I'm writing a bash script to get some podcasts. The problem is that some of the podcast numbers are one digits while others are two/three digits, therefore I need to pad them to make them all 3 digits.
I tried the following:
n=1
n = printf %03d $n
wget http://aolradio.podcast.aol.com/sn/SN-$n.mp3
but the variable 'n' doesn't stay padded permanently. How can I make it permanent?
Use backticks to assign the result of the printf command (``):
n=1
wget http://aolradio.podcast.aol.com/sn/SN-`printf %03d $n`.mp3
EDIT: Note that i removed one line which was not really necessary.
If you want to assign the output of 'printf %...' to n, you could
use
n=`printf %03d $n`
and after that, use the $n variable substitution you used before.
Seems you're assigning the return value of the printf command (which is its exit code), you want to assign the output of printf.
bash-3.2$ n=1
bash-3.2$ n=$(printf %03d $n)
bash-3.2$ echo $n
001
Attention though if your input string has a leading zero!
printf will still do the padding, but also convert your string to hex octal format.
# looks ok
$ echo `printf "%05d" 03`
00003
# but not for numbers over 8
$ echo `printf "%05d" 033`
00027
A solution to this seems to be printing a float instead of decimal.
The trick is omitting the decimal places with .0f.
# works with leading zero
$ echo `printf "%05.0f" 033`
00033
# as well as without
$ echo `printf "%05.0f" 33`
00033
to avoid context switching:
a="123"
b="00000${a}"
c="${b: -5}"
n=`printf '%03d' "2"`
Note spacing and backticks
As mentioned by noselad, please command substitution, i.e. $(...), is preferable as it supercedes backtics, i.e. `...`.
Much easier to work with when trying to nest several command substitutions instead of escaping, i.e. "backslashing", backtics.
This is in response to an answer given by cC Xx.
It will work only until a's value less is than 5 digits.
Consider when a=12345678.
It'll truncate the leading digits:
a="12345678"
b="00000${a}"
c="${b: -5}"
echo "$a, $b, $c"
This gives the following output:
12345678, 0000012345678, 45678
Putting an if to check value of a is less than 5 digits and then doing it could be solution:
if [[ $a -lt 9999 ]] ; then b="00000${a}" ; c="${b: -5}" ; else c=$a; fi
Just typing this here for additional information.
If you know the number of zeroes you need, you can use the string concatenation:
let pad="0"
pad+=1
echo $pad # this will print 01
Related
Let's say I have a filename:
Filename=AB123_10_001_00202.jpg
Using as much as possible bash "attribute expansion substring" ,
I would like to extract "202" or in general the number without the "_00".
If I do:
Name=${Filename%.jpg}
I get:
AB123_10_001_00202
but then, as many "_0" occur, I don't see how to proceed.
So I tried:
Number=${Name##*_0}
...which works in case of the last digits are 12, 123 or 1234 for example. But if a "_0" is in-between some digit like 202, I only get "2".
Removing the leading zeroes is tricky using parameter expansion. You could remove them by interpreting the number:
Filename=AB123_10_001_00202.jpg
Name="${Filename%.jpg}"
PaddedNumber="${Name##*_}"
(( Number = "10#$PaddedNumber" ))
Alternatively, use bash's matching operator
Filename=AB123_10_001_00202.jpg
Regex='.*_0*([0-9]+)'
[[ "$Filename" =~ $Regex ]]
Number="${BASH_REMATCH[1]}"
Thank you for the one who posted this answer but removed it.
It worked perfectly well. Just in case someone get stuck as I was.
$ fname=AB123_10_001_00202.jpg
$ str=${fname%.jpg}
$ echo $fname
AB123_10_001_00202
$ shopt -s extglob
$ printf -v var '%s\n' "${str##*_*(0)}"
$ echo $var
202
I have created a hex to ASCII converter for strings in bash. The application I'm on changes characters (anything but [0-9],[A-Z],[a-z]) , in a string to its corresponding %hexadecimal. Eg: / changes to %2F in a string
I want to retain the ASCII characters as it is. Below is my code:
NAME=%2fhome%40%21%23
C_NAME=""
for (( i=0; i<${#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#!#
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 $count chars. long.
is there some way to put variable in ALPHABETICAL range of cycle?
This doesnt work.
read -p "Where I should start?" start #there will be entered one small letter
for aaa in {$start..z}; do #how put variable $start in range?
...
done
Thanks for reply.
Use eval to expand the variable:
$ s=t
$ eval echo {$s..z}
t u v w x y z
Your example then becomes:
read -p "Where I should start?" start #there will be entered one small letter
for aaa in $(eval echo {$start..z}); do
echo $aaa
done
Since you have user input to eval, you may want to check the value of start as being a single lower case character first:
read -p "Where I should start?" start #there will be entered one small letter
if [[ $start =~ ^[a-y]$ ]]; then
for aaa in $(eval echo {$start..z}); do
echo $aaa
done
else
echo "Need to use a letter 'a-y'"
fi
You can read more about Bash brace expansion here
Unfortunately, you can't put variables inside {start..end} ranges in bash.
This does what you want:
until [[ $s == "_" ]]; do echo $s && s=$(tr "a-z" "b-z_" <<<$s); done
It uses tr to translate each character to the next one. "_" is the character after "z".
