PHP：语法错误帮助；

Question

Can someone help me figure out why I'm getting a syntax error with this function:

function removeFromArray(&$array, $key){
        foreach($array as $j=>$i){
            if($i == $key){
                $array = array_values(unset($array[$j])); //error on this line says expected ;
                return true;
                break;
            }
        }
}

Any help most appreciated!

Jonesy

Answer 1

Remove array_values . It seems you just want to remove one value and unset is already doing the job:

function removeFromArray(&$array, $key){
    foreach($array as $j=>$i){
        if($i == $key){
            unset($array[$j]);
            return true;
        }
    }
}

More about unset .

Demo

---

Side note:

• The code after a return is not executed anymore, so break is unnecessary.
• $key is a misleading variable name here. Better would be $value .

---

Update: If you want to reindex the values of the array (in case you have a numeric array), you have to do it in two steps (as unset does not return a value):

unset($array[$j]);
$array = array_values($array);

Demo

Answer 2

You're trying to use the unset function inside array_values? What exactly are you expecting to happen here?

You should be able to just use: unset($array[$j]);

As you've passed the array in by reference, this should be sufficient to remove it. No need to play with array values.

Answer 3

The problem is the unset. array_values expect an array as parameter, but unset does not have any return value.

Answer 4

I see what you're trying to do, I suggest you use this instead:

function removeFromArray(&$array, $key){
        foreach($array as $j=>$i){
            if($i == $key){
                unset($array[$j]);
            }
        }
}

You don't actually need to return anything. unset is a void function.

Answer 5

取消设置不会返回任何内容：

void unset ( mixed $var [, mixed $var [, mixed $... ]] )