I want to replace the value between http:// and :80 dynamically everytime with a IP address that I am fetching. I have written below script to get the value and replace the placeholder 'localhost' with the IP address. But I want it to replace value betwen http:// and :80 and not by matching with localhost as it may change.
Line to substitute value in:
"UIApiPath": "http://localhost:80/uisvc/v1/api",
"serviceApiPath": "http://localhost:80/servicesvc/v1/api",
"DBApiPath": "http://locahost:80/DBsvc/v1/api"
Existing script for fetching and replacing value by matching 'localhost':
#!/bin/bash
DEPLOYMENT_NAME=$1
# Getting the pod name from the deployment object
GETTING_POD_NAME=$(kubectl describe deploy $1 | grep 'NewReplicaSet:' | awk '{print $2}')
# Getting the ip-address from pod description
echo ''
GETTING_IP_ADDRESS=$(kubectl describe pod $GETTING_POD_NAME | sed -n '4p' | awk '{print $2}')
# Getting the correct ip-address
SERVER_IP=${GETTING_IP_ADDRESS#*/}
echo $SERVER_IP
# Path of book-store-app configmap file:
APP_CONFIGMAP_FILE_PATH=deployment/java-apps/app-configmap.yaml
# Configure the ip-address in the configmap
sed -i "s/localhost/$SERVER_IP/g" $APP_CONFIGMAP_FILE_PATH
Simple sed command based on // string:
sed "s#//[^:/]\+:80/#//$SERVER_IP:80/#" -i $APP_CONFIGMAP_FILE_PATH
Could do the job! (While $SERVER_IP variable don't hold any # character.)
Explanation
Note: Cited paragraph are copied from info sed pages.
s is normal replacment command under sed
The syntax of the 's' command is 's/REGEXP/REPLACEMENT/FLAGS'.
...
# is used as command separator;
The '/' characters may be uniformly replaced by any other single
character within any given 's' command...
[^:/] mean any character, but no slashes / and no colons :.
A "bracket expression" is a list of characters enclosed by '[' and ']'.
It matches any single character in that list; if the first character of
the list is the caret '^', then it matches any character *not* in the
list.
* mean any number (even zero) of previous character (no slashes nor colons)
'*'
Matches a sequence of zero or more instances of matches for the
preceding regular expression, which must be an ordinary character,
a special character preceded by '\', a '.', a grouped regexp (see
below), or a bracket expression.
...
'\+'
As '*', but matches one or more.
I insist: Please read carefully info sed!
Related
I am trying to get an IP from a string by using sed grouping and I can't figure out what is wrong with this.
input:
echo "instream(10.20.213.11#40266):" | sed -E 's/.*([0-9]+\.[0-9]+\.[0-9]+\.[0-9]+).*/\1/'
output:
0.20.213.11
Why is the first number missing?
You can replace .* that matches greedily to a .*[^0-9] requiring a non-digit char before the subsequent digit matching regex part:
sed -E 's/.*[^0-9]([0-9]+\.[0-9]+\.[0-9]+\.[0-9]+).*/\1/'
It will work since IP addresses in your string are not at the start of the string.
See the online demo:
#!/bin/bash
echo "instream(10.20.213.11#40266):" | \
sed -E 's/.*[^0-9]([0-9]+\.[0-9]+\.[0-9]+\.[0-9]+).*/\1/'
# => 10.20.213.11
If your IP addresses can be at the string start, you can use
sed -E 's/(.*[^0-9]|^)([0-9]+(\.[0-9]+){3}).*/\2/'
See this online demo. Here, (.*[^0-9]|^) matches either any text up to the right-most non-digit or start of string. Now, the IP address matching pattern will land in Group 2, hence is the use of \2 in the replacement (RHS).
If your sed supports word boundaries, consider using word boundaries:
sed -E 's/.*\b([0-9]+(\.[0-9]+){3})\b.*/\1/'
See this online demo, too.
I need to remove the hyphen '-' character only when it matches the pattern 'space-[A-Z]' or '[A-Z]-space'. (Assuming all letters are uppercase, and space could be a space, or newline)
sample.txt
I AM EMPTY-HANDED AND I- WA-
-ANT SOME COO- COOKIES
I want the output to be
I AM EMPTY-HANDED AND I WA
ANT SOME COO COOKIES
I've looked around for answers using sed and awk and perl, but I could only find answers relating to removing all characters between two patterns or specific strings, but not a specific character between [A-Z] and space.
Thanks heaps!!
If perl is your option, would you try the following:
perl -pe 's/(^|(?<=\s))-(?=[A-Z])//g; s/(?<=[A-Z])-((?=\s)|$)//g' sample.txt
(?<=\s) is a zero-width lookbehind assertion which matches leading
whitespace without including it in the matched substring.
