To delete particular characters from a combination list.
printf "%s\n" {a..c}{a..d} | sed 's/^cc//' | tr -s '\n'
I used the code above to delete a particular line of character from combination. Is there a way I can do it without sed, awk, grep or bc. Can I get it done with a single line of code in the script?
If you have stored your values in an array, e.g.:
arr=({a..c}{a..d})
Then you may filter your array elements with a string substitution:
printf -- '%s\n' "${arr[#]/%cc/}"
The syntax ${arr[#]/%cc/} tells to parse all elements from the array arr and substitute %cc with nothing. The % character indicates the beginning of the string, similar to ^ in sed, thus %cc means "every string beginning with cc".
Related
I need to replace one variable with another variable in a multiple strings.
For example:
string1="One,two"
string2="three.four"
string3="five:six"
y=";"
for str in string1 string2 string3; do
x="$(echo "$str" | sed 's/[a-zA-Z]//g')" # extracting a character between letters
sed 's/$x/$y/'$str # I tried this, but it does not work at all.
echo "$str"
done
Expecting output:
One;two
three;four
five;six
In my output, nothing changes:
One,two
three.four
five:six
You can use bash's substitution operator instead of sed. And simply replace anything that isn't a letter with $y.
#!/bin/bash
string1="One,two"
string2="three.four"
string3="five:six"
y=";"
for str in "$string1" "$string2" "$string3"; do
x=${str//[^a-zA-Z]+/$y}
echo "$x"
done
Output is:
One;two
three;four
five;six
Note that your general approach wouldn't work if the input string has muliple delimiters, e.g. One,two,three. When you remove all the letters you get ,,, but that doesn't appear anywhere in the string.
Addressing issues with OP's current code:
referencing variables requires a leading $, preferably a pair of {}, and (usually) double quotes (eg, to insure embedded spaces are considered as part of the variable's value)
sed can take as input a) a stream of text on stdin, b) a file, c) process substitution or d) a here-document/here-string
when building a sed script that includes variable refences the sed script must be wrapped in double quotes (not single quotes)
Pulling all of this into OP's current code we get:
string1="One,two"
string2="three.four"
string3="five:six"
y=";"
for str in "${string1}" "${string2}" "${string3}"; do # proper references of the 3x "stringX" variables
x="$(echo "$str" | sed 's/[a-zA-Z]//g')"
sed "s/$x/$y/" <<< "${str}" # feeding "str" as here-string to sed; allowing variables "x/y" to be expanded in the sed script
echo "$str"
done
This generates:
One;two # generated by the 2nd sed call
One,two # generated by the echo
;hree.four # generated by the 2nd sed call
three.four # generated by the echo
five;six # generated by the 2nd sed call
five:six # generated by the echo
OK, so we're now getting some output but there are obviously some issues:
the results of the 2nd sed call are being sent to stdout/terminal as opposed to being captured in a variable (presumably the str variable - per the follow-on echo ???)
for string2 we find that x=. which when plugged into the 2nd sed call becomes sed "s/./;/"; from here the . matches the first character it finds which in this case is the 1st t in string2, so the output becomes ;hree.four (and the . is not replaced)
dynamically building sed scripts without knowing what's in x (and y) becomes tricky without some additional coding; instead it's typically easier to use parameter substitution to perform the replacements for us
in this particular case we can replace both sed calls with a single parameter substitution (which also eliminates the expensive overhead of two subprocesses for the $(echo ... | sed ...) call)
Making a few changes to OP's current code we can try:
string1="One,two"
string2="three.four"
string3="five:six"
y=";"
for str in "${string1}" "${string2}" "${string3}"; do
x="${str//[^a-zA-Z]/${y}}" # parameter substitution; replace everything *but* a letter with the contents of variable "y"
echo "${str} => ${x}" # display old and new strings
done
This generates:
One,two => One;two
three.four => three;four
five:six => five;six
I have file like this
TT;12-11-18;text;abc;def;word
AA;12-11-18;tee;abc;def;gih;word
TA;12-11-18;teet abc;def;word
TT;12-11-18;tdd;abc;def;gih;jkl;word
I want output like this
TT;12-11-18;text;abc;def;word
TA;12-11-18;teet abc;def;word
I want to get word if it occur at position 5 after date 12-11-18. I do not want this occurrence if its found after this position that is at 6th or 7th position. Count of position start from date 12-11-18
I want tried this command
cat file.txt|grep "word" -n1
This print all occurrence in which this pattern word is matched. How should I solve my problem?
