I need to automate a backup solution which stores files in folders such as YYYYMMDD.nn.
Every day few files would be backed up like this so the resulting folder names could be 20141002.01, 20141002.2 ... 20141002.10. My current script works for YYYYMMDD.n but when n is more than 9 sorting and picking up the last folder doesn't work because 20141002.10 is above 20141002.9 hens switching to YYYYMMDD.nn format and the approach of separating the nn, stripping leading zeros, then incrementing, and adding leading zeros if needed.
I have a function which checks the last folder for today's date and creates the next one.
createNextProcessedFolder() {
local LastFolderName=`ls -1 ${ProcessedListsDir} | grep ${CurrentDate} | tail -n 1`
n=`echo ${LastFolderName} | sed -r 's/^.{9}//'`
n="$((10#$n))"
nextFolderName=${CurrentDate}.$((if[[ $(( ${n}+1 )) < 10 ]];then n="0$((${n}+1))";else n="$(( ${n}+1 ))"; fi))
mkdir ${ProcessedListsDir}/${nextFolderName}
if [[ -d ${ProcessedListsDir}/${nextFolderName} ]]
then
echo "New folder ${nextFolderName} was created"
else
echo "Error: ${nextFolderName} was not created"
fi
Location="${ProcessedListsDir}/${nextFolderName}"
}
So when I try to run this I get an error like:
line 21: if[[ 1 < 10 ]];then n="01";else n="1"; fi: syntax error: invalid arithmetic operator (error token is ";then n="01";else n="1"; fi")
Line 21 is:
nextFolderName=${CurrentDate}.$((if[[ $(( ${n}+1 )) < 10 ]];then n="0$((${n}+1))";else n="$(( ${n}+1 ))"; fi))
I'm sure there will be more errors after this one but I would really appreciate if somebody helped me with this.
You cannot use $((...)) for command substitution as it needs to be $(...)
You need spaces before and after [[ and ]]. You can also use ((...)) in BASH:
Try this:
(( (n+1) < 10 )) && n="0$((n++))" || ((n++))
nextFolderName="${CurrentDate}.${n}"
For completeness, another solution is:
n=$( printf "%02d" $n )
The 02 before the d means prepend with 0s up to 2 digits. Or:
nextFolderName="${CurrentDate}."$( printf "%02d" "$n" )
So my problem was with incrementing a number witch was extracted from a string with a leading zero and then returning the incremented number with a leading zero if smaller than 10. The solution I ended up using can be represented with the below script.
I guess it can't be shorter than that
n=$1
(( ((n++)) < 10 )) && n="0$n"
echo $n
Something I didn't expect is that I don't have to strip leading zeros from n using this, n++ does it while incrementing :-)
Thanks again anubhava for pointing me in the right direction.
Related
I have created a hex to ASCII converter for strings in bash. The application I'm on changes characters (anything but [0-9],[A-Z],[a-z]) , in a string to its corresponding %hexadecimal. Eg: / changes to %2F in a string
I want to retain the ASCII characters as it is. Below is my code:
NAME=%2fhome%40%21%23
C_NAME=""
for (( i=0; i<${#NAME}; i++ )); do
CHK=$(echo "{NAME:$i:1}" | grep -v "\%" &> /dev/null;echo $?)
if [[ ${CHK} -eq 0 ]]; then
C_NAME=`echo "$C_NAME${NAME:$i:1}"`
else
HEX=`echo "${NAME:$i:3}" | sed "s/%//"`
C_NAME=`echo -n "$C_NAME";printf "\x$HEX"`
continue 2
fi
done
echo "$C_NAME"
OUTPUT:
/2fhome#40!21#23
EXPECTED:
/home#!#
So basically the conversion is happening, but not in place. Its retaining the hex values as well, which tells me the continue 2 statement is probably not working as I expect in my code. Any workarounds please.
You only have one loop so I assume you expected that continue 2 skips the current and next iteration of the current loop, however, the documentation help continue clearly states
continue [n]
[...]
If N is specified, resumes the Nth enclosing loop.
There is no built-in to skip the current and also the next iteration of the current loop, but in your case you can use (( i += 2 )) instead of continue 2.
