How to delete a line from a variable?

Question

Inside a function, saved multiple lines of external command output to a variable.
How to delete a line from the variable? Like first line.

I can run let tmp = split(variable, "\n") to assign the variable to a tmp list, then delete an item. Not sure if there is any easy way.

Answer 1

You may try this substitute command to delete, e.g the 3rd line

substitute(var, '\v^((.{-}\n){2}).{-}\n(.*)', '\1\3', '')

The regex works as follows

• ^((.{-}\n){2}) matches the first 2 lines and capture it in group \1
• .{-}\n matches the next line
• (.*) matches the rest of the variable and capture it in group \3

\{-} matches 0 or more of the preceding atom, as few as possible

Answer 2

It depends on how you get the result of your external command but you might be interested in :h systemlist().

It executes the command given as parameter and returns a list containing every lines of the output. You can then use :h list-functions on the result or simply listVar[1:].

I think that is more robust and portable than matching the lines and doing some substitutions.

---

Here is an example:

" Get the result of ls has a list
let output = systemlist('ls')
echo output    " ['autoload', 'CODE_OF_CONDUCT.md', 'CONTRIBUTING.md', 'doc', 'plugin', 'README.md']

" Remove first line
let output = output[1:]
echo output    " ['CODE_OF_CONDUCT.md', 'CONTRIBUTING.md', 'doc', 'plugin', 'README.md']

" Remove README.md
call filter(output, 'v:val != "README.md"')
echo output    " ['CODE_OF_CONDUCT.md', 'CONTRIBUTING.md', 'doc', 'plugin']

" Remove all *.md files
call filter(output, 'v:val !~ ".md"')
echo output    " ['doc', 'plugin']

See

• :h filter()
• :h expr-!=
• :h expr-!~