I have a small bash file which I intend to use to determine my current ping vs my average ping.
#!/bin/bash
output=($(ping -qc 1 google.com | tail -n 1))
echo "`cut -d/ -f1 <<< "${output[3]}"`-20" | bc
This outputs my ping - 20 ms, which is the number I want. However, I also want to prepend a + if the number is positive and append "ms".
This brings me to my overarching problem: Bash syntax regarding escaping and such heavy "indenting" is kind of flaky.
While I'll be satisfied with an answer of how to do what I wanted, I'd like a link to, or explanation of how exactly bash syntax works dealing with this sort of thing.
output=($(ping -qc 1 google.com | tail -n 1))
echo "${output[3]}" | awk -F/ '{printf "%+fms\n", $1-20}'
The + modifier in printf tells it to print the sign, whether it's positive or negative.
And since we're using awk, there's no need to use cut or bc to get a field or do arithmetic.
Escaping is pretty awful in bash if you use the deprecated `..` style command expansion. In this case, you have to escape any backticks, which means you also have to escape any other escapes. $(..) nests a lot better, since it doesn't add another layer of escaping.
In any case, I'd just do it directly:
ping -qc 1 google.com.org | awk -F'[=/ ]+' '{n=$6}
END { v=(n-20); if(v>0) printf("+"); print v}'
Here's my take on it, recognizing that the result from bc can be treated as a string:
output=($(ping -qc 1 google.com | tail -n 1))
output=$(echo "`cut -d/ -f1 <<< "${output[3]}"`-20" | bc)' ms'
[[ "$output" != -* ]] && output="+$output"
echo "$output"
Bash cannot handle floating point numbers. A workaround is to use awk like this:
#!/bin/bash
output=($(ping -qc 1 google.com | tail -n 1))
echo "`cut -d/ -f1 <<< "${output[3]}"`-20" | bc | awk '{if ($1 >= 0) printf "+%fms\n", $1; else printf "%fms\n", $1}'
Note that this does not print anything if the result of bc is not positive
Output:
$ ./testping.sh
+18.209000ms
Related
I want to create an array from a list of words. so, i'm using this code:
for i in {1..$count}
do
array[$i]=$(cat file.txt | cut -d',' -f3 | sort -r | uniq | tail -n ${i})
done
but it fails... in tail -n ${i}
I already tried tail -n $i, tail -n $(i) but can't pass tail the value of i
Any ideas?
It fails because you cannot use a variable in range directive in shell i.e. {1..10} is fine but {1..$n} is not.
While using BASH you can use ((...)) operators:
for ((i=1; i<=count; i++)); do
array[$i]=$(cut -d',' -f3 file.txt | sort -r | uniq | tail -n $i)
done
Also note removal of useless use of cat from your command.
Your range is not evaluated the way you are thinking, e.g.:
$ x=10
$ echo {1..$x}
{1..10}
You're better off just using a for loop:
for ((i = 1; i <= count; i++))
do
# ...
done
Just to elaborate on previous answers, this occurs because the 'brace expansion' is the first part of bash's parsing, and never gets repeated: when the braces are expanded, the '$count' is just a piece of text and so the braces are left as is. Then, when '$count' is expanded to a number, the brace expansion never runs again. See here.
If you wanted for some reason to force this brace expansion to happen again, you can use 'eval':
replace the {1..$count} with $(eval echo {1..${count}})
Better, in your case, to do as anubhava suggests.
Instead of reading the file numerous times, use the built-in mapfile command:
mapfile -t array < <(cut -d, -f3 file.txt | sort -r | uniq)
I have a String like this
//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf
and want to get last part of
00000000957481f9-08d035805a5c94bf
Let's say you have
text="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
If you know the position, i.e. in this case the 9th, you can go with
echo "$text" | cut -d'/' -f9
However, if this is dynamic and your want to split at "/", it's safer to go with:
echo "${text##*/}"
This removes everything from the beginning to the last occurrence of "/" and should be the shortest form to do it.
For more information on this see: Bash Reference manual
For more information on cut see: cut man page
The tool basename does exactly that:
$ basename //ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf
00000000957481f9-08d035805a5c94bf
I would use bash string function:
$ string="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
$ echo "${string##*/}"
00000000957481f9-08d035805a5c94bf
But following are some other options:
$ awk -F'/' '$0=$NF' <<< "$string"
00000000957481f9-08d035805a5c94bf
$ sed 's#.*/##g' <<< "$string"
00000000957481f9-08d035805a5c94bf
Note: <<< is herestring notation. They do not create a subshell, however, they are NOT portable to POSIX sh (as implemented by shells such as ash or dash).
In case you want more than just the last part of the path,
you could do something like this:
echo $PWD | rev | cut -d'/' -f1-2 | rev
You can use this BASH regex:
s='//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf'
[[ "$s" =~ [^/]+$ ]] && echo "${BASH_REMATCH[0]}"
00000000957481f9-08d035805a5c94bf
This can be done easily in awk:
string="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
echo "${string}" | awk -v FS="/" '{ print $NF }'
Use "/" as field separator and print the last field.
