how to delete letters from a string variable in bash? - bash

For example
var=" |"
and
var2="hello"
How do I delete number of space characters that equals to the length of var2?
So var will now be " |" instead of " |"
I thought of doing something with ${#var2} minus var but I don't know how to delete specific characters.

Try
var=${var:${#var2}}
${var:${#var2}} expands to the characters in $var from index ${#var2} onwards. See Extracting parts of strings (BashFAQ/100 (How do I do string manipulation in bash?)).

Here's a fun one:
echo "${var/${var2//?/?}/}"
${var2//?/?} replaces every character in $var2 with the character "?"
Then we use that string ("?????") as a pattern against $var, and we replace it with an empty string

You can remove one space at a time while the $var is longer:
#!/bin/bash
var=" |"
var2=hello
expect=" |"
while (( ${#var} > ${#var2} )) ; do
var=${var#\ }
done
[[ $var = $expect ]] && echo Ok
or you can calculate the number of characters to remove and use the offset parameter expansion to skip them:
remove_length=$(( ${#var} - ${#var2} ))
var=${var:$remove_length}

Use parameter expansion:
#!/bin/bash
s1=' |' # 6 spaces
s2='hello' # 5 characters
(( to_remove=${#s2}+1 ))
echo "${s1:$to_remove}" # 5 characters removed
# ' |'

Related

Changing alternative character from lower to upper and upper to low - Unix shell script

How to convert the alternative character of a string passed to script, if it is lower then it should be converted to upper and if it is upper then to lower??
read -p " Enter string" str
for i in `seq 0 ${#str}`
do
#echo $i
rem=$(($i % 2 ))
if [ $rem -eq 0 ]
then
echo ${str:$i:1}
else
fr=${str:$i:1}
if [[ "$fr" =~ [A-Z] ]]
then
echo ${str:$i:1} | tr '[:upper:]' '[:lower:]'
elif [[ "$fr" =~ [a-z] ]]
then
echo ${str:$i:1} | tr '[:lower:]' '[:upper:]'
else
echo ""
fi
fi
done
Your question is a bit challenging given that it is tagged shell and not as a question pertaining to an advanced shell like bash or zsh. In POSIX shell, you have no string indexes, no C-style for loop, and no [[ .. ]] operator to use character class pattern matching.
However, with a bit of awkward creativity, the old expr and POSIX string and arithmetic operations, and limiting your character strings to ASCII characters, you can iterate over a string changing uppercase to lowercase and lowercase and uppercase while leaving all other characters unchanged.
I wouldn't recommend the approach if you have an advanced shell available, but if you are limited to POSIX shell, as your question is tagged, it will work, but don't expect it to be super-fast...
#!/bin/sh
a=${1:-"This Is My 10TH String"} ## input and output strings
b=
i=1 ## counter and string length
len=$(expr length "$a")
asciiA=$(printf "%d" "'A") ## ASCII values for A,Z,a,z
asciiZ=$(printf "%d" "'Z")
asciia=$(printf "%d" "'a")
asciiz=$(printf "%d" "'z")
echo "input : $a" ## output original string
while [ "$i" -le "$len" ]; do ## loop over each character
c=$(expr substr "$a" "$i" "1") ## extract char from string
asciic=$(printf "%d" "'$c") ## convert to ASCII value
## check if asciic is [A-Za-z]
if [ "$asciiA" -le "$asciic" -a "$asciic" -le "$asciiZ" ] ||
[ "$asciia" -le "$asciic" -a "$asciic" -le "$asciiz" ]
then ## toggle the sign bit (bit-6)
b="${b}$(printf "\x$(printf "%x" $((asciic ^ 1 << 5)))\n")"
else
b="$b$c" ## otherwise copy as is
fi
i=$(expr $i + 1)
done
echo "output: $b" ## output resluting string
The case change is affected by relying on a simple bit-toggle of the case-bit (bit-6) in the ASCII value of each upper or lower case character to change it from lower to upper or vice-versa. (and note, you can exchange the printf and bit-shift for tr of asciic as an alternative)
Example Use/Output
$ sh togglecase.sh
input : This Is My 10TH String
output: tHIS iS mY 10th sTRING
When you want to swab every second characters case, try this:
read -p " Enter string " str
for i in `seq 0 ${#str}`; do
rem=$(($i % 2 ))
if [ $rem -eq 0 ]
then
printf "%s" "${str:$i:1}"
else
fr=${str:$i:1}
printf "%s" "$(tr '[:upper:][:lower:]' '[:lower:][:upper:]' <<< "${str:$i:1}")"
fi
done
echo
EDIT: Second solution
Switch case of str and merge the old and new string.
#!/bin/bash
str="part is lowercase & PART IS UPPERCASE"
str2=$(tr '[:upper:][:lower:]' '[:lower:][:upper:]' <<< "${str}")
str_chopped=$(sed -r 's/(.)./\1\n/g' <<< "${str}");
# Will have 1 additional char for odd length str
# str2_chopped_incorrect=$(sed -r 's/.(.)/\1\n/g' <<< "${str2}");
str2_chopped=$(fold -w2 <<< "${str2}" | sed -nr 's/.(.)/\1/p' );
paste -d '\n' <(echo "${str_chopped}") <(echo "${str2_chopped}") | tr -d '\n'; echo

