# Print all outputs of Jupyter Cell

**URL:** https://discourse.julialang.org/t/print-all-outputs-of-jupyter-cell/134095
**Category:** General Usage
**Tags:** jupyter
**Created:** [November 24, 2025, 8:16pm UTC](https://discourse.julialang.org/t/print-all-outputs-of-jupyter-cell/134095 "2025-11-24T20:16:41Z")
**Posts on this page:** 2
**Page:** 1

<div class="post-metadata">

### Author: ![Ismam](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/ismam/32/26140_2.png) [@Ismam](https://discourse.julialang.org/u/Ismam)
#### Post date: [November 24, 2025, 8:16pm UTC](https://discourse.julialang.org/t/print-all-outputs-of-jupyter-cell/134095/1 "2025-11-24T20:16:41Z")

</div>

Is there a Julia equivalent to the below python code which enables Jupyter notebooks to print all outputs.

```julia-auto
from IPython.core.interactiveshell import InteractiveShell
InteractiveShell.ast_node_interactivity = "all"

```

I know I can manually use the display function for each output, but I find that too cumbersome.

---

<div class="post-metadata">

### Author: ![stevengj](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/stevengj/32/71_2.png) [@stevengj](https://discourse.julialang.org/u/stevengj)
#### Post date: [November 24, 2025, 10:13pm UTC](https://discourse.julialang.org/t/print-all-outputs-of-jupyter-cell/134095/2 "2025-11-24T22:13:26Z")

</div>

Nothing built-in. But you could define something like

```julia-auto
_displayall(ex::Expr) = Meta.isexpr(ex, :block) ? Expr(:block, map(_displayall, ex.args)...) : :($(Base.display)($ex))
_displayall(ex::LineNumberNode) = ex
_displayall(ex) = :($(Base.display)($ex))
macro displayall(ex)
    :($(esc(_displayall(ex))); nothing)
end

```

which allows you to do:

```julia-auto
@displayall begin
    # block of code, display(...) called on the result of each statement
end

```