Random Range on the X Axis

EDIT: Thanks guy you helped me a lot.
By the way maybe you can help me in my other thread:
http://forum.unity3d.com/threads/random-number-only-once.424469/
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Hey guys!
This is my code:
clickButton.GetComponent<RectTransform> ().localPosition = new Vector3 ((Random.Range (-12.0F, 12.0F)) * 0.5f, Random.Range (-10.0F, 10.0F), clickButton.position.z);

This is what im trying to do:
I have an UI image that i want randomally move on my screen everytime i click on it.
Right now what I have is that The image moves on the X Axis between 12 to -12, and on the Y Axis between 10 to -10.

I want to change that, I want to move the image like in the image and not like I explained to you earlier..
I want to move the image on the X Axis between 6 to 12 and between -6 to -12.
For an example:
Random.Range (-6.0F, -12.0F) AND Random.Range (6.0F, 12.0F) BOT IN THE SAME X aXIS SCRIPT.
This is an image for demonstration:


Thanks in advance

 

I think you can just randomize the sign and get the results you want.

Pseudocode:

Code (csharp):

1. float x = Random.Range(6.0f, 12.0f);
2. x *= Random.Range(0, 2) == 0 ? -1 : 1;

 

Can you please explain me this cause i didn't get anything from this line

 

There's bound to be better ways to write this, but could this work?

Code (CSharp):

1.     private float x;
2.  
3.     void ClickOnImage() {
4.         int posOrNegValue = Random.Range(0,2);
5.  
6.         if (posOrNegValue == 0){
7.             x = Random.Range(6f, 12f);
8.         }
9.         if (posOrNegValue == 1) {
10.             x = Random.Range(-6f, -12f);
11.         }
12.  
13.         transform.position = new Vector3(x, 0, 0);
14.     }

This way, you will get either 0 or 1 everytime you click. If you get 0, you random on the positive value, and if you get 1, you random on the negative value.

 

Thanks i will give it a try man

 

It's a ternary operator. It will roll 0 or 1 and compare it to 0. If the result is 0 it will multiply x by -1 otherwise it will multiply x by 1. The result will be a range from -6 to -12 or 6 to 12.

A slightly better way to write Mothil's solution:

Code (csharp):

1. float x = Random.Range(0, 2) == 0 ? Random.Range(-6.0f, -12.0f) : Random.Range(6.0f, 12.0f);

 

Your code works perfectlly !

 

Thanks guy you helped me a lot.
By the way maybe you can help me in my other thread:
http://forum.unity3d.com/threads/random-number-only-once.424469/