Concatenating digits from a string in sh

Assuming that I have a string like this one:
string="1 0 . # 1 1 ? 2 2 4"
Is it possible to concatenate digits that are next to each other?
So that string be like: 10 . # 11 ? 224 ?
I found only basic things how to distinguish integers from other characters and how to "connect" them. But I have no idea how to iterate properly.
num=""
for char in $string; do
if [ $char -eq $char 2>/dev/null ] ; then
num=$num$char
Here's an almost pure-shell implementation -- transforming the string into a character per line and using a BashFAQ #1 while read loop.
string="1 0 . # 1 1 ? 2 2 4"
output=''
# replace spaces with newlines for easier handling
string=$(printf '%s\n' "$string" | tr ' ' '\n')
last_was_number=0
printf '%s\n' "$string" | {
while read -r char; do
if [ "$char" -eq "$char" ] 2>/dev/null; then # it's a number
if [ "$last_was_number" -eq "1" ]; then
output="$output$char"
last_was_number=1
continue
fi
last_was_number=1
else
last_was_number=0
fi
output="$output $char"
done
printf '%s\n' "$output"
}
To complement Charles Duffy's helpful, POSIX-compliant sh solution with a more concise perl alternative:
Note: perl is not part of POSIX, but it is preinstalled on most modern Unix-like platforms.
$ printf '%s\n' "1 0 . # 1 1 ? 2 2 4" | perl -pe 's/\d( \d)+/$& =~ s| ||gr/eg'
10 . # 11 ? 224
The outer substitution, s/\d( \d)+/.../eg, globally (g) finds runs of at least 2 adjacent digits (\d( \d)+), and replaces each run with the result of the expression (e) specified as the replacement string (represented as ... here).
The expression in the inner substitution, $& =~ s| ||gr, whose result is used as the replacement string, removes all spaces from each run of adjacent digits:
$& represents what the outer regex matched - the run of adjacent digits.
=~ applies the s call on the RHS to the LHS, i.e., $& (without this, the s call would implicitly apply to the entire input string, $_).
s| ||gr replaces all (g) instances of <space> from the value of the value of $& and returns (r) the result, effectively removing all spaces.
Note that | is used arbitrarily as the delimiter character for the s call, so as to avoid a clash with the customary / delimiter used by the outer s call.
POSIX compliant one-liner with sed:
string="1 0 . # 1 1 ? 2 2 4"
printf '%s\n' "$string" | sed -e ':b' -e ' s/\([0-9]\) \([0-9]\)/\1\2/g; tb'
It just iteratively removes the any space between two digits until there aren't any more, resulting in:
10 . # 11 ? 224
Here is my solution:
string="1 0 . # 1 1 ? 2 2 4"
array=(${string/// })
arraylength=${#array[#]}
pattern="[0-9]"
i=0
while true; do
str=""
start=$i
if [ $i -eq $arraylength ]; then
break;
fi
for (( j=$start; j<${arraylength}; j++ )) do
curr=${array[$j]}
i=$((i + 1))
if [[ $curr =~ $pattern ]]; then
str="$str$curr"
else
break
fi
done
echo $str
done