Using the structure of your script with some simplifications and corrections:
#!/bin/bash
name=%2fhome%40%21%23
c_name=""
for (( i=0; i<${#name}; i++ )); do
c=${name:i:1}
if [[ $c != % ]]; then
c_name=$c_name$c
else
hex=${name:i+1:2}
printf -v c_name "%s\x$hex" "$c_name"
(( i += 2 )) # stolen from Dudi Boy's answer
fi
done
echo "$c_name"
Always use lower case or mixed case variables to avoid the chance of name collisions with shell or environment variables
Always use $() instead of backticks
Most of the echo commands you use aren't necessary
You can avoid using sed and grep
Variables should never be included in the format string of printf but it can't be avoided easily here (you could use echo -e "\x$hex" instead though)
You can do math inside parameter expansions
% doesn't need to be escaped in your grep command
You could eliminate the $hex variable if you used its value directly:
printf -v c_name "%s\x${name:i+1:2}" "$c_name"
I really enjoyed your exercise and decided to solve it with awk (my current study).
Hope you like it as well.
cat script.awk
BEGIN {RS = "%[[:xdigit:]]+"} { # redefine record separtor to RegEx (gawk specific)
decNum = strtonum("0x"substr(RT, 2)); # remove prefix # from record separator, convert hex num to dec
outputStr = outputStr""$0""sprintf("%c", decNum); # reconstruct output string
}
END {print outputStr}
The output
echo %2fhome%40%21%23 |awk -f script.awk
/home#!#
Write a script that expects a file as its first argument. Some lines of the
file will consist of integers 0 - 1000.
The script should select the lines matching the previous criteria and print out their average to stdout (average of n integers is their sum divided by n).
And the file given looks like this:
22
78907
77 88 99 0000
need 11 gallons of water
0
roses are red
11
Example output:
11
Explanation: (22 + 11 + 0) / 3 = 11
I have tried already with this code:
#!/bin/bash
sum=0
ind=0
while IFS='' read -r line || [[ -n "$line" ]]; do
if [[ $line =~ ^[a-zA-Z\ ]+$ ]]
then
${sum}=${sum}+${#line}
${ind}=${ind}+1
echo ${sum}
fi
done < "$1"
value=${sum}/${ind}
echo ${value}
the print of this code is always 0/0 and some errors like:
./test1: line 9: 0=0+13: command not found
./test1: line 10: 0=0+1: command not found
Any ideas?
Part of the issue with your script is answered here.. Your variable assignments are incorrect. You only use the $ to refer to a variable that has already been assigned. The assignment process drops the dollar sign.
The other issue you're having is that your arithmetic is not being expressed within an arithmetic expression.
Note that you can use use arithmetic expansion to handle your variables:
if [[ $line =~ ^[a-zA-Z\ ]+$ ]]; then
(( sum += ${#line} ))
(( ind++ ))
printf '%s\n' "$sum"
fi
and later ...
value="$(( sum / ind ))"
printf '%s\n' "$value"
Beware that bash can only deal with integer math, floats are truncated. For more advanced math, consider using bc or dc (which are not built in to bash, they are separate tools that may need to be installed on your system) or another language like awk or perl which can do the same thing with better performance and more precise math.
That said, you can "fake" a couple of decimal places with a few extra lines of code and string manipulation, if you really need to:
$ sum=100; ind=7
$ printf -v x '%d' "$((${sum}00/${ind}))"
$ printf '%d.%d\n' "${x%??}" "${x:$((${#x}-2))}"
14.28
The first printf has division which multiplies the dividend by 100 (by adding two zeroes after it). The resultant quotient is then split with the second printf to insert the decimal point. This is a hack. Use tools that support real math.
OK so Ive been at this for a couple days,im new to this whole bash UNIX system thing i just got into it but I am trying to write a script where the user inputs an integer and the script will take that integer and print out a triangle using the integer that was inputted as a base and decreasing until it reaches zero. An example would be:
reverse_triangle.bash 4
****
***
**
*
so this is what I have so far but when I run it nothing happens I have no idea what is wrong
#!/bin/bash
input=$1
count=1
for (( i=$input; i>=$count;i-- ))
do
for (( j=1; j>=i; j++ ))
do
echo -n "*"
done
echo
done
exit 0
when I try to run it nothing happens it just goes to the next line. help would be greatly appreciated :)
As I said in a comment, your test is wrong: you need
for (( j=1; j<=i; j++ ))
instead of
for (( j=1; j>=i; j++ ))
Otherwise, this loop is only executed when i=1, and it becomes an infinite loop.
