How can I do pass an argument directly to other invoke outside of my Main Flowchart?

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1

Good afternoon. Today I have a doubt, if is possible to pass any argument directly to another invoke that is outside of my Main Flowchart, but the invoke that needs the argument is inside in other invoke, but the only way that I can pass the value is this:

1- Invoke #1 that it has the argument that I need to pass, this argument it going to Main Flowchart.
2. In the Main Flowchart, I’m captured the value but, I need to pass this value to next Invoke#2. The Invoke#2 contains inside, the invoke#3, this one needs the value from the Invoke#2, but Invoke#3 it is not in Main flowchart, else it is in invoke#2, as an invoke apart of diagram.

I don’t know if exist, another activity that permit pass argument directly, or redirect from the Invoke that has the value that I need, without pass for each activities and adding another invoke inside of my MainFlowchart. I feel that not is efficient use arguments in each invokes.

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1 Like
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Arguments are meant to the same @efrain.caceres

The solution is

  1. In the main workflow, declare a variable of type which you want the argument to be
  2. In the first invoke workflow, declare variable and the flow should be in. So, assign the main workflow variable to that argument in first workflow
  3. And also declare a variable flow out to send the data out of the workflow. By assigning the output to that variable, you can get it back. Here in the import arguments, get the value and assign it to a variable which is assigned in the main workflow.

Similarly you can do for all

3

Hello
Is it like this?

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4

Good morning, well you have the idea, in my case the variable “a” is out to main, and the invoke three needs the variable"a". But I said my ‘invoke 3’ is not in “main invoke”, else into “invoke 2”.

5

For example the diagram of my partner @ken_infield, I need to pass the variable “a” to “invoke 3”, but my thought is create argument to each invoke, but I was created an argument “out_a” inside of"invoke 1", another argument of type “in_a”. My invoke three is not in the “main” I want say into the diagram, but if is inside to “invoke2”. Well in your explanation, you say that I can create a argument “in” to cacth the value of my “invk 1” at the same time create also, an argument “out” to catching the value of “in” argument, and next to send directly to "invk 3’. Tell me if is correct.

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I only know how to do MAIN → # 2 → # 3.
sorry.

7

Well my friend don’t worry, it’s the same solution that I did. Because you can invoke methods, with regular lines of commands like a other programmer language. In uipath you need to create the variable one by one, in all involucrate parts, and you need to do a chain arguments of type “in” or “out”, for pass the value. And it’s okay, you and me are trying to learning more day by day.