I'm wanting to create an alias called dir for KSH that will show only the sub-directories in a cluttered directory with many files and directories.
I can output a single-column list of the directories, but I am having a problem converting the single-column list into a multi-column.
I want it to automatically determine the number of columns to use based on the length of the longest string and the width of the terminal window in the same way that the ls command automatically creates tabbed columns.
Say I type the following at the terminal to give me a list of all the directories in the current directory:
ls -1Fa | grep /
When I pipe this output to column -t, it doesn't seem to have any effect... I get a list of 54 directories in a single column. I've also tried the common suggestion in other posts of piping to column -t -s $'\t' or column -t -s $'\n', but it always results in a single list output. I've looked at the man pages and tried all the options, but I cannot get it to produce output similar to ls.
I've also tried piping the output to pr with the following:
alias dir="ls -1Fa | grep / | pr -T -w $COLUMNS -6"
Unfortunately, this will always output 6 columns and if the terminal window is rather skinny (i.e. $COLUMNS is something small like 50) then long directory names are truncated at the beginning of the next column.
Can somebody answer why column is not outputting the list into columns, or does somebody have a better solution using another tool such as sed, awk, perl?
Thanks for your help!
Really, all you need is
alias dir='ls -d */ .*/'
The trailing slash instructs ksh to only include directories in the glob expansion. And you get to piggy-back on how ls formats into columns.
The answer to your question is that you're using the wrong option for column:
ls -1d */ .*/ | column -c $(tput cols)
column -t pretty-prints the output, from
1
a b
foobar baz
into
1
a b
foobar baz
See the column man page
Related
I want to have a command in a variable that runs a program and specifies the output filename for it depending on the number of files exits (to work on a new file each time).
Here is what I have:
export MY_COMMAND="myprogram -o ./dir/outfile-0.txt"
However I would like to make this outfile number increases each time MY_COMMAND is being executed. You may suppose myprogram creates the file soon enough before the next call. So the number can be retrieved from the number of files exists in the directory ./dir/. I do not have access to change myprogram itself or the use of MY_COMMAND.
Thanks in advance.
Given that you can't change myprogram — its -o option will always write to the file given on the command line, and assuming that something also out of your control is running MY_COMMAND so you can't change the way that MY_COMMAND gets called, you still have control of MY_COMMAND
For the rest of this answer I'm going to change the name MY_COMMAND to callprog mostly because it's easier to type.
You can define callprog as a variable as in your example export callprog="myprogram -o ./dir/outfile-0.txt", but you could instead write a shell script and name that callprog, and a shell script can do pretty much anything you want.
So, you have a directory full of outfile-<num>.txt files and you want to output to the next non-colliding outfile-<num+1>.txt.
Your shell script can get the numbers by listing the files, cutting out only the numbers, sorting them, then take the highest number.
If we have these files in dir:
outfile-0.txt
outfile-1.txt
outfile-5.txt
outfile-10.txt
ls -1 ./dir/outfile*.txt produces the list
./dir/outfile-0.txt
./dir/outfile-1.txt
./dir/outfile-10.txt
./dir/outfile-5.txt
(using outfile and .txt means this will work even if there are other files not name outfile)
Scrape out the number by piping it through the stream editor sed … capture the number and keep only that part:
ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:'
(I'm using colon : instead of the standard slash / so I don't have to escape the directory separator in dir/outfile)
Now you just need to pick the highest number. Sort the numbers and take the top
| sort -rn | head -1
Sorting with -n is numeric, not lexigraphic sorting, -r reverses so the highest number will be first, not last.
Putting it all together, this will list the files, edit the names keeping only the numeric part, sort, and get just the first entry. You want to assign that to a variable to work with it, so it is:
high=$(ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' | sort -rn | head -1)
In the shell (I'm using bash) you can do math on that, $[high + 1] so if high is 10, the expression produces 11
You would use that as the numeric part of your filename.
The whole shell script then just needs to use that number in the filename. Here it is, with lines broken for better readability:
#!/bin/sh
high=$(ls -1 ./dir/outfile*.txt \
| sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' \
| sort -rn | head -1)
echo "myprogram -o ./dir/outfile-$[high + 1].txt"
Of course you wouldn't echo myprogram, you'd just run it.
you could do this in a bash function under your .bashrc by using wc to get the number of files in the dir and then adding 1 to the result
yourfunction () {
dir=/path/to/dir
filenum=$(expr $(ls $dir | wc -w) + 1)
myprogram -o $dir/outfile-${filenum}.txt
}
this should get the number of files in $dir and append 1 to that number to get the number you need for the filename. if you place it in your .bashrc or under .bash_aliases and source .bashrc then it should work like any other shell command
You can try exporting a function for MY_COMMAND to run.
next_outfile () {
my_program -o ./dir/outfile-${_next_number}.txt
((_next_number ++ ))
}
export -f next_outfile
export MY_COMMAND="next_outfile" _next_number=0
This relies on a "private" global variable _next_number being initialized to 0 and not otherwise modified.
Lets say I have a folder of .txt files that have a dd-MM-yyyy_HH-mm-ss time followed by _name.txt. I want to be able to sort by name first then time after. Example:
BEFORE
15-2-2010_10-01-55_greg.txt
10-2-1999_10-01-55_greg.txt
10-2-1999_10-01-55_jason.txt
AFTER
greg_1_10-2-1999_10-01-55
greg_2_15-2-2010_10-01-55
jason_1_10-2-1999_10-01-55
Edit: Apologies, from my "cp" line I was meant to copy them into another directory with a different name to them.