multiplicator operation bash issue

from stdin I read string and if it's like this:
"numberOne * numberTwo"
I have to execute the multiplicator between numberOne and numberTwo.
This is my code:
read string
regex2="^[1-9]+ \*{1,1} [1-9]+$"
if [[ $string =~ $regex2 ]]; then
val=1
val1=`echo $string|cut -d " " -f 1`
val2=`echo $string|cut -d " " -f 3`
((val=$val1*$val2))#comment
echo $val
fi
but I get two errors:
1) on the line where calculate the operation ((val=$val1*$val2)), it says syntax error : arithmetic operator invalid
2) where , by shell , I insert the input string, for example 3 * 2 on shell it prints a list of files, then I thought it was for jolly character "*", and for this reason I substuited the input string with this:
3 \* 2
but the result doesn't change
Always, always quote your expansions.
echo $string, when $string contains a * surrounded by whitespace, treats that * as a glob, replacing it with a list of filenames in the current directory. Your filenames are not likely to be part of a legitimate math operation.
Use echo "$string" instead, if you must use echo at all; printf '%s\n' "$string" is the alternative that works in corner cases where echo fails (and/or behaves in ways unspecified by POSIX).
That said, there's no legitimate reason to use cut here at all; your regex will split your string into pieces perfectly well on its own.
regex2='^([1-9][0-9]*) [*] ([1-9][0-9]*)$'
read -r string
if [[ $string =~ $regex2 ]]; then
val=$(( ${BASH_REMATCH[1]} * ${BASH_REMATCH[2]} ))
echo "$val"
fi
...and even if you couldn't do that, it would be a better practice to use read:
read val1 _ val2 <<<"$string"
echo "$(( val1 * val2 ))"

substring extraction in bash

iamnewbie: this code is inefficient but it should extract the substring, the problem is with last echo statement,need some insight.
function regex {
#this function gives the regular expression needed
echo -n \'
for (( i = 1 ; i <= $1 ; i++ ))
do
echo -n .
done
echo -n '\('
for (( i = 1 ; i <= $2 ; i++ ))
do
echo -n .
done
echo -n '\)'
echo -n \'
}
# regex function ends
echo "Enter the string:"
read stg
#variable stg holds the string entered
if [ -z "$stg" ] ; then
echo "Null string"
exit
else
echo "Length of the $stg is:"
z=`expr "$stg" : '.*' `
#variable z holds the length of given string
echo $z
fi
echo "Enter the number of trailing characters to be extracted from $stg:"
read n
m=`expr $z - $n `
#variable m holds an integer value which is equal to total length - length of characters to be extracted
x=$(regex $m $n)
echo ` expr "$stg" : "$x" `
#the echo statement(above) is just printing a newline!! But not the result
What I intend to do with this code is, if I enter "racecar" and give "3" , it should display "car" which are the last three characters. Instead of displaying "car" its just printing a newline. Please correct this code rather than giving a better one.
Although you didn't ask for a better solution, it's worth mentioning:
$ n=3
$ stg=racecar
$ echo "${stg: -n}"
car
Note that the space after the : in ${stg: -n} is required. Without the space, the parameter expansion is a default-value expansion rather than a substring expansion. With the space, it's a substring expansion; -n is interpreted as an arithmetic expression (which means that n is interpreted as $n) and since the result is a negative number, it specifies the number of characters from the end to start the substring. See the Bash manual for details.
