getXPosition
Gets the element's x position.
Since the UI elements in your app are not static but can be moved, your app may need to know where a UI element currently is placed on the screen. All UI elements (button()
, textInput()
, textLabel()
, dropdown()
, checkbox()
, radioButton()
, image()
), can be queried for their current position.
Examples
Am I off the screen?
Check whether the logo has been displayed too close to the right edge.
// Check whether the logo has been displayed too close to the right edge.
image("logo", "http://code.org/images/logo.png");
setPosition("logo", randomNumber(0,320), randomNumber(0,450));
if (isCloseToEdge()){
write("Too Close to Edge");
}
function isCloseToEdge(){
if (getXPosition("logo")>=270) {
return true;
} else{
return false;
}
}
// add the Code.org logo at a random position on the screen and display it's position.
image("logo", "http://code.org/images/logo.png");
setPosition("logo", randomNumber(0,320), randomNumber(0,450));
write(getXPosition("logo"));
Syntax
getXPosition(id);
Parameters
Name | Type | Required? | Description |
---|---|---|---|
id | string | The ID of the UI element to find the x position. Must begin with a letter, contain no spaces, and may contain letters, digits, - and _. |
Returns
Tips
- The screen default size is 320 pixels wide and 450 pixels high, but you can move a UI element off the screen by exceeding those dimensions.
- A UI element can be moved off the screen so getXPosition() can return a negative number if the element is off the screen to the left and getXPosition() can return a number greater than 320 if the element is off the screen to the right.

Found a bug in the documentation? Let us know at support@code.org.