Bash Passing variables to string

Question

So, I am trying to create an alias to open the browser in a specific URL for me when I run the pr command on my command line.

function pr() {
  repoName=(basename `git rev-parse --show-toplevel`)
  branchName=(`git rev-parse --abbrev-ref HEAD`)
  $repoName <- printing a-public
  $branchName <- printing acq-248
  open -a "Google Chrome" "https://bitbucket.org/company/$repoName/pull-requests/new?source=$branchName&event_source=branch_list"
}

So, when I run pr, I can see the $repoName and the $branchName has the right value, but the opened url does interprets it partially -> https://bitbucket.org/company/Users/v/projs/a-public/pull-requests/new?source=acq-248&event_source=branch_list

So, what would be the right way of passing these variables to concatenate into the URL I wanna open ?

Why are you tagging this question as zsh and sh and shell, if you are interested in a bash solution only? — user1934428
– user1934428, Commented Mar 5, 2020 at 8:33

Jetchisel · Accepted Answer · 2020-03-04 05:55:01Z

1

both of them are an array assignment.

repoName=(basename `git rev-parse --show-toplevel`)
branchName=(`git rev-parse --abbrev-ref HEAD`)

If you want to expand the command output in a variable use command substitution without the backticks but the $( )

repoName=$(basename $(git rev-parse --show-toplevel))
branchName=$(git rev-parse --abbrev-ref HEAD)

Try quoting the url with the %q so it can be pass to the shell as a safe argument.

var="https://bitbucket.org/company/$repoName/pull-requests/new?source=$branchName&event_source=branch_list"

Quote it shell safe.

printf -v url '%q' "$var"

Now try your command.

open -a "Google Chrome" "$url"

edited Mar 4, 2020 at 5:55

answered Mar 4, 2020 at 5:49

Jetchisel

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8 Comments

vbotio Over a year ago

You killed it man, tks. I didn't know about $() thing. working smoothly!

Jetchisel Over a year ago

Glad to be of help.

Freddy Over a year ago

You should quote the inner command substitution and use lowercase variable names. To make the variables local to the function, use the declare or local keyword: declare repo=$(basename "$(git rev-parse --show-toplevel)")

Jetchisel Over a year ago

@Freddy, right declare the variables local but without the assignment.

Jetchisel Over a year ago

Maybe I'm wrong but here it is. mywiki.wooledge.org/BashPitfalls#local_var.3D.24.28cmd.29

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