I want to sort URLs based on path/levels. How can I do it ? (It can be a online tool/PHP/bash/linux script), Anything is okay.
example.com/a
example.com/b
example.com/c
example.com/aaa/1
example.com/aaa/2
example.com/aaa/3
example.com/aaa/bbb/apath
example.com/aaa/bbb/bpath
example.com/aaa/bbb/cpath
example.com/aaa/bbb/path/1
example.com/aaa/bbb/path/2
I echo'd your inputs, reordered, into a loop.
I assigned the input line to a second key, striped the non-delimiters, and prepended the key, then printed the new line.
I piped the loop to a sort, and piped the sorted output to a sed to strip off the prepended key.
I haven't used the ${x///} construct much, so there's probably a cleaner way to do it than this conversion to underscores, and I'm looking forward to seeing it, lol
$: echo "example.com/a
example.com/c
example.com/aaa/bbb/path/1
example.com/aaa/3
example.com/aaa/bbb/apath
example.com/aaa/2
example.com/aaa/bbb/bpath
example.com/aaa/1
example.com/aaa/bbb/cpath
example.com/b
example.com/aaa/bbb/path/2" |
while read line
do key=${line//\//_}
key=${key//[^_]/}
printf "$key\t$line\n"
done | sort |sed 's/^_* //'
example.com/a
example.com/b
example.com/c
example.com/aaa/1
example.com/aaa/2
example.com/aaa/3
example.com/aaa/bbb/apath
example.com/aaa/bbb/bpath
example.com/aaa/bbb/cpath
example.com/aaa/bbb/path/1
example.com/aaa/bbb/path/2
Related
Input:-
echo "1234ABC89,234" # A
echo "0520001DEF78,66" # B
echo "46545455KRJ21,00"
From the above strings, I need to split the characters to get the alphabetic field and the number after that.
From "1234ABC89,234", the output should be:
ABC
89,234
From "0520001DEF78,66", the output should be:
DEF
78,66
I have many strings that I need to split like this.
Here is my script so far:
echo "1234ABC89,234" | cut -d',' -f1
but it gives me 1234ABC89 which isn't what I want.
Assuming that you want to discard leading digits only, and that the letters will be all upper case, the following should work:
echo "1234ABC89,234" | sed 's/^[0-9]*\([A-Z]*\)\([0-9].*\)/\1\n\2/'
This works fine with GNU sed (I have 4.2.2), but other sed implementations might not like the \n, in which case you'll need to substitute something else.
Depending on the version of sed you can try:
echo "0520001DEF78,66" | sed -E -e 's/[0-9]*([A-Z]*)([,0-9]*)/\1\n\2/'
or:
echo "0520001DEF78,66" | sed -E -e 's/[0-9]*([A-Z]*)([,0-9]*)/\1$\2/' | tr '$' '\n'
DEF
78,66
Explanation: the regular expression replaces the input with the expected output, except instead of the new-line it puts a "$" sign, that we replace to a new-line with the tr command
Where do the strings come from? Are they read from a file (or other source external to the script), or are they stored in the script? If they're in the script, you should simply reformat the data so it is easier to manage. Therefore, it is sensible to assume they come from an external data source such as a file or being piped to the script.
You could simply feed the data through sed:
sed 's/^[0-9]*\([A-Z]*\)/\1 /' |
while read alpha number
do
…process the two fields…
done
The only trick to watch there is that if you set variables in the loop, they won't necessarily be visible to the script after the done. There are ways around that problem — some of which depend on which shell you use. This much is the same in any derivative of the Bourne shell.
You said you have many strings like this, so I recommend if possible save them to a file such as input.txt:
1234ABC89,234
0520001DEF78,66
46545455KRJ21,00
On your command line, try this sed command reading input.txt as file argument:
$ sed -E 's/([0-9]+)([[:alpha:]]{3})(.+)/\2\t\3/g' input.txt
ABC 89,234
DEF 78,66
KRJ 21,00
How it works
uses -E for extended regular expressions to save on typing, otherwise for example for grouping we would have to escape \(
uses grouping ( and ), searches three groups:
firstly digits, + specifies one-or-more of digits. Oddly using [0-9] results in an extra blank space above results, so use POSIX class [[:digit:]]
the next is to search for POSIX alphabetical characters, regardless if lowercase or uppercase, and {3} specifies to search for 3 of them
the last group searches for . meaning any character, + for one or more times
\2\t\3 then returns group 2 and group 3, with a tab separator
Thus you are able to extract two separate fields per line, just separated by tab, for easier manipulation later.
