I am new to bash scripting and I have written a script to match regex and output lines to print to a file.
However, each line contains multiple columns, one of which is the timestamp column, which appears in the form YYYYMMDDHHMMSSTTT (to millisecond) as shown below.
20180301050630663,ABC,,,,,,,,,,
20180301050630664,ABC,,,,,,,,,,
20180301050630665,ABC,,,,,,,,,,
20180301050630666,ABC,,,,,,,,,,
20180301050630667,ABC,,,,,,,,,,
20180301050630668,ABC,,,,,,,,,,
20180301050630663,ABC,,,,,,,,,,
20180301050630665,ABC,,,,,,,,,,
20180301050630661,ABC,,,,,,,,,,
20180301050630662,ABC,,,,,,,,,,
My code is written as follow:
awk -F "," -v OFS=","'{if($2=="ABC"){print}}' < $i>> "$filename"
How can I modify my code such that it can sort the rows by timestamp (YYYYMMDDHHMMSSTTT) in ascending order before printing to file?
You can use a very simple sort command, e.g.
sort yourfile
If you want to insure sort only looks at the datestamp, you can tell sort to only use the first command separated field as your sorting criteria, e.g.
sort -t, -k1 yourfile
Example Use/Output
With your data save in a file named log, you could do:
$ sort -t, -k1 log
20180301050630661,ABC,,,,,,,,,,
20180301050630662,ABC,,,,,,,,,,
20180301050630663,ABC,,,,,,,,,,
20180301050630663,ABC,,,,,,,,,,
20180301050630664,ABC,,,,,,,,,,
20180301050630665,ABC,,,,,,,,,,
20180301050630665,ABC,,,,,,,,,,
20180301050630666,ABC,,,,,,,,,,
20180301050630667,ABC,,,,,,,,,,
20180301050630668,ABC,,,,,,,,,,
Let me know if you have any problems.
Just add a pipeline.
awk -F "," '$2=="ABC"' < "$i" |
sort -n >> "$filename"
In the general case, to sort on column 234. try sort -t, -k234,234n
Notice alse the quoting around "$i", like you already have around "$filename", and the simplifications of the Awk script.
If you are using gawk you can do:
$ awk -F "," -v OFS="," '$2=="ABC"{a[$1]=$0} # Filter lines that have "ABC"
END{ # set the sort method
PROCINFO["sorted_in"] = "#ind_num_asc"
for (e in a) print a[e] # traverse the array of lines
}' file
An alternative is to use sed and sort:
sed -n '/^[0-9]*,ABC,/p' file | sort -t, -k1 -n
Keep in mind that both of these methods are unrelated to the shell used. Bash is just executing the tools (sed, awk, sort, etc) that are otherwise part of the OS.
Bash itself could do the sort in pure Bash but it would be long and slow.
I got a txt file with some content looking like
stuff,stuff,2012-12-12
morestuff,morestuff,2012-09-09
evenmorestuff,yeah,2012-08-02
and I want to use cat and sort to get them reverse ordered by the date as an output on my command-line by concatenation.
not sure why you think you need to cat a file into sort, but here are 2 options
cat yourFile | sort -t, -k3r
sort -t, -k3r yourFile
To test this I did
echo "stuff,stuff,2012-12-12
morestuff,morestuff,2012-09-09
evenmorestuff,yeah,2012-08-02" \
| sort -t, -k3r
output
stuff,stuff,2012-12-12
morestuff,morestuff,2012-09-09
evenmorestuff,yeah,2012-08-02
And finally, you can overwrite your existing file using the -o option like
sort -t, -o yourFile -k3r yourFile
Thanks to #karakfa for reminding me your your requirement for reverse order sort. This is accomplished by adding an r to the key specification, hence -k3r.
IHTH
I am trying to sort files in a directory, depending on the 'date string' attached in the file name, for example files looks as below
SSA_F12_05122013.request.done
SSA_F13_12142012.request.done
SSA_F14_01062013.request.done
Where 05122013,12142012 and 01062013 represents the dates in format.
Please help me in providing a unix shell script to sort these files on the date string present in their file name(in descending and ascending order).