Something I tried to do is make a copy with the count, but it doesn't sort the files with the same name properly in terms of dates:
cd data/unfilteredNames
for filename in *.txt; do
n=${filename%.*}
n=${filename##*_}
filteredName=${n%.*}
count=0
find . -type f -name "*_$n" | while read name; do
count=$(($count+1))
cp -p $name ../filteredNames/"$filteredName"_"$count"
done
done
Not sure that the renaming of files is one of your expectation. At least for only sorting file name, you don't need to.
You can do this by only using GNU sort command:
sort -t- -k5.4 -k3.1,3.4 -k2.1,2.1 -k1.1,1.2 -k3.6,3.13 <(printf "%s\n" *.txt)
-t sets the field separator to a dash -.
-k enables to sort based on fields. As explained in man sort page, the syntax is -k<start>,<stop> where <start> or is composed of <field number>.<position>. Adding several -k option to the command allows to sort on multiple fields; the first in he command line having more precedence than the other.
For example, the first -k5.4 tells to sort based on the 5th fields with an offset of 4 characters. There isn't a stop field because this is the end of the filename.
The -k3.1,3.4 option sorts based on the 3rd field starting from offset 1 to 4.
The same principle applies to other -k options.
In your example the month field only has 1 digit. If you have files with a month coded with 2 digits, you might want to pad with 0 all month filenames. This can be done by adding to the printf statement this <(... | sed 's/-0\?\([0-9]\)/-0\1/') and change the -k 2.1,2.1 by -k2.1,2.2.
I have searched extensively on the internet about this but haven't found much details.
Problem Description:
I am using aix server.
I have a pattern.txt file that contains customer_id for 100 customers in the following sample format:
160471231
765082023
75635713
797649756
8011688321
803056646
I have a directory (/home/aswin/temp) with several files (1.txt, 2.txt, 3.txt and so on) which are pipe(|) delimited. Sample format:
797649756|1001|123270361|797649756|O|2017-09-04 23:59:59|10|123769473
803056646|1001|123345418|1237330|O|1999-02-13 00:00:00|4|1235092
64600123|1001|123885297|1239127|O|2001-08-19 00:00:00|10|1233872
75635713|1001|123644701|75635713|C|2006-11-30 00:00:00|11|12355753
424346821|1001|123471924|12329388|O|1988-05-04 00:00:00|15|123351096
427253285|1001|123179704|12358099|C|2012-05-10 18:00:00|7|12352893
What I need to do search all the strings from pattern.txt file in all files in the directory, in first column of each file and list each filename with number of matches. so if same row has more than 1 match it should be counted as 1.
So the output should be something like (only the matches in first column should count):
1.txt:4
2.txt:3
3.txt:2
4.txt:5
What I have done till now:
cd /home/aswin/temp
grep -srcFf ./pattern.txt * /dev/null >> logfile.txt
This is giving the output in the desired format, but it searching the strings in all columns and not just first column. So the output count is much more than expected.
Please help.
If you want to do that with grep, you must change the pattern.
With your command, you search for pattern in /dev/null and the output is /dev/null:0
I think you want 2>/dev/null but this is not needed because you tell -s to grep.
Your pattern file is in the same directory so grep search in it and output pattern.txt:6
All your files are in the same directory so the -r is not needed.
You put the logfile in the same directory, so the second time you run the command grep search in it and output logfile.txt:0
If you can modify the pattern file, you write each line like ^765082023|
and you rename this file without .txt
So this command give you what you look for.
grep -scf pattern *.txt >>logfile
If you can't modify the pattern file, you can use awk.
awk -F'|' '
NR==FNR{a[$0];next}
FILENAME=="pattern.txt"{next}
$1 in a {b[FILENAME]++}
END{for(i in b){print i,":",b[i]}}
' pattern.txt *.txt >>logfile.txt
If I have two strings, for example "class" and "btn", what is the linux command that would allow me to search for these two strings in the entire directory.
To be more specific, lets say I have directory that contains few folders with bunch of .php files. My goal is to be able to search throughout those .php files so that it prints out only files that contain "class" and "btn" in one line. Hopefully this clarifies things better.
Thanks,
I normally use the following to search for strings inside my source codes. It searches for string and shows the exact line number where that text appears. Very helpful for searching string in source code files. You can always pipes the output to another grep and filter outputs.
grep -rn "text_to_search" directory_name/
example:
$ grep -rn "angular" menuapp
$ grep -rn "angular" menuapp | grep some_other_string
output would be:
menuapp/public/javascripts/angular.min.js:251://# sourceMappingURL=angular.min.js.map
menuapp/public/javascripts/app.js:1:var app = angular.module("menuApp", []);
grep -r /path/to/directory 'class|btn'
grep is used to search a string in a file. With the -r flag, it searches recursively all files in a directory.
Or, alternatively using the find command to "identify" the files to be searched instead of using grep in recursive mode:
find /path/to/your/directory -type f -exec grep "text_to_search" {} \+;
I have several header files in a directory with the format imageN.hd where N is some integer. Only one of these header files contains the text 'trans'. What I am trying to do is find which image contains this expression using csh (I need to use csh for this purpose - although I can call sed or perl one-liners) and show the corresponding image.
show iN
Here is my initial unsophisticated approach which does not work.
#find number of header files in directory
set n_images = `ls | grep 'image[0-9]*.hd' | wc -l`
foreach N(`seq 1 n_images`)
if (`more image$N{.hd} | grep -i 'trans`) then
show i$N
sc c image #this command uses an alias to set the displayed image as current within the script
endif
end
I'm not sure what is wrong with the above commands but it does not return the correct image number.
Also I'm sure there is a more elegant one line perl or sed solution but I am fairly unfamiliar with both
show `grep -l trans image[0-9]*.hd | sed 's/image/i/`