Php verification code, how to deal with it

Php verification code: for ($ I = 0; & nbsp; $ I & lt; 4; & nbsp; $ I ++) {$ rand. = & nbsp; dechex (rand ();} echo & nbsp; $ rand & nbsp; The Not php verification code is displayed in the browser.
Run the following code:
For ($ I = 0; $ I <4; $ I ++ ){
$ Rand. = dechex (rand (1, 15 ));
}
Echo $ rand

The browser displays:
Notice: Undefined variable: rand in D: \ demo. php on line 68
7d7b

But I put. = if it is changed to =, only one random number is displayed, but no error is reported. how can I display four random numbers? I copied the above code online, why can't I execute it here? Thank you for your help !!
------ Solution --------------------
This is not an error. it is a prompt.
It means that your variable $ rand is used without declaration.

Solution:
1. you can declare the value before use, that is, before.
2. for all the error messages of this line on the bi page, add @ before $ rand, that is, @ $ rand .....
3. set the error prompt level globally, error_reporting (......)
4. add a judgment, isset
I personally suggest the first method...