I'm writing helm testcase in shell script.
I have passed 3 arguments to this function. in awk '{print $2}' I want it to print the 2nd column value of previous grep output. But it's taking 2nd argument passed to function as input for $2.
Can anyone please help me to solve this?
function get_selector() {
local cr_name=$1
local cr_type=$2
selector_name=$3
echo "kubectl get $cr_type $cr_name -o yaml -n {{ .Release.Namespace }} | grep $selector_name | awk '{print $2}'"
output = $(kubectl get $cr_type $cr_name -o yaml -n {{ .Release.Namespace }} | grep $selector_name | awk '{print $2}')
echo $output
}
Related
I have created a little shellscript that is capable of receiving a list of values such as "MY_VAR_NAME=var_value MY_VAR_NAME2=value2 ...", separated by spaces only. There should be also the possibility to use values such as MY_VAR_NAME='' or MY_VAR_NAME= (nothing).
These values are then used to change the value inside a environment variables file, for example, MY_VAR_NAME=var_value would make the script change the MY_VAR_NAME value inside the .env file to var_value, without changing anything else about the file.
The env file has the following configuration:
NODE_ENV=development
APP_PATH=/media
BASE_URL=http://localhost:3000
ASSETS_PATH=http://localhost:3000
USE_CDN=false
APP_PORT=3000
WEBPACK_PORT=8080
IS_CONNECTED_TO_BACKEND=false
SHOULD_BUILD=false
USE_REDUX_TOOL=false
USE_LOG_OUTPUT_AS_JSON=false
ACCESS_KEY_ID=
SECRET_ACCESS_KEY=
BUCKET_NAME=
BASE_PATH=
MIX_PANEL_KEY=
RDSTATION_KEY=
RESOURCE_KEY=
SHOULD_ENABLE_INTERCOM=false
SHOULD_ENABLE_GTM=false
SHOULD_ENABLE_UTA=false
SHOULD_ENABLE_WOOTRIC=false
I have debugged my script, and found out that this is the point where sometimes it has a problem
cat .envtemp | awk -v var_value="$VAR_VALUE" \
-v var_name="$VAR_NAME" \
-F '=' '$0 !~ var_name {print $0} $0 ~ var_name {print $1"="var_value}' | tee .envtemp
This piece of code sometimes outputs to .envtemp the proper result, while sometimes it just outputs nothing, making .envtemp empty
The complete code i am using is the following:
function change_value(){
VAR_NAME=$1
VAR_VALUE=$2
cat .envtemp | awk -v var_value="$VAR_VALUE" \
-v var_name="$VAR_NAME" \
-F '=' '$0 !~ var_name {print $0} $0 ~ var_name {print $1"="var_value}' | tee .envtemp
ls -l -a .env*
}
function manage_env(){
for VAR in $#
do
var_name=`echo $VAR | awk -F '=' '{print $1}'`
var_value=`echo $VAR | awk -F '=' '{print $2}'`
change_value $var_name $var_value
done
}
function main(){
manage_env $#
cat .envtemp > .env
exit 0
}
main $#
Here is an example script for recreating the error. It does not happen every time, and when it happens, it is not always with the same input.
#!/bin/bash
ENV_MANAGER_INPUT="NODE_ENV=production BASE_URL=http://qa.arquivei.com.br ASSETS_PATH=https://d4m6agb781hapn.cloudfront.net USE_CDN=true WEBPACK_PORT= IS_CONNECTED_TO_BACKEND=true ACCESS_KEY_ID= SECRET_ACCESS_KEY= BUCKET_NAME=frontend-assets-dev BASE_PATH=qa"
cp .env.dist .env
#Removes comment lines. The script needs a .envtemp file.
cat .env.dist | grep -v '#' | grep -v '^$' > .envtemp
./jenkins_env_manager.sh ${ENV_MANAGER_INPUT}
Have you tried use two files:
mv .envtemp .envtemp.tmp
cat .envtemp.tmp | awk ... | tee .envtemp
I'm trying to cut the below string starting on the single quote:
name1=O'Reilly
so it leaves:
name2=Reilly
That's easy from the command line with the following commands:
echo $name | cut -d\' -f
echo $name | awk -F\' '{print $2}'
However when I run these commands from a script the string remains unaltered. I've been looking into problems with using single quotes as a delimiter but couldn't find anything. Any way to solve this issue?
