I'm trying to make a list with a simple bash looping
I want this:
000000
000001
000002
They give me this:
0
1
2
My shell code:
countBEG="000000"
countEND="999999"
while [ $countBEG != $countEND ]
do
echo "$countBEG"
countBEG=$[$countBEG +1]
done
Change your echo to use printf, where you can specify format for left padding.
printf "%06d\n" "$countBEG"
This sets 6 as fixed length of the output, using zeros to fill empty spaces.
You're looking for:
seq -w "$countBEG" "$countEND"
The -w option does the padding.
The following command will produce the desired output (no need for the loop) :
printf '%06d\n' {1..999999}
Explanation :
{1..999999} is expanded by bash to the sequence of 1 to 999999
the format string '%06d\n' tells printf to display the number it is given as argument padded to 6 digits and followed by a linefeed
printf repeats this output if it is given more arguments than is defined in its format specification
Related
I have a list of files with file names that contain a substring of 6 numbers that represents HHMMSS, HH: 2 digits hour, MM: 2 digits minutes, SS: 2 digits seconds.
If the list of files is ordered, the increments should be in steps of 30 minutes, that is, the first substring should be 000000, followed by 003000, 010000, 013000, ..., 233000.
I want to check that no file is missing iterating the list of files and checking that neither of these substrings is missing. My approach:
string_check=000000
for file in ${file_list[#]}; do
if [[ ${file:22:6} == $string_check ]]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi
string_check=$((string_check+3000)) #this is the key line
done
And the previous to the last line is the key. It should be formatted to 6 digits, I know how to do that, but I want to add time like a clock, or, in more specific words, modular arithmetic modulo 60. How can that be done?
Assumptions:
all 6-digit strings are of the format xx[03]0000 (ie, has to be an even 00 or 30 minutes and no seconds)
if there are strings like xx1529 ... these will be ignored (see 2nd half of answer - use of comm - to address OP's comment about these types of strings being an error)
Instead of trying to do a bunch of mod 60 math for the MM (minutes) portion of the string, we can use a sequence generator to generate all the desired strings:
$ for string_check in {00..23}{00,30}00; do echo $string_check; done
000000
003000
010000
013000
... snip ...
230000
233000
While OP should be able to add this to the current code, I'm thinking we might go one step further and look at pre-parsing all of the filenames, pulling the 6-digit strings into an associative array (ie, the 6-digit strings act as the indexes), eg:
unset myarray
declare -A myarray
for file in ${file_list}
do
myarray[${file:22:6}]+=" ${file}" # in case multiple files have same 6-digit string
done
Using the sequence generator as the driver of our logic, we can pull this together like such:
for string_check in {00..23}{00,30}00
do
[[ -z "${myarray[${string_check}]}" ]] &&
echo "Problem: (file) '${string_check}' is missing"
done
NOTE: OP can decide if the process should finish checking all strings or if it should exit on the first missing string (per OP's current code).
One idea for using comm to compare the 2 lists of strings:
# display sequence generated strings that do not exist in the array:
comm -23 <(printf "%s\n" {00..23}{00,30}00) <(printf "%s\n" "${!myarray[#]}" | sort)
# OP has commented that strings not like 'xx[03]000]` should generate an error;
# display strings (extracted from file names) that do not exist in the sequence
comm -13 <(printf "%s\n" {00..23}{00,30}00) <(printf "%s\n" "${!myarray[#]}" | sort)
Where:
comm -23 - display only the lines from the first 'file' that do not exist in the second 'file' (ie, missing sequences of the format xx[03]000)
comm -13 - display only the lines from the second 'file' that do not exist in the first 'file' (ie, filenames with strings not of the format xx[03]000)
These lists could then be used as input to a loop, or passed to xargs, for additional processing as needed; keeping in mind the comm -13 output will display the indices of the array, while the associated contents of the array will contain the name of the original file(s) from which the 6-digit string was derived.
Doing this easy with POSIX shell and only using built-ins:
#!/usr/bin/env sh
# Print an x for each glob matched file, and store result in string_check
string_check=$(printf '%.0sx' ./*[0-2][0-9][03]000*)
# Now string_check length reflects the number of matches
if [ ${#string_check} -eq 48 ]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi
Alternatively:
#!/usr/bin/env sh
if [ "$(printf '%.0sx' ./*[0-2][0-9][03]000*)" \
= 'xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx' ]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi
I need to zero-pad a sequence of numbers in a loop in Bash. I know how to do it with
seq -f "%03g" 5
or the comparable printf approach, also
for index in {003..006}
The problem I did not find an answer to is that I need the number of digits to be a variable:
read CNT
seq -f "%0$CNTd" 3 6
Will return an error
seq: das Format »%0“ endet mit %
I have not found any way to insert a variable in a format string or any other way to produce a zero-padded sequence where the number of digits comes from a (user-provided) variable.