I have a requirement where i need to fetch first four characters from each line of file and sort them.
I tried below way. but its not sorting each line
cut -c1-4 simple_file.txt | sort -n
O/p using above:
appl
bana
uoia
Expected output:
alpp
aabn
aiou
sort isn't the right tool for the job in this case, as it used to sort lines of input, not the characters within each line.
I know you didn't tag the question with perl but here's one way you could do it:
perl -F'' -lane 'print(join "", sort #F[0..3])' file
This uses the -a switch to auto-split each line of input on the delimiter specified by -F (in this case, an empty string, so each character is its own element in the array #F). It then sorts the first 4 characters of the array using the standard string comparison order. The result is joined together on an empty string.
Try defining two helper functions:
explodeword () {
test -z "$1" && return
echo ${1:0:1}
explodeword ${1:1}
}
sortword () {
echo $(explodeword $1 | sort) | tr -d ' '
}
Then
cut -c1-4 simple_file.txt | while read -r word; do sortword $word; done
will do what you want.
The sort command is used to sort files line by line, it's not designed to sort the contents of a line. It's not impossible to make sort do what you want, but it would be a bit messy and probably inefficient.
I'd probably do this in Python, but since you might not have Python, here's a short awk command that does what you want.
awk '{split(substr($0,1,4),a,"");n=asort(a);s="";for(i=1;i<=n;i++)s=s a[i];print s}'
Just put the name of the file (or files) that you want to process at the end of the command line.
Here's some data I used to test the command:
data
this
is a
simple
test file
a
of
apple
banana
cat
uoiea
bye
And here's the output
hist
ais
imps
estt
a
fo
alpp
aabn
act
eiou
bey
Here's an ugly Python one-liner; it would look a bit nicer as a proper script rather than as a Bash command line:
python -c "import sys;print('\n'.join([''.join(sorted(s[:4])) for s in open(sys.argv[1]).read().splitlines()]))"
In contrast to the awk version, this command can only process a single file, and it reads the whole file into RAM to process it, rather than processing it line by line.
For some reason I cannot get this to output just the version of this line. I suspect it has something to do with how grep interprets the dash.
This command:
admin#DEV:~/TEMP$ sendemail
Yields the following:
sendemail-1.56 by Brandon Zehm
More output below omitted
The first line is of interest. I'm trying to store the version to variable.
TESTVAR=$(sendemail | grep '\s1.56\s')
Does anyone see what I am doing wrong? Thanks
TESTVAR is just empty. Even without TESTVAR, the output is empty.
I just tried the following too, thinking this might work.
sendemail | grep '\<1.56\>'
I just tried it again, while editing and I think I have another issue. Perhaps im not handling the output correctly. Its outputting the entire line, but I can see that grep is finding 1.56 because it highlights it in the line.
$ TESTVAR=$(echo 'sendemail-1.56 by Brandon Zehm' | grep -Eo '1.56')
$ echo $TESTVAR
1.56
The point is grep -Eo '1.56'
from grep man page:
-E, --extended-regexp
Interpret PATTERN as an extended regular expression (ERE, see below). (-E is specified by POSIX.)
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output
line.
Your regular expression doesn't match the form of the version. You have specified that the version is surrounded by spaces, yet in front of it you have a dash.
Replace the first \s with the capitalized form \S, or explicit set of characters and it should work.
I'm wondering: In your example you seem to know the version (since you grep for it), so you could just assign the version string to the variable. I assume that you want to obtain any (unknown) version string there. The regular expression for this in sed could be (using POSIX character classes):
sendemail |sed -n -r '1 s/sendemail-([[:digit:]]+\.[[:digit:]]+).*/\1/ p'
The -n suppresses the normal default output of every line; -r enables extended regular expressions; the leading 1 tells sed to only work on line 1 (I assume the version appears in the first line). I anchored the version number to the telltale string sendemail- so that potential other numbers elsewhere in that line are not matched. If the program name changes or the hyphen goes away in future versions, this wouldn't match any longer though.
