I have a string like below. str='014387650' Now i want to split this string like below and put the value in the array.
A[0]=0
A[1]=01
A[2]=014
A[3]=0143
A[4]=01438
A[5]=014387
A[6]=0143876
A[7]=01438765
A[8]=014387650
-
the question is not clear , you want the string '014387650 to put in the array A? using bash? or you want to populate every index of the array A as you have written above?– Ijaz AhmadJan 4 '16 at 5:50
-
yes i want to put the string in the array A using bash in unix platform.– user3548033Jan 4 '16 at 5:56
-
ok , check the answer , i did it for both the cases– Ijaz AhmadJan 4 '16 at 6:28
Add a comment
|
The following should work in bash:
str='014387650'
arr=()
for ((i=0; i<${#str}; i++)); do
arr+=("${arr[i-1 < 0 ? 0 : i-1]}${str:$i:1}")
done
The result:
$ printf '<%s>\n' "${arr[@]}"
<0>
<01>
<014>
<0143>
<01438>
<014387>
<0143876>
<01438765>
<014387650>
-
1I'm confused why you refer back to the previous element instead of just building up the split range on str. It's not wrong, but it seems convoluted to me.– kojiroJan 4 '16 at 12:08
Explicit declaration of an array is done using the declare built-in:
declare -a ARRAYNAME
Array variables may also be created using compound assignments in this format:
ARRAY=(value1 value2 ... valueN)
in your case:
#!/bin/bash
str='014387650'
declare -a A
for (( c=0; c<${#str}; c++ ))
do
A[c]=`echo ${str:0:$(( $c + 1 ))}`
echo "A[$c]="${A[$c]}
done
output:
A[0]=0
A[1]=01
A[2]=014
A[3]=0143
A[4]=01438
A[5]=014387
A[6]=0143876
A[7]=01438765
A[8]=014387650
if you just want to put the string in an array:
declare -a A
str=014387650
A=${str[*]}
echo ${A[*]}
014387650
echo ${A[3]}
4
-
2
-
yes , its just for showing the output , if you dont want to show the output , you dont need that Jan 4 '16 at 8:00
-
2
-
Another way:
triangle_split() {
_len=1
while [ "$_len" -le "${#1}" ]; do
printf '%.*s\n' "$_len" "$1"
: "$((_len+=1))"
done
}
IFS='
'
A=($(triangle_split 014387650))
printf '%s\n' "${A[@]}"
-
Could you please explain below line. printf '%.*s\n' "$_len" "$1" : "$((_len+=1))" Jan 4 '16 at 6:47
-
The first argument to
printf '%.*s\n'is maximum bytes to be printed, here it's$_lenvariable.$((_len+=1))increase_lenby one. So you got the first, the second, ... until you got the whole string.– cuonglmJan 4 '16 at 6:56 -
maybe weirder, but:
len=0;while printf "%.$((len+=1))s\n%.d" "$1" "0$((len<${#1}?0:8))"; do :;done 2>/dev/null– mikeservJan 4 '16 at 9:45 -
1@mikeserv: tricky, maybe
"0$((len<${#1}?0:8))"should become"0$((len<${#1}?0:$((${#1}-1))))"for more dynamic?– cuonglmJan 4 '16 at 9:53 -
1ok.
printfis spec'd for interpreting it's arguments as C integer constants. POSIX (apparently aligned w/ the ANSI standard) specs integer constants thus: A decimal constant begins with a non-zero digit, and consists of a sequence of decimal digits. An octal constant consists of the prefix '0' optionally followed by a sequence of the digits '0' to '7' only... [EINVAL] No conversion could be performed.– mikeservJan 5 '16 at 2:19
Obvious approach: let's loop downward through the array, chopping one character at a time from the tail of a string variable which is assigned to the successive elements:
#!/bin/bash
str="014387650"
while [ ${#str} -gt 0 ] ; do
A[$((${#str}-1))]=$str
str=${str%?}
done
printf "%s\n" ${A[*]}
Output:
0 01 014 0143 01438 014387 0143876 01438765 014387650
Since we are Bash-specific, we might as well rephrase the logic using a for loop:
#!/bin/bash
str="014387650"
for (( i=${#str} - 1; i >= 0; i-- )); do
A[$i]=$str
str=${str%?}
done
printf "%s\n" ${A[*]}
-
is it more obvious than the other one? i would think you would just flip the sign.– mikeservJan 5 '16 at 5:25
You might do:
set --; src=014387650 OPTIND=1 tgt=
while getopts : na -"$src"
do tgt=$tgt$OPTARG
set "$@" "A[$#]=$tgt"
done; printf %s\\n "$@"
A[0]=0
A[1]=01
A[2]=014
A[3]=0143
A[4]=01438
A[5]=014387
A[6]=0143876
A[7]=01438765
A[8]=014387650
Obvious approach: fill the array backwards, and chop the last character of the string as we go:
#!/bin/bash
str="014387650"
i=${#str}
while [ $i -gt 0 ] ; do
A[$((--i))]=$str
str=${str%?}
done
printf "%s\n" ${A[*]}
Output:
0 01 014 0143 01438 014387 0143876 01438765 014387650
Maybe this will help:
#!/bin/bash
a="014387650"
i=0; unset -v A B
while (( i<${#a} ))
do A[i]=${a:0:i+1}
B+=("A[$i]=${A[i++]}")
done
printf '%s\n' "${B[@]}"
Output:
A[0]=0
A[1]=01
A[2]=014
A[3]=0143
A[4]=01438
A[5]=014387
A[6]=0143876
A[7]=01438765
A[8]=014387650