Your solution is based on evaluating the equivalent of:
expr "$stg" : '......\(...\)'
with an appropriate number of dots. It's important to understand what the above bash syntax actually means. It invokes the command expr, passing it three arguments:
arg 1: the contents of the variable stg
arg 2: :
arg 3: ......\(...\)
Note that there are no quotes visible. That's because the quotes are part of bash syntax, not part of the argument values.
If the value of stg had enough characters, the result of the above expr invocation would be to print out the 7th, 8th and 9th character of the value of stg`. Otherwise, it would print a blank line, and fail.
But that's not what you are doing. You're creating the regular expression:
'......\(...\)'
which has single quotes in it. Since single-quotes are not special characters in a regex, they match themselves; in other words, that pattern will match a string which starts with a single quote, followed by nine arbitrary characters, followed by another single quote. And if the string does match, it will print the three characters prior to the second single-quote.
Of course, since the regular expression you make has a . for every character in the target string, it won't match the target even if the target started and begun with a single-quote, since there would be too many dots in the regex to match that.
If you don't put single quotes into the regex, then your program will work, but I have to say that few times have I seen such an intensely circuitous implementation of the substring function. If you're not trying to win an obfuscated bash competition (a difficult challenge since most production bash code is obfuscated by nature), I'd suggest you use normal bash features instead of trying to do everything with regexen.
One of those is the syntax to determine the length of a string:
$ stg=racecar
$ echo ${#stg}
7
(although, as shown at the beginning, you don't actually even need that.)
What about:
$ n=3
$ string="racecar"
$ [[ "$string" =~ (.{$n})$ ]]
$ echo ${BASH_REMATCH[1]}
car
This looks for the last n characters at the end of the line. In a script:
#!/bin/bash
read -p "Enter a string: " string
read -p "Enter the number of characters you want from the end: " n
[[ "$string" =~ (.{$n})$ ]]
echo "These are the last $n characters: ${BASH_REMATCH[1]}"
You may want to add some more error handling, but this'll do it.
I'm not sure you need loops for this task. I wrote some example to get two parameters from user and cut the word according to it.
#!/bin/bash
read -p "Enter some word? " -e stg
#variable stg holds the string entered
if [ -z "$stg" ] ; then
echo "Null string"
exit 1
fi
read -p "Enter some number to set word length? " -e cutNumber
# check that cutNumber is a number
if ! [ "$cutNumber" -eq "$cutNumber" ]; then
echo "Not a number!"
exit 1
fi
echo "Cut first n characters:"
echo ${stg:$cutNumber}
echo
echo "Show first n characters:"
echo ${stg:0:$cutNumber}
echo "Alternative get last n characters:"
echo -n "$stg" | tail -c $cutNumber
echo
Example:
Enter some word? TheRaceCar
Enter some number to set word length? 7
Cut first n characters:
Car
Show first n characters:
TheRace
Alternative get last n characters:
RaceCar

