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 ${var//&/%20%26%20}
%20%26%20test
The reason that tr command does not work is that tr treats it's 2 arguments, not as arbitrary strings, but as lists. tr finds "&" in the string; it's the first element in the first argument; it gets replaced by the corresponding first element in the second argument.
From the man page:
When the -d option is not specified:
o Each input character found in the array specified by
string1 is replaced by the character in the same rela-
tive position in the array specified by string2. When
the array specified by string2 is shorter that the one
specified by string1, the results are unspecified.
The sed and bash solutions offered are what you want.
Related
I need to replace one variable with another variable in a multiple strings.
For example:
string1="One,two"
string2="three.four"
string3="five:six"
y=";"
for str in string1 string2 string3; do
x="$(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} => ${x}" # display old and new strings
done
This generates:
One,two => One;two
three.four => three;four
five:six => five;six
To delete particular characters from a combination list.
printf "%s\n" {a..c}{a..d} | sed 's/^cc//' | tr -s '\n'
I used the code above to delete a particular line of character from combination. Is there a way I can do it without sed, awk, grep or bc. Can I get it done with a single line of code in the script?
If you have stored your values in an array, e.g.:
arr=({a..c}{a..d})
Then you may filter your array elements with a string substitution:
printf -- '%s\n' "${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".
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 have strings like following which should be parsed with only unix command (bash)
49_sftp_mac_myfile_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed
I want to trim the strings like above upto 4th underscore from end/right side. So output should be
49_sftp_mac_myfile_simul_test
Number of underscores can vary in overall string. For example, The string could be
49_sftp_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed
Output should be (after trimming up to 4th occurrence of underscore from right.
49_sftp_simul_test
Easily done using awk that decrements NF i.e. no. of fields to -4 after setting input+output field separator as underscore:
s='49_sftp_mac_myfile_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed'
awk 'BEGIN{FS=OFS="_"} {NF -= 4; $1=$1} 1' <<< "$s"
49_sftp_mac_myfile_simul_test
You can use bash's parameter expansion for that:
string="..."
echo "${string%_*_*_*_*}"
With GNU sed:
$ sed -E 's/(_[^_]*){4}$//' <<< "49_sftp_mac_myfile_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed"
49_sftp_mac_myfile_simul_test
From the end of line, removes 4 occurrences of _ followed by non _ characters.
Perl one-liner
echo $your-string | perl -lne '$n++ while /_/g; print join "_",((split/_/)[-$n-1..-5])'
input
49_sftp_mac_myfile_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed
the output
49_sftp_mac_myfile_simul_test
input
49_sftp_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed
the output
49_sftp_simul_test
Not the fastest but maybe the easiest to remember and funiest:
echo "49_sftp_mac_myfile_simul_test_9999_4000000000000001_2017-02-06_15-15-26.49.csv.failed"|
rev | cut -d"_" -f5- | rev
Given a list of paths separated by a single space:
/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b
I want to remove the prefix /home/me/src/ so that the result is:
test vendor/a vendor/b
For a single path I would do: ${PATH#/home/me/src/} but how do I apply it to this series?
You can use // to replace all occurrences of substring. Replace it with null string to remove them.
$ path="/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b"
$ echo ${path//\/home\/me\/src\/}
test vendor/a vendor/b
Reference: ${parameter/pattern/string} in Bash reference manual
Using shell parameter expansion doesn't seem to be the solution for this, since it would remove everything up to / from a given point is useful, as nu11p01n73R's answer reveals.
For clarity, I would use sed with the syntax sed 's#pattern#replacement#g':
$ str="/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b"
$ sed 's#/home/me/src/##g' <<< "$str"
test vendor/a vendor/b
Like always a grep solution from my side :
echo 'your string' | grep -Po '^/([^ /]*/)+\K.+'
Please note that the above regex do this for any string like /x/y/z/test ... But if you are interested only in replacing /home/me/src/, try the following :
echo 'your string' | grep -Po '^/home/me/src/\K.+' --color