Get current branch name - bitbucket-pipelines

I'm running a script (bitbucket_pipelines.yml) and on one of the steps I need to know the current branch name, How can I get it?
I saw there is a predefined BITBUCKET_BRANCH variable, but I'm having troubles to print it so I can see its content.
I tried to do:
...
step:
script:
- echo $BITBUCKET_BRANCH
but when pipelines runs all I see is
echo $BITBUCKET_BRANCH
How can I really see the content of this variable?

I found that Bb Pipelines are sometimes picky when dealing with variables. Try changing this to echo "$BITBUCKET_BRANCH". Also, enclosing the whole line in single quotes might help.

#Shvalb, the question should be how to display the value of a variable in bitbucket pipeline.
I deal with bitbucket support on this matter before.
I want to echo a repo/pipeline variable to see the value and it is not showing correctly.
In my case, it was my repo variable conflict with my deployment/pipeline variable. However, from the support, I understand bitbucket is using search and replace the screen value to "hide" the actual value of the variable with direct echo.
in order to see the value, you can use
echo $VAR > /tmpfile
cat /tmpfile
It was the trick I used before but I am not sure whether it will still work.

Related

Azure DevOps Yaml: Gaining secret variable out of Azure KeyVault Task from Variable

I'm trying to obtain a secret out of my KeyVault.
The variable name is secretVar.
Obtaining the secret like this: $(secretVar) works fine however I would like to retrieve it from a variable like this:
I keep getting command not found and I've no idea why this shouldn't be working.
So the name of the secret I want to extract is inside a bash variable. For this question I've simplified the problem but in my real use case I have a bash for loop which loops through secret names and inside the for loop I want to extract the appropriate value from the KeyVault with the corresponding secret name like this:
for secretname in secrets; do
echo $($secretname) # This should contain the value of the secret but gives command not found
done
If anyone has an idea what could be happening, any help is very appreciated.
Thanks in Advance!
Look at the syntax you're using.
variable=secretVar
You are creating an environment variable with the literal value secretVar
Then you try to execute the value of the variable $variable with $($variable). So it tries to run the command secretVar, which obviously doesn't exist, and you get an error message.
The syntax you're looking for is
variable=$(secretVar)
just like you used in the first echo command in the script.
If you don't want to run the variable value as a command, the syntax would be $variable, not $($variable)
$variable is the syntax for a Bash environment variable.
$(variable) is the syntax for referencing Azure DevOps variables.
First of all, the script keyword is a shortcut for the command-line task. The task runs a script using cmd.exe on Windows and Bash on other platforms. You need to pay attention to the agent you are using.
If you want to set variables in scripts, you can use task.setvariable logging command. For example:
- script: |
echo $(secretvar)
echo "##vso[task.setvariable variable=variable]$(secretvar)"
- script: |
echo $(variable)
You can find more detailed information in this document.

Azure DevOps - pipeline variables - special char issue $$

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

How to stop GitLab from evaluating $ inside variable value?

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 $

Return a String from a Windows Batch file

I want to find the target branch when a pull request is submitted on GitHub, in my Jenkins pipeline. To achieve this I am doing the following:
I am invoking a windows batch file from my Jenkinsfile, which in turn invokes a nodejs script. This script internally invokes GitHub APIs to get the target branch which is to be set on some variable in Jenkinsfile(code snippet given below):
Jenkinsfile
env.TARGET_BRANCH = bat "GetTargetBranchFromGit.bat ${env.BRANCH_NAME}"
BatchFile:
node getTargetBranchForPR.js %1
But unfortunately, the variable env.TARGET_BRANCH is not getting set to the target branch even though the nodejs script gets the right value. I am in fact not able to return the value from the batch file. Could someone please help me here?
#npocmaka mention is the right way: How to do I get the output of a shell command executed using into a variable from Jenkinsfile (groovy)?
Accodring to Jenkins' documentation.
returnStdout (optional) If checked, standard output from the task is
returned as the step value as a String, rather than being printed to
the build log. (Standard error, if any, will still be printed to the
log.) You will often want to call .trim() on the result to strip off a
trailing newline.
So your code should look like
env.TARGET_BRANCH = bat( script: "GetTargetBranchFromGit.bat ${env.BRANCH_NAME}",
returnStdout: true
).trim()
If you get back more than expected you probably need to parse it.

returning values in a bash function

I'm working with a growing bash script and within this script I have a number of functions. One of these functions is supposed to return a variables value, but I am running into some issues with the syntax. Below is an example of the code.
ShowTags() {
local tag=0
read tag
echo "$tag"
}
selected_tag=$(ShowTags)
echo "$selected_tag"
pulled this code from a Linux Journal article, but the problem is it doesn't seem to work, or perhaps it does and im missing something. Essentially whenever the function is called the script hangs up and does not output anything, I need to CTRL+C to drop back to CLI.
The article in question is below.
http://www.linuxjournal.com/content/return-values-bash-functions
So my question is this the proper way to return a value? Is there a better or more dependable way of doing this? And if there is please give me an example so I can figure this out without using global variables.
EDIT:
The behavior of this is really getting to me now. I am using the following script.
ShowTags() {
echo "hi"
local tag=0
read tag
echo "$tag"
}
selected_tag=$(ShowTags)
echo "$selected_tag
Basically what happens is bash will act as if the read command is taking place before the echo tag at the top of the function. As soon as I pass something to read though it will run the top echo, and complete the rest of the script. I am not sure why this is happening. This is exactly what is happening in my main script.
Change echo "hi" to echo "hi" >/dev/tty.
The reason you're not seeing it immediately is that $(ShowTags) captures all the standard output of the function, and that gets assigned to selected_tag. So you don't see any of it until you echo that variable.
By redirecting the prompt to /dev/tty, it's always displayed immediately on the terminal, not sent to the function's stdout, so it doesn't get captured by the command substitution.
You are trying to define a function with Name { ... ]. You have to use name() { ... }:
ShowTags() { # add ()
local tag=0
read tag
echo "$tag"
} # End with }
selected_tag=$(ShowTags)
echo "$selected_tag"
It now lets the user type in a string and have it written back:
$ bash myscript
hello world # <- my input
hello world # script's output
You can add a prompt with read -p "Enter tag: " tag to make it more obvious when to write your input.
As #thatotherguy pointed out, your function declaration syntax is off; but I suspect that's a transcription error, as if it was wrong in the script you'd get different problems. I think what's going on is that the read tag command in the function is trying to read a value from standard input (by default that's the terminal), and pausing until you type something in. I'm not sure what it's intended to do, but as written I'd expect it to pause indefinitely until something's typed in.
Solution: either type something in, or use something other than read. You could also add a prompt (read -p "Enter a tag: " tag) to make it more clear what's going on.
BTW, I have a couple of objections to the linux journal article you linked. These aren't relevant to your script, but things you should be aware of.
First, the function keyword is a nonstandard bashism, and I recommend against using it. myfunc() ... is sufficient to introduce a function definition.
Second, and more serious, the article recommends using eval in an unsafe way. Actually, it's really hard to use eval safely (see BashFAQ #48). You can improve it a great deal just by changing the quoting, and even more by not using eval at all:
eval $__resultvar="'$myresult'" # BAD, can evaluate parts of $myresult as executable code
eval $__resultvar='"$myresult"' # better, is only vulnerable to executing $__resultvar
declare $__resultvar="$myresult" # better still
See BashFAQ #6 for more options and discussion.

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