Now if you want another way to solve that, in a much better way:
#!/bin/bash
[[ $1 = +([[:digit:]]) ]] || { printf >&2 'Argument must be a number\n'; exit 1; }
number=$((10#$1))
for ((;number>=1;--number)); do
printf -v spn '%*s' "$number"
printf '%s\n' "${spn// /*}"
done
Why is it better? first off, we check that the argument is really a number. Without this, your code is subject to arbitrary code injection. Also, we make sure that the number is understood in radix 10 with 10#$1. Otherwise, an argument like 09 would raise an error.
We don't really need an extra variable for the loop, the provided argument is good enough. Now the trick: to print n times a pattern, a cool method is to store n spaces in a variable with printf: %*s will expand to n spaces, where n is the corresponding argument found by printf.
For example:
printf '%s%*s%s\n' hello 42 world
would print:
hello world
(with 42 spaces).
Editor's note: %*s will NOT generally expand to n spaces, as evidenced by above output, which contains 37 spaces.
Instead, the argument that * is mapped to,42, is the field width for the sfield, which maps to the following argument,world, causing string world to be left-space-padded to a length of 42; since world has a character count of 5, 37 spaces are used for padding.
To make the example work as intended, use printf '%s%*s%s\n' hello 42 '' world - note the empty string argument following 42, which ensures that the entire field is made up of padding, i.e., spaces (you'd get the same effect if no arguments followed 42).
With printf's -v option, we can store any string formatted by printf into a variable; here we're storing $number spaces in spn. Finally, we replace all spaces by the character *, using the expansion ${spn// /*}.
Yet another possibility:
#!/bin/bash
[[ $1 = +([[:digit:]]) ]] || { printf >&2 'Argument must be a number\n'; exit 1; }
printf -v s '%*s' $((10#1))
s=${s// /*}
while [[ $s ]]; do
printf '%s\n' "$s"
s=${s%?}
done
This time we construct the variable s that contains a bunch of * (number given by user), using the previous technique. Then we have a while loop that loops while s is non empty. At each iteration we print the content of s and we remove a character with the expansion ${s%?} that removes the last character of s.
Building on gniourf_gniourf's helpful answer:
The following is simpler and performs significantly better:
#!/bin/bash
count=$1 # (... number-validation code omitted for brevity)
# Create the 1st line, composed of $count '*' chars, and store in var. $line.
printf -v line '%.s*' $(seq $count)
# Count from $count down to 1.
while (( count-- )); do
# Print a *substring* of the 1st line based on the current value of $count.
printf "%.${count}s\n" "$line"
done
printf -v line '*%.s' $(seq $count) is a trick that prints * $count times, thanks to %.s* resulting in * for each argument supplied, irrespective of the arguments' values (thanks to %.s, which effectively ignores its argument). $(seq $count) expands to $count arguments, resulting in a string composed of $count * chars. overall, which - thanks to -v line, is stored in variable $line.
printf "%.${count}s\n" "$line" prints a substring from the beginning of $line that is $count chars. long.
Hi I am trying to print/echo line numbers that are multiple of 5. I am doing this in shell script. I am getting errors and unable to proceed. below is the script
#!/bin/bash
x=0
y=$wc -l $1
while [ $x -le $y ]
do
sed -n `$x`p $1
x=$(( $x + 5 ))
done
When executing above script i get below errors
#./echo5.sh sample.h
./echo5.sh: line 3: -l: command not found
./echo5.sh: line 4: [: 0: unary operator expected
Please help me with this issue.
For efficiency, you don't want to be invoking sed multiple times on your file just to select a particular line. You want to read through the file once, filtering out the lines you don't want.
#!/bin/bash
i=0
while IFS= read -r line; do
(( ++i % 5 == 0 )) && echo "$line"
done < "$1"
Demo:
$ i=0; while read line; do (( ++i % 5 == 0 )) && echo "$line"; done < <(seq 42)
5
10
15
20
25
30
35
40
A funny pure Bash possibility:
#!/bin/bash
mapfile ary < "$1"
printf "%.0s%.0s%.0s%.0s%s" "${ary[#]}"
This slurps the file into an array ary, which each line of the file in a field of the array. Then printf takes care of printing one every 5 lines: %.0s takes a field, but does nothing, and %s prints the field. Since mapfile is used without the -t option, the newlines are included in the array. Of course this really slurps the file into memory, so it might not be good for huge files. For large files you can use a callback with mapfile:
#!/bin/bash
callback() {
printf '%s' "$2"
ary=()
}
mapfile -c 5 -C callback ary < "$1"
We're removing all the elements of the array during the callback, so that the array doesn't grow too large, and the printing is done on the fly, as the file is read.