How to split one string into multiple strings separated by at least one space in bash shell?

I have a string containing many words with at least one space between each two. How can I split the string into individual words so I can loop through them?
The string is passed as an argument. E.g. ${2} == "cat cat file". How can I loop through it?
Also, how can I check if a string contains spaces?
I like the conversion to an array, to be able to access individual elements:
sentence="this is a story"
stringarray=($sentence)
now you can access individual elements directly (it starts with 0):
echo ${stringarray[0]}
or convert back to string in order to loop:
for i in "${stringarray[#]}"
do
:
# do whatever on $i
done
Of course looping through the string directly was answered before, but that answer had the the disadvantage to not keep track of the individual elements for later use:
for i in $sentence
do
:
# do whatever on $i
done
See also Bash Array Reference.
Did you try just passing the string variable to a for loop? Bash, for one, will split on whitespace automatically.
sentence="This is a sentence."
for word in $sentence
do
echo $word
done
This
is
a
sentence.
Probably the easiest and most secure way in BASH 3 and above is:
var="string to split"
read -ra arr <<<"$var"
(where arr is the array which takes the split parts of the string) or, if there might be newlines in the input and you want more than just the first line:
var="string to split"
read -ra arr -d '' <<<"$var"
(please note the space in -d ''; it cannot be omitted), but this might give you an unexpected newline from <<<"$var" (as this implicitly adds an LF at the end).
Example:
touch NOPE
var="* a *"
read -ra arr <<<"$var"
for a in "${arr[#]}"; do echo "[$a]"; done
Outputs the expected
[*]
[a]
[*]
as this solution (in contrast to all previous solutions here) is not prone to unexpected and often uncontrollable shell globbing.
Also this gives you the full power of IFS as you probably want:
Example:
IFS=: read -ra arr < <(grep "^$USER:" /etc/passwd)
for a in "${arr[#]}"; do echo "[$a]"; done
Outputs something like:
[tino]
[x]
[1000]
[1000]
[Valentin Hilbig]
[/home/tino]
[/bin/bash]
As you can see, spaces can be preserved this way, too:
IFS=: read -ra arr <<<' split : this '
for a in "${arr[#]}"; do echo "[$a]"; done
outputs
[ split ]
[ this ]
Please note that the handling of IFS in BASH is a subject on its own, so do your tests; some interesting topics on this:
unset IFS: Ignores runs of SPC, TAB, NL and on line starts and ends
IFS='': No field separation, just reads everything
IFS=' ': Runs of SPC (and SPC only)
Some last examples:
var=$'\n\nthis is\n\n\na test\n\n'
IFS=$'\n' read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[#]}"; do let i++; echo "$i [$a]"; done
outputs
1 [this is]
2 [a test]
while
unset IFS
var=$'\n\nthis is\n\n\na test\n\n'
read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[#]}"; do let i++; echo "$i [$a]"; done
outputs
1 [this]
2 [is]
3 [a]
4 [test]
BTW:
If you are not used to $'ANSI-ESCAPED-STRING' get used to it; it's a timesaver.
If you do not include -r (like in read -a arr <<<"$var") then read does backslash escapes. This is left as exercise for the reader.
For the second question:
To test for something in a string I usually stick to case, as this can check for multiple cases at once (note: case only executes the first match, if you need fallthrough use multiple case statements), and this need is quite often the case (pun intended):
case "$var" in
'') empty_var;; # variable is empty
*' '*) have_space "$var";; # have SPC
*[[:space:]]*) have_whitespace "$var";; # have whitespaces like TAB
*[^-+.,A-Za-z0-9]*) have_nonalnum "$var";; # non-alphanum-chars found
*[-+.,]*) have_punctuation "$var";; # some punctuation chars found
*) default_case "$var";; # if all above does not match
esac
So you can set the return value to check for SPC like this:
case "$var" in (*' '*) true;; (*) false;; esac
Why case? Because it usually is a bit more readable than regex sequences, and thanks to Shell metacharacters it handles 99% of all needs very well.
Just use the shells "set" built-in. For example,
set $text
After that, individual words in $text will be in $1, $2, $3, etc. For robustness, one usually does
set -- junk $text
shift
to handle the case where $text is empty or start with a dash. For example:
text="This is a test"
set -- junk $text
shift
for word; do
echo "[$word]"
done
This prints
[This]
[is]
[a]
[test]
$ echo "This is a sentence." | tr -s " " "\012"
This
is
a
sentence.
For checking for spaces, use grep:
$ echo "This is a sentence." | grep " " > /dev/null
$ echo $?
0
$ echo "Thisisasentence." | grep " " > /dev/null
$ echo $?
1
echo $WORDS | xargs -n1 echo
This outputs every word, you can process that list as you see fit afterwards.
(A) To split a sentence into its words (space separated) you can simply use the default IFS by using
array=( $string )
Example running the following snippet
#!/bin/bash
sentence="this is the \"sentence\" 'you' want to split"
words=( $sentence )
len="${#words[#]}"
echo "words counted: $len"
printf "%s\n" "${words[#]}" ## print array
will output
words counted: 8
this
is
the
"sentence"
'you'
want
to
split
As you can see you can use single or double quotes too without any problem
Notes:
-- this is basically the same of mob's answer, but in this way you store the array for any further needing. If you only need a single loop, you can use his answer, which is one line shorter :)
-- please refer to this question for alternate methods to split a string based on delimiter.
(B) To check for a character in a string you can also use a regular expression match.
Example to check for the presence of a space character you can use:
regex='\s{1,}'
if [[ "$sentence" =~ $regex ]]
then
echo "Space here!";
fi
For checking spaces just with bash:
[[ "$str" = "${str% *}" ]] && echo "no spaces" || echo "has spaces"
$ echo foo bar baz | sed 's/ /\n/g'
foo
bar
baz
For my use case, the best option was:
grep -oP '\w+' file
Basically this is a regular expression that matches contiguous non-whitespace characters. This means that any type and any amount of whitespace won't match. The -o parameter outputs each word matches on a different line.
Another take on this (using Perl):
$ echo foo bar baz | perl -nE 'say for split /\s/'
foo
bar
baz

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