I think you want seq, but did you know the * operator in printf?
printf "%0*d\n" ${CNT} 5
A variable name (CNT) should be enclosed in curly braces when it is followed by a character (d) which is not to be interpreted as part of its name,
seq doesn't support %d, you should use %g.
$ read -r CNT
$ seq -f "%0${CNT}g" 3 6
00003
00004
00005
00006
I'm a bit stuck with something. I have a for loop like this:
#!/bin/bash
for i in {10..15}
do
I want to obtain the last digit of the number, so if i is 12, I want to get 2. I'm having difficulties with the syntax though. I've read that I should convert it into a character array, but when I do something like:
j=${i[#]}
echo $j
I don't get 1 0 1 1 1 2 and so on...I get 10, 11, 12...How do I get the numbers to be split up so I can get the last one of i, when I don't always know how many digits will make up i (ex. it may be 1, or 10, or a 100, etc.)?
Trick is to treat $i like a string.
for i in {10..15}; do j="${i: -1}"; echo $j; done
Of course, you do not need to assign to a variable if you don't want to:
for i in {10..15}; do echo "${i: -1}"; done
This answer which uses GNU shell parameter expansion is the most sensible method, I guess.
However, you can also use the double parenthesis construct which allows C-style manipulation of variables in Bash.
for i in {10..15}
do
(( j = i % 10 )) # modulo 10 always gives the ones' digit
echo $j
done
This awk command could solve your problem:
awk '{print substr($0,length,1)}' test_file
I'm assuming that the numbers are saved in a file test_file
If you want to use for loop:
for i in `cat test_1`
do
echo $i |tail -c 2
done
I am trying to create a Bash script that
- prints a random word
- if a number is supplied as the first command line argument then it will select from only words with that many characters.
This is my go at the first section (print a random word):
C=$(sed -n "$RANDOM p" /usr/share/dict/words)
echo $C
I am really stuck with the second section. Can anyone help?
might help someone coming from ryans tutorial
#!/bin/bash
charlen=$1
grep -E "^.{$charlen}$" $PWD/words.txt | shuf -n 1
you have to use a while loop to read every single line of that file and check if the length of a word equals the specified number ( including apostrophes ). In my o.s it is 99171 line ( i.e the file).
#!/usr/bin/env bash
readWords() {
declare -i int="$1"
(( int == 0 )) && {
printf "%s\n" "$int is 0, cant find 0 words"
return 1
}
while read getWords;do
if [[ ${#getWords} -eq $int ]];then
printf "%s\n" "$getWords"
fi
done < /usr/share/dict/words
}
readWords 20
this function takes a single argument. the declare command coerces the argument into an integer, if the argument is a string , it coerces it into a number which is 0 . Since we don't have 0 words if the specified argument ( number ) is 0 ( or a string coerced to 0 ) return from the function.
Read every single line in /usr/share/dict/words, get the length of each line with ${#getWords} ( $# >> gives the length of a string/commandline parameters/array size ) check if it equals the specified argument ( number )
A loop is not required, you can do something like
CH=$1; # how many characters the word must have
WordFile=/usr/share/dict/words; # file to read from
# find how many words that matches that length
TOTW=$(grep -Ec "^.{$CH}$" $WordFile);
# pick a random one, if you expect more than 32767 hits you
# need to do something like ($RANDOM+1)*($RANDOM+1)
RWORD=$(($RANDOM%$TOTW+1));
#show that word
grep -E "^.{$CH}$" $WordFile|sed -n "$RWORD p"
Depending on things you probably need to add checks for things like that $1 is a reasonable number, the file exist, that TOTW is >0 and so on.
This code would achieve what you want:
awk -v n="$1" 'length($0) == n' /usr/share/dict/words > /tmp/wordsHolder
shuf -n 1 /tmp/wordsHolder
Some comments: by using "$RANDOM" (as you did on your original script attempt), one would generate an integer on the range 0 - 32767, which could be more (or less) than the number of words (lines) available, given the desired number of characters on a word -- thus, potential for errors here.
To avoid that, we are using a shuf syntax that will retrieve a (sub)randomly picked word (line) on the file using its entire range (from line 1 - last line of file).
I'm working with an existing script which was written a bit messily. Setting up a loop with all of the spaghetti code could make a bigger headache than I want to deal with in the near term. Maybe when I have more time I can clean it up but for now, I'm just looking for a simple fix.
