How to get Coordinates of AGVNetwork through Code.

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question

SudheerReddy asked Apr 18, '20 tannerp answered May 5, '20

How to get Coordinates of AGVNetwork through Code.

As shown in below image I drawn simple AGV paths. I would like to get the coordinates of all paths through code. How to achieve this?

Any feedback on this.

Attaching the model for reference.

AGVCoord.fsm

Software Version:

FlexSim 20.1.0

flexsim 20.1.0

agvcoord.fsm (26.6 KiB)

agvcoord.jpg (48.8 KiB)

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tannerp commented · May 04, 2020 at 07:11 PM

Hi @Sudheer R, was Jörg Vogel's answer helpful? If so, please click the red "Accept" button on their answer. If not, feel free to comment back with more information and we'll provide further support. Thanks!

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SudheerReddy tannerp commented · May 05, 2020 at 03:18 PM

@tanner.p I'm trying to find coordinates of all control points through script. Could you please guide me

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Answer 1 · 2020-05-05T15:55:03Z

tannerp answered May 5, '20

In Jörg's answer, he points out where to find the coordinates in the tree. You can use a sampling tool in the Script window to get the values of these nodes as shown in the gif below. The "spatialx" and "spatialy" values make up the start point of the path, then the "spatialsx" and other values are vectors that give the magnitude (length) of the path in that direction.

The code is as follows:

Model.find("AGVNetwork/Path1").as(Object).location.x //Start location (x)

Model.find("AGVNetwork/Path1").as(Object).size.x //Vector magnitude (x)

Hopefully this helps.

agv-path-coordinates.gif (186.7 KiB)

· 2

SudheerReddy commented · May 05, 2020 at 04:02 PM

@tanner.p thanks for your answer. Like if I have more than 10 control points instead of using sampler every time any loop logic?

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tannerp SudheerReddy commented · May 05, 2020 at 04:30 PM

Yes, you can implement a loop if you'd like. Just parse through each of the AGVNetwork subnodes. Something like this:

for(int i=1; i<Model.find("AGVNetwork").subnodes.length; i++) {
   print(Model.find("AGVNetwork").subnodes[i].as(Object).location.x);
}

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