That does not change the string the variable expands to, it just outputs the result of string manipulation.
If you want to create a new reference for variable name, use command substitution to save the result of cut/awk operation as variable name:
% name="O'Reilly"
% echo "$name" | awk -F\' '{print $2}'
Reilly
% name=$(echo "$name" | awk -F\' '{print $2}')
% echo "$name"
Reilly
On the other hand, if you want to declare the input as one (name1), and save the output as a different variable (name2):
% name1="O'Reilly"
% name2=$(echo "$name1" | awk -F\' '{print $2}')
% echo "$name2"
Reilly
This might be easier to get using Parameter expansion though:
$ name="O'Reilly"
$ echo "${name#*\'}"
Reilly
$ name="${name#*\'}"
$ echo "$name"
Reilly
Problem Description
In my Unix ksh system, I am having an environment variable A on doing
echo ${A}
I am getting the desired value which is hello.
I can check this value in
env | grep ${A}
output: A=hello
or in
printenv | grep ${A}
output: A=hello
Now I have a file file which contains the list of environment variables and I have to fetch the corresponding value.
Now what I tried just for only first value of the file.
env | grep $(cat file | awk 'NR==1{print $1}') --shows nothing
b=$(cat file | awk 'NR==1{print $1}')
env | grep echo $b
b=cat TEMP_ES_HOST_MAP | awk -F"|" 'NR==1{print $2 }'
echo $b
c=eval $b
c=echo $b
Nothing seems to be working.
Thank you
You can use xargs:
awk -F '[$()]+' '{print $1$2}' envfile | xargs printenv
Where:
cat envfile
$(LANG)
$LINES
USER
HISTFILESIZE
If you are looking for the variable named A in the output from env and printenv then using grep ${A} is incorrect (and unsafe/does not work for variables of more than one word).
What you want for that is:
env | grep ^A=
printenv | grep ^A=
So assuming your file looks like this:
VAR1
VAR2
OTHER_VAR
and you want to search for those you could use something like this (assuming you have process substitution):
env | grep -f <(awk '{print "^"$0"="}' varfile)
Assuming you don't have process substitution (or you would rather a simpler solution, I do) you can do this:
env | awk -F = 'NR==FNR{a[$1]++; next} $1 in a' varfile -
Actually this should work too and is even simpler (assuming an awk with ENVIRON):
awk '$1 in ENVIRON {print $1"="ENVIRON[$1]}' varfile
My bash script is:
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=echo"$output" | awk '{print $1}'
echo $score
The above script prints just a newline in my console whereas my required output is
$ curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*
?)<\/title>.*/\1/p' | awk '{print $1}'
SA
So, why am I not getting the output from my bash script whereas it works fine in terminal am I using echo"$output" in the wrong way.
#!/bin/bash
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=$( echo "$output" | awk '{ print $1 }' )
echo "$score"
Score variable was probably empty, since your syntax was wrong.
I'm trying to pass a variable to nawk in a bash script, but it's not actually printing the $commentValue variable's contents. Everything works great, except the last part of the printf statement. Thanks!
echo -n "Service Name: "
read serviceName
echo -n "Comment: "
read commentValue
for check in $(grep "CURRENT SERVICE STATE" $nagiosLog |grep -w "$serviceName" | nawk -F": " '{print $2}' |sort -u ) ; do
echo $check | nawk -F";" -v now=$now '{ printf( "[%u]=ACKNOWLEDGE_SVC_PROBLEM;"$1";"$2";2;1;0;admin;$commentValue"\n", now)}' >> $nagiosCommand
done
$commentValue is inside an invocation to nawk, so it is considered as a variable in nawk, not a variable in bash. Since you do not have such a variable in nawk, you won't get anything there. You should first pass the variable "inside" nawk using the -v switch just like you did for the now variable; i.e.:
... | nawk -F";" -v now=$now -v "commentValue=$commentValue"
Note the quotes - they are required in case $commentValue contains whitespace.