The script deals with virtual disks on a xen server. It reads multipath output and asks if particular LUNs should be formatted in any way based on specific criteria. However, rather than taking that disk path and inserting it, already formatted, into a configuration file, it simply presents every line in the format
'phy:/dev/mapper/UUID,xvd?,w',
UUID, of course, is an actual UUID.
The script actually presents each of the found LUNs in this format expecting the user to copy and paste them into the config file replacing each ? with a letter in sequence. This is tedious at best.
There are several ways to increment a number in bash. Among others:
var=$((var+1))
((var+=1))
((var++))
Is there a way to do the same with characters which doesn't involve looping over the entire alphabet such that I could easily "increment" the disk assignment from xvda to xvdb, etc?
To do an "increment" on a letter, define the function:
incr() { LC_CTYPE=C printf "\\$(printf '%03o' "$(($(printf '%d' "'$1")+1))")"; }
Now, observe:
$ echo $(incr a)
b
$ echo $(incr b)
c
$ echo $(incr c)
d
Because, this increments up through ASCII, incr z becomes {.
How it works
The first step is to convert a letter to its ASCII numeric value. For example, a is 97:
$ printf '%d' "'a"
97
The next step is to increment that:
$ echo "$((97+1))"
98
Or:
$ echo "$(($(printf '%d' "'a")+1))"
98
The last step is convert the new incremented number back to a letter:
$ LC_CTYPE=C printf "\\$(printf '%03o' "98")"
b
Or:
$ LC_CTYPE=C printf "\\$(printf '%03o' "$(($(printf '%d' "'a")+1))")"
b
Alternative
With bash, we can define an associative array to hold the next character:
$ declare -A Incr; last=a; for next in {b..z}; do Incr[$last]=$next; last=$next; done; Incr[z]=a
Or, if you prefer code spread out over multiple lines:
declare -A Incr
last=a
for next in {b..z}
do
Incr[$last]=$next
last=$next
done
Incr[z]=a
With this array, characters can be incremented via:
$ echo "${Incr[a]}"
b
$ echo "${Incr[b]}"
c
$ echo "${Incr[c]}"
d
In this version, the increment of z loops back to a:
$ echo "${Incr[z]}"
a
How about an array with entries A-Z assigned to indexes 1-26?
IFS=':' read -r -a alpharray <<< ":A:B:C:D:E:F:G:H:I:J:K:L:M:N:O:P:Q:R:S:T:U:V:W:X:Y:Z"
This has 1=A, 2=B, etc. If you want 0=A, 1=B, and so on, remove the first colon.
IFS=':' read -r -a alpharray <<< "A:B:C:D:E:F:G:H:I:J:K:L:M:N:O:P:Q:R:S:T:U:V:W:X:Y:Z"
Then later, where you actually need the letter;
var=$((var+1))
'phy:/dev/mapper/UUID,xvd${alpharray[$var]},w',
The only problem is that if you end up running past 26 letters, you'll start getting blanks returned from the array.
Use a Bash 4 Range
You can use a Bash 4 feature that lets you specify a range within a sequence expression. For example:
for letter in {a..z}; do
echo "phy:/dev/mapper/UUID,xvd${letter},w"
done
See also Ranges in the Bash Wiki.
Here's a function that will return the next letter in the range a-z. An input of 'z' returns 'a'.
nextl(){
((num=(36#$(printf '%c' $1)-9) % 26+97));
printf '%b\n' '\x'$(printf "%x" $num);
}
It treats the first letter of the input as a base 36 integer, subtracts 9, and returns the character whose ordinal number is 'a' plus that value mod 26.
Use Jot
While the Bash range option uses built-ins, you can also use a utility like the BSD jot utility. This is available on macOS by default, but your mileage may vary on Linux systems. For example, you'll need to install athena-jot on Debian.
More Loops
One trick here is to pre-populate a Bash array and then use an index variable to grab your desired output from the array. For example:
letters=( "" $(jot -w %c 26 a) )
for idx in 1 26; do
echo ${letters[$idx]}
done
A Loop-Free Alternative
Note that you don't have to increment the counter in a loop. You can do it other ways, too. Consider the following, which will increment any letter passed to the function without having to prepopulate an array:
increment_var () {
local new_var=$(jot -nw %c 2 "$1" | tail -1)
if [[ "$new_var" == "{" ]]; then
echo "Error: You can't increment past 'z'" >&2
exit 1
fi
echo -n "$new_var"
}
var="c"
var=$(increment_var "$var")
echo "$var"
This is probably closer to what the OP wants, but it certainly seems more complex and less elegant than the original loop recommended elsewhere. However, your mileage may vary, and it's good to have options!