PHP有些帮助理解这个！isset

Question

I am trying to understand this simple code, if someone could explain this one section of code to a beginner. Please bear with me on the simple code and trying to understand this.

I understand what the loops are doing, it's just the !isset($value[$values]) part that's tripping me up.

Can someone explain to me what's going on with the !isset and why it's not set?

Example

$array1 = [1, 2, 3];

$array2 = [4, 5, 6];

$array3 = [7, 8, 9];

$value = [];

foreach ([$array1, $array2, $array3] as $arr) {
    foreach ($arr as $values) {

        if (!isset($value[$values])) { // Why is this part of the code not set?
         ...
        }

    }
}

From my understanding is:

1) $value[] Value is an empty array

2) $value[$values] You are passing in $values (1, 2, 3, 4, 5 ...) as the $key of the $value array

3) When I var_dump $value[$values] inside of the inner foreach loop I get Notice: Undefined index: Why am I getting this? The index should be the numbers right?

4) Lastly, why is $value[$values] considered not set?

Answer 1

$value is an empty array. Therefore, any keys that you are calling within that array will not be set, until you assign a value to that key.

$value[$values] is passing [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ], so you are calling $value[1], $value[2], etc.

You are getting undefined index because the index is undefined. PHP handles this well and issues a Notice. Other languages (Java, for example) would issue a Null Pointer Exception.

And lastly, because $value is an empty array, and that key is not set, isset() returns false. And !isset() would then return true, thus the code would execute.