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Question by Razputin · Feb 26, 2014 at 04:20 AM · guiinputkeycodeonkeydown

Make GUI appear on key down

How do I make a GUI appear when a button is pressed? I want to present the player with a list of options they can choose from at any given time.

     void OnGUI(){
         if(Input.GetKeyDown(KeyCode.LeftShift)){
             GUI.Button(new Rect(10,200,1200,100), "Forage");
             GUI.Button(new Rect(10,400,1200,100), "Rest");
             GUI.Button(new Rect(10,600,1200,100), "Wait");
 
         }
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Answer by fafase · Feb 26, 2014 at 05:01 AM

Input should not be used in OnGUI, the alternative is to use Event or Input in Update with a boolean:

 bool guiOn = false;
 void Update(){
    if(Input.GetKeyDown(KeyCode.LeftShift))guiOn = !guiOn;
 }
 void OnGUI(){
    if(guiOn){
        // Your code
    }
 }

using event:

 bool guiOn = false;
 void OnGUI(){
    Event e = Event.current;
    if (e.keyCode == KeyCode.LeftShift)guiOn = !guiOn;
    if(guiOn){// Your code}
 }
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avatar image Razputin · Feb 26, 2014 at 05:23 AM 0
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Oh that's pretty interesting, thanks for the info

avatar image fafase · Feb 26, 2014 at 06:43 AM 0
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The reason comes from how both are handled, Input is based on the operating system while the OnGUI can be run once or more per frame so it is not related to the frame rate. All in all, Input in Update, event in OnGUI (event is not available outside OnGUI).

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