You can try this...
echo //ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf |awk -F "/" '{print $NF}'
I want to compare two string variables and print the characters that are the same for both. I'm not really sure how to do this, I was thinking of using comm or diff but I'm not really sure the right parameters to print only matching characters. also they say they take in files and these are strings. Can anyone help?
Input:
a=$(echo "abghrsy")
b=$(echo "cgmnorstuvz")
Output:
"grs"
You don't need to do that much work to assign $a and $b shell variables, you can just...
a=abghrsy
b=cdgmrstuvz
Now, there is a classic computer science problem called the longest common subsequence1 that is similar to yours.
However, if you just want the common characters, one way would let Ruby do the work...
$ ruby -e "puts ('$a'.chars.to_a & '$b'.chars.to_a).join"
1. Not to be confused with the different longest common substring problem.
Use Character Classes with GNU Grep
The isn't a widely-applicable solution, but it fits your particular use case quite well. The idea is to use the first variable as a character class to match against the second string. For example:
a='abghrsy'
b='cgmnorstuvz'
echo "$b" | grep --only-matching "[$a]" | xargs | tr --delete ' '
This produces grs as you expect. Note that the use of xargs and tr is simply to remove the newlines and spaces from the output; you can certainly handle this some other way if you prefer.
Set Intersection
What you're really looking for is a set intersection, though. While you can "wing it" in the shell, you'd be better off using a language like Ruby, Python, or Perl to do this.
A Ruby One-Liner
If you need to integrate with an existing shell script, a simple Ruby one-liner that uses Bash variables could be called like this inside your current script:
a='abghrsy'
b='cgmnorstuvz'
ruby -e "puts ('$a'.split(//) & '$b'.split(//)).join"
A Ruby Script
You could certainly make things more elegant by doing the whole thing in Ruby instead.
string1_chars = 'abghrsy'.split //
string2_chars = 'cgmnorstuvz'.split //
intersection = string1_chars & string2_chars
puts intersection.join
This certainly seems more readable and robust to me, but your mileage may vary. At least now you have some options to choose from.
Nice question +1.
You can use an awk trick to get this done.
a=abghrsy
b=cdgmrstuvz
comm -12 <(echo $a|awk -F"\0" '{for (i=1; i<=NF; i++) print $i}') <(echo $b|awk -F"\0" '{for (i=1; i<=NF; i++) print $i}')|tr -d '\n'
OUTPUT:
grs
Note use of awk -F"\0" that breaks input string character by character into different awk fiedls. Rest is pretty straightforward use of comm and tr.
PS: If you input string is not sorted then you need to pipe awk's output to sort or do sort of an array inside awk.
UPDATE: awk only solution (without comm):
echo "$a;$b" | awk -F"\0" '{scnd=0; for (i=1; i<=NF; i++) {if ($i!=";") {if (!scnd) arr1[$i]=$i; else if ($i in arr1) arr2[$i]=$i} else scnd=1}} END { for (a in arr2) printf("%s", a)}'
This assumes semicolon doesn't appear in your string (you can use any other character if that's not the case).
UPDATE 2: I think simplest solution is using grep -o
(thanks to answer from #CodeGnome)
echo "$b" | grep -o "[$a]" | tr -d '\n'
Using gnu coreutils(inspired by #DigitalRoss)..
a="abghrsy"
b="cgmnorstuvz"
echo "$(comm -12 <(echo "$a" | fold -w1 | sort | uniq) <(echo "$b" | fold -w1 | sort | uniq) | tr -d '\n')"
will print grs. I assumed you only want uniq characters.
UPDATE:
Modified for dash..
#!/bin/dash
string1=$(printf "$1" | fold -w1 | sort | uniq | tr -d '\n');
string2=$(printf "$2" | fold -w1 | sort | uniq | tr -d '\n');
while [ "$string1" != "" ]; do
c1=$(printf '%s\n' "$string1" | cut -c 1-1 )
string2=$(printf "$2" | fold -w1 | sort | uniq | tr -d '\n');
while [ "$string2" != "" ]; do
c2=$(printf '%s\n' "$string2" | cut -c 1-1 )
if [ "$c1" = "$c2" ]; then
echo "$c1\c"
fi
string2=$(printf '%s\n' "$string2" | cut -c 2- )
done
string1=$(printf '%s\n' "$string1" | cut -c 2- )
done
echo;
Note: I am just a beginner. There might be a better way of doing this.
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133
In string "12345", out string "54321". Preferably without third party tools and regex.
I know you said "without third-party tools", but sometimes a tool is just too obviously the right one, plus it's installed on most Linux systems by default:
[madhatta#risby tmp]$ echo 12345 | rev
54321
See rev's man page for more.