Another funny possibility, in the spirit of glenn jackmann's solution, yet without a counter (and still pure Bash):
#!/bin/bash
while read && read && read && read && IFS= read -r line; do
printf '%s\n' "$line"
done < "$1"
Use sed.
sed -n '0~5p' $1
This prints every fifth line in the file starting from 0
Also
y=$wc -l $1
wont work
y=$(wc -l < $1)
You need to create a subshell as bash will see the spaces as the end of the assignment, also if you just want the number its best to redirect the file into wc.
Dont know what you were trying to do with this ?
x=$(( $x + 5 ))
Guessing you were trying to use let, so id suggest looking up the syntax for that command. It would look more like
(( x = x + 5 ))
Hope this helps
There are cleaner ways to do it, but what you're looking for is this.
#!/bin/bash
x=5
y=`wc -l $1`
y=`echo $y | cut -f1 -d\ `
while [ "$y" -gt "$x" ]
do
sed -n "${x}p" "$1"
x=$(( $x + 5 ))
done
Initialize x to 5, since there is no "line zero" in your file $1.
Also, wc -l $1 will display the number of line counts, followed by the name of the file. Use cut to strip the file name out and keep just the first word.
In conditionals, a value of zero can be interpreted as "true" in Bash.
You should not have space between your $x and your p in your sed command. You can put them right next to each other using curly braces.
You can do this quite succinctly using awk:
awk 'NR % 5 == 0' "$1"
NR is the record number (line number in this case). Whenever it is a multiple of 5, the expression is true, so the line is printed.
You might also like the even shorter but slightly less readable:
awk '!(NR%5)' "$1"
which does the same thing.
I am trying to match first few letters of a file.
for entry in `ls`; do
echo $entry
done
With the above code I get the name of all the files.
I have a few files with similar name at the start:
Beaglebone-v1
Beaglebone-v3
Beaglebone-v2
How can I compare $entry with Beaglebone* and then extract the latest version file name?
If you want to loop over all Beaglebone-* files:
for entry in Beaglebone-* ; do
echo $entry
done
if you just need the file with the latest version, you can depend on the fact that ls sorts your names alphabetically, so you could just do:
LATEST_FILE_NAME=$(ls Beaglebone-* | tail -n 1)
which will just take the last one alphabetically.
To deal with larger numbers, you could use numeric comparison like this:
stem="Beaglebone-v"
for file in $stem*; do
ver=${file#"$stem"} # cut away stem to get version number
(( ver > max )) && max=$ver # conditionally assign `ver` to `max`
done
echo "$stem$max"
Testing it out:
bash-4.3$ ls Beaglebone-v*
Beaglebone-v1 Beaglebone-v10 Beaglebone-v2 Beaglebone-v3
bash-4.3$ stem="Beaglebone-v" &&
for file in $stem*
do
ver=${file#"$stem"}
(( ver > max )) && max=$ver
done; echo "$stem$max"
Beaglebone-v10
You can store the filenames matching the pattern in an array and then pick the last element of the array.
shopt -s nullglob
arr=( Beaglebone-* )
if (( ${#arr[#]} > 0 ))
then
latest="${arr[ (( ${#arr[#]} - 1 )) ]}"
echo "$latest"
fi
You need to enable nullglob so that if there are no files matching the pattern, you will get an empty array rather than the pattern itself.
If version numbers can go beyond single digits,
function version_numbers {
typeset w; for w in $1-v*; do echo ${w#$1-v}; done
}
version_numbers "Beaglebone" | sort -n | tail -1
Or, adding function max:
# read a stream of numbers, from stdin (one per line)
# and return the largest value
function max
{
typeset _max n
read _max || return
while read n; do
((_max < n)) && _max=$n
done
echo $_max
}
We can now do the whole thing without external commands:
version_numbers Beaglebone | max
Note that max will fail horribly if any one line fails the numerical comparison.