Thanks in advance.
Hmmm... why call on heavyweights like awk and Perl when sort itself has the capability to define what exactly to sort by?
ls SSA_F*.request.done | sort -k 1.13,1.16 -k 1.9,1.10 -k 1.11,1.12
Each -k option defines a "sort key":
-k 1.13,1.16
This defines a sort key ranging from field 1, column 13 to field 1, column 16. (A field is by default delimited by whitespace, which your filenames don't have.)
If your filenames are varying in length, defining the underscore as field separator (using the -t option) and then addressing columns in the third field would be the way to go.
Refer to man sort for details. Use the -r option to sort in descending order.
one way with awk and sort:
ls -1|awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|sort|awk '$0=$NF'
if we break it down:
ls -1|
awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|
sort|
awk '$0=$NF'
the ls -1 just example. I think you have your way to get the file list, one per line.
test a little bit:
kent$ echo "SSA_F13_12142012.request.done
SSA_F12_05122013.request.done
SSA_F14_01062013.request.done"|awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|
sort|
awk '$0=$NF'
SSA_F13_12142012.request.done
SSA_F14_01062013.request.done
SSA_F12_05122013.request.done
ls -lrt *.done | perl -lane '#a=split /_|\./,$F[scalar(#F)-1];$a[2]=~s/(..)(..)(....)/$3$2$1/g;print $a[2]." ".$_' | sort -rn | awk '{$1=""}1'
ls *.done | perl -pe 's/^.*_(..)(..)(....)/$3$2$1$&/' | sort -rn | cut -b9-
this would do +
I am new to shell scripting so i need some help need how to go about with this problem.
I have a directory which contains files in the following format. The files are in a diretory called /incoming/external/data
AA_20100806.dat
AA_20100807.dat
AA_20100808.dat
AA_20100809.dat
AA_20100810.dat
AA_20100811.dat
AA_20100812.dat
As you can see the filename of the file includes a timestamp. i.e. [RANGE]_[YYYYMMDD].dat
What i need to do is find out which of these files has the newest date using the timestamp on the filename not the system timestamp and store the filename in a variable and move it to another directory and move the rest to a different directory.
For those who just want an answer, here it is:
ls | sort -n -t _ -k 2 | tail -1
Here's the thought process that led me here.
I'm going to assume the [RANGE] portion could be anything.
Start with what we know.
Working Directory: /incoming/external/data
Format of the Files: [RANGE]_[YYYYMMDD].dat
We need to find the most recent [YYYYMMDD] file in the directory, and we need to store that filename.
Available tools (I'm only listing the relevant tools for this problem ... identifying them becomes easier with practice):
ls
sed
awk (or nawk)
sort
tail
I guess we don't need sed, since we can work with the entire output of ls command. Using ls, awk, sort, and tail we can get the correct file like so (bear in mind that you'll have to check the syntax against what your OS will accept):
NEWESTFILE=`ls | awk -F_ '{print $1 $2}' | sort -n -k 2,2 | tail -1`
Then it's just a matter of putting the underscore back in, which shouldn't be too hard.
EDIT: I had a little time, so I got around to fixing the command, at least for use in Solaris.
Here's the convoluted first pass (this assumes that ALL files in the directory are in the same format: [RANGE]_[yyyymmdd].dat). I'm betting there are better ways to do this, but this works with my own test data (in fact, I found a better way just now; see below):
ls | awk -F_ '{print $1 " " $2}' | sort -n -k 2 | tail -1 | sed 's/ /_/'
... while writing this out, I discovered that you can just do this:
ls | sort -n -t _ -k 2 | tail -1
I'll break it down into parts.
ls
Simple enough ... gets the directory listing, just filenames. Now I can pipe that into the next command.
awk -F_ '{print $1 " " $2}'
This is the AWK command. it allows you to take an input line and modify it in a specific way. Here, all I'm doing is specifying that awk should break the input wherever there is an underscord (_). I do this with the -F option. This gives me two halves of each filename. I then tell awk to output the first half ($1), followed by a space (" ")
, followed by the second half ($2). Note that the space was the part that was missing from my initial suggestion. Also, this is unnecessary, since you can specify a separator in the sort command below.