Simple:
var="12345"
copy=${var}
len=${#copy}
for((i=$len-1;i>=0;i--)); do rev="$rev${copy:$i:1}"; done
echo "var: $var, rev: $rev"
Output:
$ bash rev
var: 12345, rev: 54321
Presume that a variable 'var' has the value '123'
var="123"
Reverse the string and store in a new variable 'rav':
rav=$(echo $var | rev)
You'll see the 'rav' has the value of '321' using echo.
echo $rav
rev | tail -r (BSD) or rev | tac (GNU) also reverse lines:
$ rev <<< $'12\n34' | tail -r
43
21
$ rev <<< $'12\n34' | gtac
43
21
If LC_CTYPE is C, rev reverses the bytes of multibyte characters:
$ LC_CTYPE=C rev <<< あの
��め�
$ export LC_ALL=C; LC_ALL=en_US.UTF-8 rev <<< あの
のあ
A bash solution improving over #osdyng answer (my edit was not accepted):
var="12345" rev=""
for(( i=0 ; i<${#var} ; i++ )); do rev="${var:i:1}$rev"; done
echo "var: $var, rev: $rev"
Or an even simpler (bash) loop:
var=$1 len="${#var}" i=0 rev=""
while (( i<len )); do rev="${var:i++:1}$rev"; done
echo "var: $var, rev: $rev"
A POSIX solution:
var="12345" rev="" i=1
while [ "$i" -le "${#var}" ]
do rev="$(echo "$var" | awk -v i="$i" '{print(substr($0,i,1))}')$rev"
: $(( i+=1 ))
done
echo "var: $var, rev: $rev"
Note: This works on multi byte strings. Cut solutions will work only in ASCII (1 byte) strings.
Some simple methods of reversing a string
echo '!!!esreveR si sihT' | grep -o . | tac | tr -d '\n' ; echo
echo '!!!esreveR si sihT' | fold -w 1 | tac | tr -d '\n' ; echo
Convert to hex values then reverse
echo '!!!esreveR si sihT' | xxd -p | grep -o .. | tac | xxd -r -p ; echo
echo '!!!esreveR si sihT' | xxd -p | fold -w 2 | tac | xxd -r -p ; echo
This reverses the string "in place":
a=12345
len=${#a}
for ((i=1;i<len;i++)); do a=$a${a: -i*2:1}; done; a=${a:len-1}
echo $a
or the third line could be:
for ((i=0;i<len;i++)); do a=${a:i*2:1}$a; done; a=${a:0:len}
or
for ((i=1;i<len;i++)); do a=${a:0:len-i-1}${a: -i:i+1}${a:len-i-1:1}; done
For those without rev (recommended), there is the following simple awk solution that splits fields on the null string (every character is a separate field) and prints in reverse:
awk -F '' '{ for(i=NF; i; i--) printf("%c", $i); print "" }'
The above awk code is POSIX compliant. As a compliant awk implementation is guaranteed to be on every POSIX compliant OS, the solution should thus not be thought of as "3rd-party." This code will likely be more concise and understandable than a pure POSIX sh (or bash) solution.
(; I do not know if you consider the null string to -F a regex... ;)
If var=12345:
bash for((i=0;i<${#var};i++)); do rev="$rev${var:~i:1}"; done
sh c=$var; while [ "$c" ]; do rev=$rev${c#"${c%?}"}; c=${c%?}; done
echo "var: $var, rev: $rev"
Run it:
$ rev
var: 12345, rev: 54321
This can of course be shortened, but it should be simple to understand: the final print adds the newline.
echo 12345 | awk '{for (i = length($0); i > 0; i--) {printf("%s", substr($0, i, 1));} print "";}'
Nobody appears to have posted a sed solution, so here's one that works in non-GNU sed (so I wouldn't consider it "3rd party"). It does capture single characters using the regex ., but that's the only regex.
In two stages:
$ echo 123456 | sed $'s/./&\\\n/g' | sed -ne $'x;H;${x;s/\\n//g;p;}'
654321
This uses bash format substitution to include newlines in the scripts (since the question is tagged bash). It works by first separating the input string into one line per character, and then by inserting each character into the beginning of the hold buffer.
x swaps the hold space and the pattern space, and
H H appends the (current) pattern space to the hold space.
So for every character, we place that character into the hold space, then append the old hold space to it, thus reversing the input. The final command removes the newlines in order to reconstruct the original string.
This should work for any single string, but it will concatenate multi-line input into a single output string.
Here is another simpler awk solution:
awk 'BEGIN{FS=""} {for (i=NF; i>0; i--) s=s $i; print s}' <<< '123456'
654321
Try Perl:
echo 12345 | perl -nle 'print scalar reverse $_'
Source: Perl one-liners
read word
reve=`echo "$word" | awk '{for(i=length($0); i>0;i--) printf (substr($0,i,1));}'`
echo "$reve"