Try this(GNU awk):
awk -F"[; ]" '/12-11-18/ && $6=="word"' file
Or sed one:
sed -n '/12-11-18;\([^; ]*[; ]\)\{3\}word/p' file
Or grep with basically the same regex(different escape):
grep -E "12-11-18;([^; ]*[; ]){3}word" file
[^; ] means any character that's not ; or (space).
* means match any repetition of former character/group.
-- [^; ]* means any length string that don't contain ; or space, the ^ in [^; ] is to negate.
[; ] means ; or space, either one occurance.
() is to group those above together.
{3} is to match three repetitives of former chracter/group.
As a whole ([^; ]*[; ]){3} means ;/space separated three fields included the delimiters.
As #kvantour points out, if there could be multiple spaces at one place they could be faulty.
To consider multiple spaces as one separator, then:
awk -F"(;| +)" '/12-11-18/ && $6=="word"'
and
grep -E "12-11-18;([^; ]*(;| +)){3}word"
or GNU sed (posix/bsd/osx sed does not support |):
sed -rn '/12-11-18;([^; ]*(;| +)){3}word/p'
I am trying to trim the data above the line from a file, where line containing some string by passing variable to it
varfile=$(cat variable.txt)
echo "$varfile"
if [ -z "$varfile" ]; then
echo "null"
else
echo "data"
sed "1,/$varfile/d" fileee.txt
fi
Here I am taking a string from variable.txt file and trying to find that text in fileee.txt file and removing all the data above the line
EX: variable.txt has 3
I am finding 3 in fileee.txt and removing data above three
INPUT:
1
2
3
4
OUTPUT:
3
4
I suppose the issue here is that you want to remove all lines before the match, but not the matching line itself?
One way, with GNU sed, is to explicitly add a print for the matching line first:
pattrn=3
seq 1 4 | sed -e "/$pattrn/p;1,/$pattrn/d"
Though this will duplicate any further lines that match the pattern.
Better, invert the sense of the match:
seq 1 4 | sed -ne "/$pattrn/,\$p"
That is, don't print by default (-n), but print (p) anything from a match to the end ($, escaped because of the double-quoted string)
Even better would be to use awk:
pattrn=3
seq 1 4 | awk -vpat="$pattrn" '$0 ~ pat {p=1} p'
This sets a flag on the line where the whole line ($0) matches the pattern (~ is a regex match), then prints the lines whenever that flag is set.
The awk solution is also better in that special characters in the pattern don't cause issues (at least not as many); in the sed case, if the pattern contains a slash /, it will terminate the regex in the sed code, and cause syntax errors or allow for code injection.
I used seq from GNU coreutils here only to make up the sequence of numbers for input.
Let's say I have hell0 w0rld, I want it to become hell00 w0rld.
I tried sed s/0/00/, but that only replaces 0, it wouldn't work for he1lo wor1d(he11lo wor1d), what can I do so that it replaces any first digit, instead of just 0?
Since you don't want to match just 0, but any digit, you want to use [0-9]. This stands for "any one of the digits 0-9". You put this in parentheses to "capture" it, and in the replacement string, you can add backrefences:
$ sed 's/\([0-9]\)/\1\1/' <<< "he1lo wor1d"
he11lo wor1d
If you want to repeat the first number (as per the title) and not just digit, you append \+ to your character class. This stands for "one or more of these":
$ sed 's/\([0-9]\+\)/\1\1/' <<< "he12o wor1d"
he1212o wor1d
An alternative to the backreferences \1, which match the capture group /(.../), would be to use &, which stands for the complete match, i.e.,
sed 's/[0-9]/&&/' <<< "he1lo wor1d"
and
sed 's/[0-9]\+/&&/' <<< "he12lo wor1d"
where the /(.../) are not needed any longer.
I know I can translate upper to lower case letters by
echo 'linux' | tr "a-z" "A-Z"
how would I translate or replace an occurrence of & with %20%26%20. Maybe something like this
echo '&' | tr "&" "%20%26%20"
sed -i 's/&/%20%26%20/g' inputfile
will edit the file in place.
you can just use the shell.
$ var="&test"
$ echo ${var//&/%20%26%20}
%20%26%20test
The reason that tr command does not work is that tr treats it's 2 arguments, not as arbitrary strings, but as lists. tr finds "&" in the string; it's the first element in the first argument; it gets replaced by the corresponding first element in the second argument.
From the man page:
When the -d option is not specified:
o Each input character found in the array specified by
string1 is replaced by the character in the same rela-
tive position in the array specified by string2. When
the array specified by string2 is shorter that the one
specified by string1, the results are unspecified.
The sed and bash solutions offered are what you want.