(?=[A-Z]) is a zero-width lookahead assertion which matches trailing
character between A and Z without including it in the matched substring.
As a result, only the dash characters which match the pattern above are
removed from the original text.
The second statement s/..//g is the flipped version of the first one.
Could you please try following.
awk '{for(i=1;i<=NF;i++){if($i ~ /^-[a-zA-Z]+$|^[a-zA-Z]+-$/){sub(/-/,"",$i)}}} 1' Input_file
Adding a non-one liner form of solution:
awk '
{
for(i=1;i<=NF;i++){
if($i ~ /^-[a-zA-Z]+$|^[a-zA-Z]+-$/){
sub(/-/,"",$i)
}
}
}
1
' Input_file
Output will be as follows.
I AM EMPTY-HANDED AND I WA
ANT SOME COO COOKIES
If you can provide Extended Regular Expressions to sed (generally with the -E or -r option), then you can shorten your sed expression to:
sed -E 's/(^|\s)-(\w)/\1\2/g;s/(\w)-(\s|$)/\1\2/g' file
Where the basic form is sed -E 's/find1/replace1/g;s/find2/replace2/g' file which can also be written as separate expressions sed -E -e 's/find1/replace1/g' -e 's/find2/replace2/g' (your choice).
The details of s/find1/replace1/g are:
find1 is
(^|\s) locate and capture at the beginning or whitespace,
followed by the '-' hyphen,
then capture the next \w (word-character); and
replace1 is simply \1\2 reinsert both captures with the first two backreferences.
The next substitution expression is similar, except now you are looking for the hyphen followed by a whitespace or at the end. So you have:
find2 being
a capture of \w (word-character),
followed by the hyphen,
followed by a capture of either a following space or the end (\s|$), then
replace2 is the same as before, just reinsert the captured characters using backreferences.
In each case the g indicates a global replace of all occurrences.
(note: the \w word-character also includes the '_' (underscore), so while unlikely you would have a hyphen and underscore together, if you do, you need to use the [A-Za-z] list instead of \w)
Example Use/Output
In your case, then output is:
$ sed -E 's/(^|\s)-(\w)/\1\2/g;s/(\w)-(\s|$)/\1\2/g' file
I AM EMPTY-HANDED AND I WA
ANT SOME COO COOKIES
remove the hyphen '-' character only when it matches the pattern 'space-[A-Z]' or '[A-Z]-space'. Assuming all letters are uppercase, and space could be a space, or newline
It's:
sed 's/\( \|^\)-\([A-Z]\)/\1\2/g; s/\([A-Z]\)-\( \|$\)/\1\2/g'
s - substitute
/
\( \|^\) - space or beginning of the line
- - hyphen...
\(A-Z]\) - a single upper case character
/
\1\2 - The \1 is replaced by the first \(...\) thing. So it is replaced by a space or nothing. \2 is replaced by the single upper case character found. Effectively - is removed.
/
g apply the regex globally
; - separate two s commands
s
Same as above. The $ means end of the line.
awk '{sub(/ -/,"");sub(/^-|-$/,"");sub(/- /," ")}1' file
I AM EMPTY-HANDED AND I WA
ANT SOME COO COOKIES
i found many similar questions about my issue but i still don't find the correct one for me.
I need to grep for the content of a variable plus a dot but it doesn't run escaping the dot after the variable. For example:
The file content is
item.
newitem.
My variable content is item. and i want to grep for the exact word, therefore I must use -w and not -F but with the command I can't obtain the correct output:
cat file | grep -w "$variable\."
Do you have suggestions please?
Hi, I have to rectify my scenario. My file contains some FQDN and for some reasons I have to look for hostname. with the dot.
Unfortunatelly the grep -wF doesn't run:
My file is
hostname1.domain.com
hostname2.domain.com
and the command
cat file | grep -wF hostname1.
doesn't show any output. I have to find another solution and I'm not sure that grep could help.
If $variable contains item., you're searching for item.\. which is not what you want. In fact, you want -F which interprets the pattern literally, not as a regular expression.
var=item.
echo $'item.\nnewitem.' | grep -F "$var"
Try:
grep "\b$word\."
\b: word boundary
\.: the dot itself is a word boundary
Following awk solution may help you in same.
awk -v var="item." '$0==var' Input_file
You are dereferencing variable and append \. to it, which results in calling
cat file | grep -w "item.\.".
Since grep accepts files as parameter, calling grep "item\." file should do.
from man grep
-w, --word-regexp
Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent
character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore.
and
The Backslash Character and Special Expressions
The symbols \< and \> respectively match the empty string at the beginning and end of a word. The symbol \b matches the empty string at the edge of a word, and \B matches the empty string
provided it's not at the edge of a word. The symbol \w is a synonym for [[:alnum:]] and \W is a synonym for [^[:alnum:]].
as the last character is a . it must be followed by a non word [A-Za-z0-9_] however the next character is d
grep '\<hostname1\.'
should work as \< ensures previous chracter is not a word constituent.
