I have command that gives me following output:
'First' : 'abc',
'Second' :'xyz',
'Third' :'lmn'
Requirement here is to convert this output into valid json format.
So I replaced all ' to " using sed :
<command> | sed "s/'/\"/g"
"First" : "abc",
"Second" :"xyz",
"Third" :"lmn"
Now I also need to add { in the begining and end of the output how can I do that.
Any other thoughts are also welcome.
sed -z "s/[[:space:]]*'\([^']*\)'[[:space:]]*:[[:space:]]*'\([^']*\)'[[:space:]]*/"'"\1":"\2",/g; s/,$//; s/^/{/; s/$/}/'
First match the '<this>' : '<and this>'
Then convert each such sequences into "<this>":"<and this>",
Remove trailing comma.
Add { } in front of it.
-z is a GNU extension to parse it all as one line. Alternatively you could remove newlines before passing to sed.
|sed -e '1s/^/{/' -e "s/'/"/g" -e '$s/$/}/' does the work.
Related
I have a array declared in a file and i want to replace/update array with latest values.
below is the array saved in a file.
declare -A IMAGES_OVERRIDE
IMAGES_OVERRIDE=(
[service1]='gcr.io/test-project/image1:latest'
[service2]='gcr.io/test-project/image2:9.5.16'
[service3]='gcr.io/test-project/data/image3:latest'
)
Now I want to update service2 with latest image gcr.io/test-project/image2:10.0.1 and save into file.
I tried like below
sed -i 's/[service2]=.*/[service2]='gcr.io/test-project/image2:10.0.1'/' ./override
but I am getting below error.
sed: -e expression #1, char 35: unknown option to `s'
Same command is working me for other script but that is not array.
Simply:
> sed -i "s#\[service2\]=.*#[service2]='gcr.io/test-project/image2:10.0.1'#" ./override
Notes:
Use " instead of ' around sed expression (your sed script contains ');
Use # instead of / to limit each part of sed replace expression (your new token contains /);
Use \ before each [ and ] in RE expression ([ and ] are special RE characters);
I am trying to insert a variable value into file from Jenkinsfile using shell script in the section
Variable value is dynamic. I am using sed.
Sed is working fine but it is not retaining the white spaces that the variable have at the beginning.
ex:
The value of >> repoName is " somename"
stage('trying sed command') {
steps {
script {
sh """
#!/bin/bash -xel
repo='${repoName}'
echo "\$repo"
`sed -i "5i \$repo" filename`
cat ecr.tf
"""
}
}
}
current output:
names [
"xyz",
"ABC",
somename
"text"
]
Expected output:
names [
"xyz",
"ABC",
somename
"text"
]
How do i retain the spaces infront of the variable passing from sed
With
$ cat filename
names [
"xyz",
"ABC",
"text"
]
$ repo=somename
we can do:
sed -E "3s/^([[:blank:]]*).*/&\\n\\1${repo},/" filename
names [
"xyz",
"ABC",
somename,
"text"
]
That uses capturing parentheses to grab the indentation from the previous line.
if $repo might contain a value with slashes, you can tell the shell to escape them with this (eye-opening) expansion
repo='some/name'
sed -E "3s/^([[:blank:]]*).*/&\\n\\1${repo//\//\\\/},/" filename
names [
"xyz",
"ABC",
some/name,
"text"
]
I used 1 sed statement to add the content first to the file and then another sed statement for just adding spaces. This fixed my issue. All day i was trying to fit in one command did not work probably from Jenkins and shell usage. But using 2 sed commands as a workaround i was able to finish my task
The i command skips over spaces after the i command to find the text to insert. You can put the text on a new line, with a backslash before the newline, to have the initial whitespace preserved.
stage('trying sed command') {
steps {
script {
sh """
#!/bin/bash -xel
repo='${repoName}'
echo "\$repo"
`sed -i "5i \\\\\\
\$repo" filename`
cat ecr.tf
"""
}
}
}
I've tested this from a regular shell command line, I hope it will also work in the Jenkins recipe.
remove from the string 89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz.tar.gz only the .tar.gz part and the result should be 89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz.
It can also happen some files with this extension:
91xhq8vkxlkdfpmfg566qahrwkh01c7n0scpdsr4p4vf6.tbz.tar.bz2 and others with double extension tar.tbz tar.zip and so on ...
In case .tar.zip the result must be nomearchivio.tar in the case 91xhq8vkxlkdfpmfg566qahrwkh01c7n0scpdsr4p4vf6.tbz.tar.bz2 must be 91xhq8vkxlkdfpmfg566qahrwkh01c7n0scpdsr4p4vf6.tbz
I use this :
nameFile= "89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz.tar.gz"
name=${nameFile%.*}
and the result is :
echo $name
89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz.tar
can you help me? Thanks
P.S. note that there are also other points within the file name.
Since you know exactly what you want to remove, just write it in full:
name=${nameFile%.tar.gz}
Or to remove the last two "extensions" .*.*:
name=${nameFile%.*.*}
You could use sed and remove the last 7 characters
echo $nameFile |sed 's/.\{7\}$//'
You could give a try to awk, for example:
echo 89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz.tar.gz | awk -F '\.tgz' '{print $1".tgz"}'
It will output:
89dde7.rqsnhq34h.fmu8s1vn0i94hl.tgz
For other files:
echo "01c7n0scpdsr4p4vf6.tbz.tar.bz2" | awk -F '\.tbz' '{print $1".tbz"}'
It will output:
01c7n0scpdsr4p4vf6.tbz
In this case, awk is using as a delimiter -F '\.tbz' your pattern, .tgz or tbz and then prints all items found at the left + your desired extension.
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
I'm trying find and remove strings like:
[1126604244001,85.00], [1122204245002,85.00], [1221104246003,85.00],
[1222204247004,85.00], [1823304248005,85.00], [1424404249006,85.00],
85.00 = constans. I mean [xxxxxxxxxxxxx,85.00],
In notepad++ is simple:
find: "[^........].............,85.00]" and replace:""
I wolud like to use awk or sed to remove string automaticly without importing it to notepad++.
ok, I have file
temp.txt
[1126604244001,17.00], [1126604244001,17.00], [1126604244001,17.00],
[1126604244001,85.00], [1122204245002,85.00], [1221104246003,85.00],
[1222204247004,85.00], [1823304248005,85.00], [1424404249006,85.00], [1126604244001,17.00], [1126604244001,17.00],
My desire output
temp.txt
[1126604244001,17.00],[1126604244001,17.00],[1126604244001,17.00],[1126604244001,17.00],[1126604244001,17.00],
Thx in advance!
With sed, simply:
sed 's/\[[^]]*,85.00\],[[:space:]]*//g' filename
With this, everything that matches the regex \[[^]]*,85.00\],[[:space:]]* is removed. The regex matches [ followed by an arbitrary number of characters that are not ], followed by ,85.00], and optionally spaces; the only syntactically tricky bit is the [^]] character set which matches all characters other than ].
Alternatively with awk:
awk -v RS='],' -v ORS='],' '!/,85.00$/' filename
This splits the input into records delimited by ], and prints only those that don't end with ,85.00.
egrep -v '[^0-9]85\.00]' YourFile
remove (not empty) line with your pattern