How to avoid warning array to string converstion in php when you passing param as string or array to-CodePudding

I am having difficulty passing array as an argument. I am getting an error string to array conversion. the $optional variable sometimes need to get a string as a parameter and sometimes an array. Please check my code below. Many Thanks

  function getLists($str, array $items,$optional=null, ){

        $items1 = array_map(function($x) use ($optional) { return "$x $optional"; }, $items);
        $itemsCount = count($items1);
        $sentence = '';
        if ($itemsCount == 1) {
            $sentence = $items[0] . '.';
        } else {
            if($optional == null)
            {
                $partial = array_slice($items, 0, $itemsCount-1);
                $sentence = implode(', ', $partial) . ' and ' . $items[$itemsCount-1]; 

            }
            if(is_string($optional))
            {
                $partial = array_slice($items1, 0, $itemsCount-1);
                $sentence =   implode(', ', $partial). ' and ' . $items1[$itemsCount-1];
            }
        else
        {
            $partial = array_slice($items1, 0, $itemsCount-1);
            $sentence =   implode(', ', $partial). ' and ' . $items1[$itemsCount-1];
        }              
        }
                    
        return $str.': '.$sentence.'.';
    }

Here are what I am trying to do, the following two are working correctly

getList("What do you want",array("chocolate","icecream","shake"));
getList("Ice Cream has",array("chocolate","vanilla","mango"),"flavour");

But when I replace try to pass [] as parameter then I got an error array to string conversion warning

getList("The color required",array("green","red","blue"),['chilli','onion','berry']);

So when I pass this parameter and my output should be like: I am not getting the correct output that should be like:

The color required: green chilli,red onion and blue berry.

Instead I am getting:

The color required: green Array, red Array and blue Array.

CodePudding user response：

I suggest you to use the gettype function to test the type of $optional before applying any logic https://www.php.net/manual/fr/function.gettype.php

CodePudding user response：

In the first line of the body of your function you have a following statement:

return "$x $optional";

As you said, the $optional variable can be a string or an array. If it's a string there is no problem here. Your string will be inserted there. But if it's an array, then it's a syntax error. You're trying to put the whole array there. Instead you should put just one element of that array.

What you should do, is to replace the first line ($items1 = ...) of your function's body to:

$items1 = [];
if( is_array($optional) ) {
  for($i=0 ; $i < count($items) ; $i  ) {
    $items1[] = "$items[$i] $optional[$i]";
  }
} else {
  $items1 = array_map(function($x) use ($optional) { return "$x $optional"; }, $items);
}