You can dynamically construct the search pattern and then call grep
rexp='^hostname1\.'
grep "$rexp" file.txt
The single quotes tell bash not to interpret special characters in the variable. Double quotes tell bash to allow replacing $rexp with its value. The caret ( ^ ) in the expression tells grep to look for lines starting with 'hostname1.'
I've a CSV-file with a few hundred lines and a lot (not all) of these lines contains data (Klas/Lesgroep:;;T2B1) which I want to extract.
i.e. ;;;;;;Klas/Lesgroep:;;T2B1;;;;;;;;;;
I want to delete the semicolons which are in front of Klas/Lesgroep but the number of semicolons is variable. How can I delete these semicolons in Bash ?
I'm not a native speaking Englishman so I hope it's clear to you
To remove any nonempty run of ; chars. that come directly before literal Klas/Lesgroep:
With GNU or BSD/macOS sed:
$ sed -E 's|;+(Klas/Lesgroep)|\1|' <<< ";;;;;;Klas/Lesgroep:;;T2B1;;;;;;;;;;"
Klas/Lesgroep:;;T2B1;;;;;;;;;;
The s function performs string substitution (replacement):
The 1st argument is a regex (regular expression) that specifies what part of the line to match,
and the 2nd arguments specifies what to replace the matching part with.
Note how I've chosen | as the regex/argument delimiter instead of the customary /, because that allows unescaped use of / chars. inside the regex.
;+ matches one or more directly adjacent ; chars.
(Klas/Lesgroep) matches literal Klas/Lesgroep and by enclosing it in (...) - making it a capture group - the match is remembered and can be referenced as \1 - the 1st capture group in the regex - in the replacement argument to s.
The net effect is that all ; chars. directly preceding Klas/Lesgroep are removed.
POSIX-compliant form:
$ sed 's|;\{1,\}\(Klas/Lesgroep\)|\1|' <<< ";;;;;;Klas/Lesgroep:;;T2B1;;;;;;;;;;"
Klas/Lesgroep:;;T2B1;;;;;;;;;;
POSIX requires the less powerful and antiquated BRE syntax, where duplication symbol + must be emulated as \{1,\}, and, generally, metacharacters (, ), {, } must be \-escaped.
With sed you can search for lines starting with at least one semi-colon followed by Klas/Lesgroep and, if found, substitute leading ; with nothing:
$ sed '/;;*Klas\/Lesgroep/s/^;*//g' <<< ";;;;;;Klas/Lesgroep:;;T2B1;;;;;;;;;;"
Klas/Lesgroep:;;T2B1;;;;;;;;;;
To remove all ";" from a file , we can use sed command . sed is used for modifying the files.
$ sed 's/find/replace/g' file
The substitute flag /g (global replacement) specifies the sed command to replace all the occurrences of the string in the line.
So to remove ";" just find and replace it with nothing.
sed 's/;//g' file.csv
I am trying to use sed to extract some assignments being made in a text file. My text file looks like ...
color1=blue
color2=orange
name1.first=Ahmed
name2.first=Sam
name3.first=
name4.first=
name5.first=
name6.first=
Currently, I am using sed to print all the strings after the name#.first's ...
sed 's/name.*.first=//' file
But of course, this also prints all of the lines with no assignment ...
Ahmed
Sam
# I'm just putting this comment here to illustrate the extra carriage returns above; please ignore it
Is there any way I can get sed to ignore the lines with blank or whitespace only assignments and store this to an array? The number of assigned name#.first's is not known, nor are the number of assignments of each type in general.
This is a slight variation on sputnick's answer:
sed -n '/^name[0-9]\.first=\(.\+\)/ s//\1/p'
The first part (/^name[0-9]\.first=\(.\+\)/) selects the lines you want to pass to the s/// command. The empty pattern in the s command re-uses the previous regular expression and the replacement portion (\1) replaces the entire match with the contents of the first parenthesized part of the regex. Use the -n and p flags to control which lines are printed.
sed -n 's/^name[0-9]\.\w\+=\(\w\+\)/\1/p' file
Output
Ahmed
Sam
Explainations
the -n switch suppress the default behavior of sed : printing all lines
s/// is the skeleton for a substitution
^ match the beginning of a line
name literal string
[0-9] a digit alone
\.\w\+ a literal dot (without backslash means any character) followed by a word character [a-zA-Z0-9_] al least one : \+
( ) is a capturing group and \1 is the captured group