Value as a string from a void function

Processing Libraries
1

Hello,i have these two functions in the end of a code and i call into the draw() the function displayinformation() to show in the screen of Android the characteristics(Name, age,…) everytime i select a specific choice from a list. My problem is that after this procedure i want to send an OscMessage with this characteristics and i do not know how can i add the Information from the void Information function as a string.

void displayinformation(){
  if (selection.equals("1")) {
    Information("John", 30);
  }
  if (selection.equals("2")) {
    Information("Melanie", 40,);
  } 
}

void Information (String name, int age) {
  textSize(60);
  fill(0); 
  text("Name:"+name, 800,500);
  fill(0);
  text("Age:"+age, 850, 600);
}
2

For example i want to put in my code this function but i can not use information as a Sting value:

OscMessage myMessage = new OscMessage ( "Characteristics" + information );
      myMessage.add(information);
    oscP5.send(myMessage, remoteLocation);
3

If you need something out of a function, it can’t return void (nothing)! :wink: You could return a String.
And method (function) names should start lower-case by convention.

String information (String name, int age) {
  textSize(60);
  fill(0); 
  text("Name:"+name, 800,500);
  fill(0);
  text("Age:"+age, 850, 600);
  return "Name: " + name + "\nAge: " + age;
}
4

I have also tried the string function but when i put the name as as string in the function myMessage.add(name) into draw the console shows me that the variable name does not exist.

5

what do you mean with ‘‘should start lower-case by convention’’ ???

6

https://www.oracle.com/technetwork/java/javase/documentation/codeconventions-135099.html

2 Likes
7

ok i understood what does this mean and i am doing all these above, but the problem still exists…
Please help me…

8

Is there a way that i can add in the OscMessage the function displayinformation()???
i just want to send wireless in my PC via oscP5 library whatever is into display() method into draw().
Something like :

OscMessage myMessage = new OscMessage ( "Characteristics" + information );
      myMessage.add(displayinformation);
    oscP5.send(myMessage, remoteLocation);
9

not this way, you use in the creation of “myMessage”
information
displayinformation
in the wrong way.

  • displayinformation does not return a string
  • information might if you call it correctly
    if using @neilcsmith example:
    information(“Anna”,77)
    could do the job by returning “Name: Anna\nAge: 77”
    and put it also to the canvas window.
    this mix might not be wanted.
10

if i use the @neilcsmith example the problem still exists as i said
i fix the code like this:

void displayinformation(){
  if (selection.equals("1")) {
    Information("John", 30);
  }
  if (selection.equals("2")) {
    Information("Melanie", 40,);
  } 
}

String information (String name, int age) {
  textSize(60);
  fill(0); 
  text("Name:"+name, 800,500);
  fill(0);
  text("Age:"+age, 850, 600);
  return "Name: " + name + "\nAge: " + age;
}
}
11

i cant change the format of displayInformation() to string because i call this method into draw() to show me the information(nam, age ,…) in the Android screen.

12

so, why you do this?
and not
myMessage.add(information(“Anna”,77));

13

i know that this wrong but i want to explain you that i want to send with OscMessage whatever is inside the displayinformation, because everytime i change the selection there is another name… I dont want to send one specific name, i want everytime i choose selection 1 to send the name John, everytime i choose the selection 2 to send the name Melanie(etc…)

14

so buffer it:

String thisOne = "";


void displayinformation(){
  if (selection.equals("1")) {
    thisOne = information("John", 30);
  }
  if (selection.equals("2")) {
    thisOne = information("Melanie", 40);
  } 
}

//....
myMessage.add(thisOne);
1 Like
15

Finally worked fine…Thank you so much @kll …It was so simple and easy eventually…
Ι am so grateful!!! :wink::blush::grinning::grin:

16

One more question based on these:
after this procedure i want to take only the value of age and make an if statement inside draw() with the value of age but i don’t know how…
Because in the example above i declared as

String thisOne = “”;
both name and age and if i write into draw() for example if info<40 (i mean if age<40) there is an error about : The operator <= is undefined for the argument type(s) String, int

Any ideas???
@kll @neilcsmith

17

in that case it makes no sense to remember it as a string, like i suggested last time.

String thisOne = "";
String thisName = "";
int thisAge=0;

void displayinformation(){
  if (selection.equals("1")) {
    //thisOne = information("John", 30);
   thisName="John";
   thisAge=30;
  }
  if (selection.equals("2")) {
   // thisOne = information("Melanie", 40);
   thisName="Melanie";
   thisAge=40;

  } 
}

//....
thisOne = information(thisName,thisAge);
myMessage.add(thisOne);


//... draw
if ( thisAge > 40 ) { .. }
1 Like
18

You are perfect…Thank you so so much…You have helped me unconditionally…:grinning::blush::wink::heart_eyes: