I have an inline script where one of the command is the below one. How can i Replace LINUX_PASSWD with a variable in Azure release.
I have added LINUX_PASSWD and FILENAME as variable in Azure release pipeline but they are empty after release
sed -i 's/password/$(LINUX_PASSWD)/g' FILENAME
If you want to replace your variables with their values, use $(varable_name) template. In your case:
sed -i 's/password/$(LINUX_PASSWD)/g' $(FILENAME)
Check the documentation: Understand variable syntax, Using custom variables
To use custom variables in your build and release tasks, simply enclose the variable name in parentheses and precede it with a $ character. For example, if you have a variable named adminUserName, you can insert the current value of that variable into a parameter of a task as $(adminUserName).
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I'm having 2 Variables inside a variable group and their values are as shown below:
cr 2200
tr cd1200
I would like to assign these two variables to a third variable cd as shown below:
cd '2200;cd1200;abc.txt'
I'm trying to use below script but it is showing too many arguments.
Can someone please help in this.
I'm using below code:
cd = \'"$cr";"$tr";abc.txt\'
echo "$cd"
I need output as:
cd = '2200;cd1200;abc.txt'
Using Azure macro syntax ($(<varName>)):
$cd = '$(cr);$(tr);abc.txt'
Note:
PowerShell always needs the $ sigil when accessing a variable - even when assigning to it (unlike in POSIX-compatible shells such as Bash).
Azure's macro syntax - which textually expands references to Azure variables up front, before PowerShell sees the code, is not be confused with PowerShell's subexpression operator ($(...))
How can I assigned the $NewIP variable precalculated in this step to a DevOps pipeline variable called $pipeline_ip?
You should use logging command if you want to assign powershell variable to Azure DevOps variable
echo "##vso[task.setvariable variable=pipeline_ip;]$NewIP"
Update after clarification:
If you use syntax like:
$NewIP = $(pipeline_ip)
Then $(pipeline_ip) would be replaced with the value before script will be executed.
And if you use syntax like
$NewIP = $env:PIPELINE_IP
then you will refer to environment variable and since all DevOps variables are mapped (except secret variables - here you need to express this excplicitly) it would also work.
However, these are two ways of doing that.
You can use two methods:
$NewIP = $(pipeline_ip) Macro syntax variables
$NewIP = $env:PIPELINE_IP Set variables in pipeline
I am using DevOps pipeline to build and deploy to different environments
For one environment I am encountering this issue where i am using a Pipeline Variable with $$ in the value
For Example:
Password pipeline variable with value = $omeCla$$Password
When i deploy it fails and when i check the logs the password is displayed as $omeCla$Password. So basically when $$ are together it drops one $
For all variable i am using regex __VaraibleValue__ and its working fine
I have tried:
$omeCla$\$Password to try and escape and it displays as $omeCla$\$Password . So basically \ doesn't work.
I tried '$omeCla$$Password' to try and escape and it displays as '$omeCla$Password'
I want to keep this value as a normal pipeline variable before review
So basically how can I escape this?
Or should I add a Secret Token here in the replace token task (see screenshot below)? and then make the pipeline variable secret? If so, what should I set for Secret Token? Also, in app.config in my repo what should I use instead of the regex __VariableName__ that I use for normal variables?
The solution was to use 4 $. So if you have $$ together you need to add $$$$
Example: $someCla$$$$Password
#JaneMa-MSFT as requested
https://developercommunity.visualstudio.com/content/problem/1296808/azure-pipeline-how-to-escape-special-characters-in.html
I am passing GitLab variable while running CI/CD pipeline as below.
type - variable
key - password
value - {"a": "abc$def#pqr"}
I am reading it in some GitLab pipeline stage as below.
echo $password
It is showing as below.
{"a":"abc#pqr"}
But I want it to show as below.
{"a":"abc$def#pqr"}
I don't want it to evaluate $def as blank
Note:
I tried with \ escaping and with single quotes too.
I need this to be in json kind of format itself for further use.
Works fine on using double $$ instead of single $
I have a small little script that deals with pipelines on Jenkins. It needs to be able to grab a file from a folder named after the pipeline name.
Most pipeline names follow this pattern: {Name}Pipeline/{Name}Pipeline.properties
However, a few pipelines have a three-digit version number appended, like so: {Name}Pipeline122/{Name}Pipeline122.properties
In my script, I have a line that stores the path to this properties file in a variable: APP_PROPERTY_FILE=/path/to/file/${NAME}Pipeline/${NAME}Pipeline.properties
Herein lies the problem! How can I allow bash to match pipeline names without the version number AND pipeline names with the version number?
Thanks!
I believe that the user want to select a file that has optional element (3 digit number), and store the file name into shell variable.
Two challenges: (1) regular assignment var=/path/to/something* do not perform pathname expansion and (2) regular pattern matching do not support optional elements.
Possible solutions are 'if-then-else' or using extended globs. Both solutions assumed that one of the files exists.
APP_PROPERTY_FILE=/path/to/file/${NAME}Pipeline/${NAME}Pipeline.properties
if [ ! -f "$APP_PROPERY_FILE" ] ; then
APP_PROPERTY_FILE=$(echo /path/to/file/${NAME}Pipeline/${NAME}Pipeline[0-9][-0-9][0-9].properties)
Using extglob can also work.
APP_PROPERTY_FILE=$(shopt -s extglob ; echo /path/to/file/${NAME}Pipeline/${NAME}Pipeline?([0-9][-0-9][0-9]).properties ; echo $1)