For example:
$ s=t
$ until [[ $s == "_" ]]; do echo $s && s=$(tr "a-z" "b-z_" <<<$s); done
t
u
v
w
x
y
z
If you don't mind using Perl, you could use this:
perl -le 'print for shift .. "z"' $s
It uses .. to create a list between the first argument on the command line and "z".
A slightly more esoteric way to do it in bash would be:
for ((i=$(LC_CTYPE=C printf '%d' "'$s"); i<=122; ++i)); do
printf "\\$(printf '%03o' $i)\n"
done
The for loop goes from the ASCII character number of the variable $s to "z", which is ASCII character 122. The format specifier the inner printf converts the character number to octal, padding it with zeros up to three characters long. The outer printf then interprets this as an escape sequence and prints the character. Credit goes to Greg's wiki for the code used to convert ASCII characters to their values.
Of course you could just use eval to expand the variable, the advantage being that the code required to do so is much shorter. However, executing arbitrary strings that have been read in to your script is arguably a bit of a security hole.
x=t
for I in $(sed -nr "s/.*($x.*)/\1/;s/ /\n/g;p" <(echo {a..z}))
do
# do something with $I
done
Output:
t
u
v
w
x
y
z
I would avoid the use of eval.
for aaa in {a..z}; do
[[ $aaa < $start ]] && continue
...
done
The overhead of comparing $aaa to $start should be negligible, especially compare to the cost of starting a separate process to compute the range.
I was trying to make a small script that calculates a binary number to decimal. It works like this:
-Gets '1' or '0' digits as SEPARATE parameters. e.g. "./bin2dec 1 0 0 0 1 1 1".
-For each parameter digit: if it is '1', multiplies it with the corresponding power of 2 (in the above case, the most left '1' will be 64), then adds it in a 'sum' variable.
Here's the code (it is wrong in the noted point):
#!/bin/bash
p=$((2**($#-1))) #Finds the power of two for the first parameter.
sum=0 #The sum variable, to be used for adding the powers of two in it.
for (( i=1; i<=$#; i++ )) #Counts all the way from 1, to the total number of parameters.
do
if [ $i -eq 1 ] # *THIS IS THE WRONG POINT* If parameter content equals '1'...
then
sum=$(($sum+$p)) #...add the current power of two in 'sum'.
fi
p=$(($p/2)) #Divides the power with 2, so to be used on the next parameter.
done
echo $sum #When finished show the 'sum' content, which is supposed to be the decimal equivalent.
My question is in the noted point (line #10, including blank lines). There, I’m trying to check if EACH parameter's content equals 1. How can I do this using a variable?
For example, $1 is the first parameter, $2 is the 2nd and so on. I want it to be like $i, where 'i' is the variable that is increased by one each time so that it matches the next parameter.
Among other things, I tried this: '$(echo "$"$i)' but didn't work.
I know my question is complicated and I tried hard to make it as clear as I could.
Any help?
How about this?
#!/usr/bin/bash
res=0
while [[ $# -gt 0 ]]
do
res=$(($res * 2 + $1))
shift
done
echo $res
$ ./bin2dec 1 0 0 1
9
shift is a command that moves every parameter passed to the script to the left, decreasing it's value by 1, so that the value of $2 is now at $1 after a shift. You can actually a number with shift as well, to indicate how far to shift the variables, so shift is like shift 1, and shift 2 would give $1 the value that used to be at $3.
How about this one?
#!/bin/bash
p=$((2**$#))
while(($#)); do
((sum+=$1*(p/=2)))
shift
done
echo "$sum"
or this one (that goes in the other direction):
#!/bin/bash
while(($#)); do
((sum=2*sum+$1))
shift
done
echo "$sum"
Note that there are no error checkings.
Please consider fedorqui's comment: use echo $((2# binary number )) to convert from binary to decimal.
Also note that this will overflow easily if you give too many arguments.
If you want something that doesn't overflow, consider using dc:
dc <<< "2i101010p"
(setting input radix to 2 with 2i and putting 101010 on the stack and printing it).
Or if you like bc better:
bc <<< "ibase=2;101010"
Note that these need the binary number to be entered as one argument, and not as you wanted with all digits separated. If you really need all digits to be separated, you could also use these funny methods:
Pure Bash.
#!/bin/bash
echo $((2#$(printf '%s' "$#")))
With dc.
#!/bin/bash
dc <<< "2i$(printf '%s' "$#")p"
With bc.
#!/bin/bash
bc <<< "ibase=2;$(printf '%s' "$#")"
Abusing IFS, with dc.
#!/bin/bash
( IFS=; echo "2i$*p" ) | dc
Abusing IFS, with bc.
#!/bin/bash
( IFS=; echo "ibase=2;$*" ) | bc
So we still haven't answered your question (the one in your comment of the OP): And to find out if and how can we refer to a parameter using a variable. It's done in Bash with indirect expansion. You can read about it in the Shell Parameter Expansion section of the Bash reference manual. The best thing is to show you in a terminal:
$ a="hello"
$ b=a
$ echo "$b"
a
$ echo "${!b}"
hello
neat, eh? It's similar with positional parameters:
$ set param1 param2 param3
$ echo "$1, $2, $3"
param1, param2, param3
$ i=2
$ echo "$i"
2
$ echo "${!i}"
param2
Hope this helps!
To reference variables indirectly by name, use ${!name}:
i=2
echo "${!i} is now equivalent to $2"
Try this one also:
#!/bin/bash -
( IFS=''; echo $((2#$*)) )