How to get single item or filter items?

Questions
1

Hi,

When I use a function node to return items[0] I get no results but when i use items.reverse() it works so I am ensuring it is an array. How can I apply filter and only pass selected items to next node ? And how can i choose only the first item to be sent to the next node?

1 Like
2

Welcome to the community!

There are multiple ways that can be done. Here some examples:

You can simply copy the JSON, click once into the n8n window and the paste (CTRL+v)

3

Thanks a lot Jan ! Now I understand how the JSON is transferred between the nodes.

4

Great! I have it on my to-do list since a long time to create documentation for the Function-Nodes. Until they are in place problems like that are sadly to expect.

5

Will be looking forward to that. I am also interested in how we can manipulate data from 2 separate source in one function node.

6

Do you mean two different inputs or from previous nodes and not just the direct input?

7

Yes, accessing previous node that has execution data and also if possible using 2 separate inputs(currently using merge node with append for this)

8

Accessing previous nodes is possible already. Do sadly not remember the exact syntax right now. Think was something like node("name") but have to check once I am in front of my computer again. Will then write you.

Multiple inputs are not possible yet. But for that you can simply use a Merge-Node in “pass through” in combination with the syntax I write you later.

9

Ok could check now. It is possible like that in the Function-Node:

// $node returns always the data of the first item
items[0].json.nodeVariable = $node["Init Data"].data["myNumber"];
items[1].json.nodeVariable = $node["Init Data"].data["myNumber"];
items[2].json.nodeVariable = $node["Init Data"].data["myNumber"];

// $item allows to access the data of the item with the given index
items[0].json.itemVariable = $item(0).$node["Init Data"].data["myNumber"];
items[1].json.itemVariable = $item(1).$node["Init Data"].data["myNumber"];
items[2].json.itemVariable = $item(2).$node["Init Data"].data["myNumber"];

return items;

Here in a workflow to make that a little bit clearer:

10

Did that solve the problem you had?

11

Hi Jan, sorry for extremely late reply. Yes it did thank you very much

12

No problem. Great to hear! Have a nice day!