the underscore. Thus the above example would be renamed:
blahblah_p0851.dbf
asdf_l0754.dbf
All help is appreciated.
I am running ubuntu.
Thanks DJ
Here is a solution in pure bash:
for file in *.dbf; do
ext=${file##*.};num=${file%%_*};name=${file%.*};name=${name#*_}
mv $file $name$num"."$ext;
done
Broken down with comments:
for file in *.dbf
do
ext=${file##*.} # Capture the extension
num=${file%%_*} # Capture the number
name=${file%.*} # Step 1: Capture the name
name=${name#*_} # Step 2: Capture the name
mv "$file" "$name$num.$ext" # move the files to new name
done
You can use the rename command:
rename 's/([0-9]{4})_([[:alpha:]]*)_.*.dbf/$2_$1.dbf/' *
You can use sed also
$sed -r 's/([^_]+)_([^.]+)/\2\1/g'
Using this way the given name is splitted and modified as per your requirement.
(or)
Use this script and pass the file names as argument it will move the file name as per the requirement.
#!/bin/sh
if [ $# -ne 1 ] ; then
echo "Usage : <sh filename> <arguments>"
exit ;
fi
for file in $*
do
mv $file `echo $file | sed -r 's/([^_]+)_([^.]+)/\2\1/g' `
done