Now the output is split into [RANGE] [yyyymmdd].dat on each line. Now we can sort this:
sort -n -k 2
This takes the input and sorts it based on the 2nd field. The sort command uses whitespace as a separator by default. While writing this update, I found the documentation for sort, which allows you to specify the separator, so AWK and SED are unnecessary. Take the ls and pipe it through the following sort:
sort -n -t _ -k 2
This achieves the same result. Now you only want the last file, so:
tail -1
If you used awk to separate the file (which is just adding extra complexity, so don't do it sheepish), you can replace the space with an underscore again with sed:
sed 's/ /_/'
Some good info here, but I'm sure most people aren't going to read down to the bottom like this.
This should work:
newest=$(ls | sort -t _ -k 2,2 | tail -n 1)
others=($(ls | sort -t _ -k 2,2 | head -n -1))
mv "$newest" newdir
mv "${others[#]}" otherdir
It won't work if there are spaces in the filenames although you could modify the IFS variable to affect that.
Try:
$ ls -lr
Hope it helps.
Use:
ls -r -1 AA_*.dat | head -n 1
(assuming there are no other files matching AA_*.dat)
ls -1 AA* |sort -r|tail -1
Due to the naming convention of the files, alphabetical order is the same as date order. I'm pretty sure that in bash '*' expands out alphabetically (but can not find any evidence in the manual page), ls certainly does, so the file with the newest date, would be the last one alphabetically.
Therefore, in bash
mv $(ls | tail -1) first-directory
mv * second-directory
Should do the trick.
If you want to be more specific about the choice of file, then replace * with something else - for example AA_*.dat
My solution to this is similar to others, but a little simpler.
ls -tr | tail -1
What is actually does is to rely on ls to sort the output, then uses tail to get the last listed file name.
This solution will not work if the filename you require has a leading dot (e.g. .profile).
This solution does work if the file name contains a space.
I'm writing a small shell script that needs to reverse the lines of a text file. Is there a standard filter command to do this sort of thing?
My specific application is that I'm getting a list of Git commit identifiers, and I want to process them in reverse order:
git log --pretty=oneline work...master | grep -v DEBUG: | cut -d' ' -f1 | reverse
The best I've come up with is to implement reverse like this:
... | cat -b | sort -rn | cut -f2-
This uses cat to number every line, then sort to sort them in descending numeric order (which ends up reversing the whole file), then cut to remove the unneeded line number.
The above works for my application, but may fail in the general case because cat -b only numbers nonblank lines.
Is there a better, more general way to do this?
In GNU coreutils, there's tac(1)
There is a command for your purpose:
tail -r file.txt
Prints the lines of file.txt in reverse order!
The -r flag is non-standard, may not work on all systems, works e.g. on macOS.
Beware: Amount of lines limited. Works mostly, but when working with huge files be careful and check.
Answer is not 42 but tac.
Edit: Slower but more memory consuming using sed
sed 'x;1!H;$!d;x'
and even longer
perl -e'print reverse<>'
Similar to the sed example above, using perl - maybe more memorable (depending on how your brain is wired):
perl -e 'print reverse <>'
cat -b only numbers nonblank lines"
If that's the only issue you want to avoid, then why not use "cat -n" to number all the lines?
: "#(#)$Id: reverse.sh,v 1.2 1997/06/02 21:45:00 johnl Exp $"
#
# Reverse the order of the lines in each file
awk ' { printf("%d:%s\n", NR, $0);}' $* |
sort -t: +0nr -1 |
sed 's/^[0-9][0-9]*://'
Works like a charm for me...
In this case, just use --reverse:
$ git log --reverse --pretty=oneline work...master | grep -v DEBUG: | cut -d' ' -f1
rev <name of your text file.txt>
You can even do this:
echo <whatever you want to type>|rev
awk '{a[i++]=$0}END{for(;i-->0;)print a[i]}'
More faster than sed and compatible for embed devices like openwrt.