Both the grep solution above and this one have the disadvantage to read the whole output which (as emails go these days) may be long. In addition, grep would find all other lines in the program's output which contain the pattern (if it's indeed emails, somebody might discuss this problem in them, with examples!). If it's indeed the first line, piping through head -1 first would be efficient and prudent.
jayadevan#jayadevan-Vostro-2520:~$ echo $sendmail
sendemail-1.56 by Brandon Zehm
jayadevan#jayadevan-Vostro-2520:~$ echo $sendmail | cut -f2 -d "-" | cut -f1 -d" "
1.56
Say I have this text file. I wanted to grab the lines that have a movie in them, match the title, print, and sort based on title. I got most of the way with:
File.open("features/sort_movie_list.feature","r").each {|line| puts [$1] if line =~ /\|\s+([A-Z0-9][a-zA-Z0-9: ]+)\s+\|/}
but I didn't know how to sort from there. I got some of the way with grep:
egrep -o "\|\s([A-Z0-9][A-Za-z0-9: ]+)\s+\|" sort_movie_list.feature
but wasn't sure how to print the bracketed match only. How would you do it so that the output is as below?
2001: A Space Odyssey
Aladdin
Amelie
Chicken Run
...
In your ruby example you could just stuff them into an array, sort the array and print them out.
With egrep you could pipe the results to sort:
egrep -o "\|\s([A-Z0-9][A-Za-z0-9: ]+)\s+\|" sort_movie_list.feature | sort
The given regexps do not catch Movie titles with comma (','), dots ('.') or other weird stuff (like german Umlauts 'ä', exclamation ("Mars Attacks!"!!)) inside.
Therefore, I would exploit the given format and assume that the pipe shall not come up in the movie title list.
Therefore, either cut the file in parts, where the pipe ("|") is the field delimiter and chose the second field, like:
grep "|" movies.txt | cut -f 2 -d"|" | sort
(The grep "|" is to omit lines without movies, otherwise cut -f 2 -d "|" movies.txt would do).
or in ruby use something along the lines
line.split("|")[1]
Note that in both cases you can get rid of whitespaces afterwards.
Ok, I've figured it out for Ruby now. Thanks to David for the suggesting to use an array.
movie_titles = []
File.open("features/sort_movie_list.feature","r").each {|line| movie_titles.push($1.strip) if line =~ /\|\s+([A-Z0-9][a-zA-Z0-9: ]+)\s+\|/}
puts movie_titles.sort
If someone has a more efficient/succinct answer, I'm always open to learning more.
I've never used sed apart from the few hours trying to solve this. I have a config file with parameters like:
test.us.param=value
test.eu.param=value
prod.us.param=value
prod.eu.param=value
I need to parse these and output this if REGIONID is US:
test.param=value
prod.param=value
Any help on how to do this (with sed or otherwise) would be great.
This works for me:
sed -n 's/\.us\././p'
i.e. if the ".us." can be replaced by a dot, print the result.
If there are hundreds and hundreds of lines it might be more efficient to first search for lines containing .us. and then do the string replacement... AWK is another good choice or pipe grep into sed
cat INPUT_FILE | grep "\.us\." | sed 's/\.us\./\./g'
Of course if '.us.' can be in the value this isn't sufficient.
You could also do with with the address syntax (technically you can embed the second sed into the first statement as well just can't remember syntax)
sed -n '/\(prod\|test\).us.[^=]*=/p' FILE | sed 's/\.us\./\./g'
We should probably do something cleaner. If the format is always environment.region.param we could look at forcing this only to occur on the text PRIOR to the equal sign.
sed -n 's/^\([^,]*\)\.us\.\([^=]\)=/\1.\2=/g'
This will only work on lines starting with any number of chars followed by '.' then 'us', then '.' and then anynumber prior to '=' sign. This way we won't potentially modify '.us.' if found within a "value"