How to add a variable with each array element in php? [duplicate]

Question

I want to add variables with each array item but I do not have clue what to do? Here is my code. Can anyone help me, please? Thanks

$str = "Ice Cream has";
$optional= "flavor";
$items = array("vanilla","chocolate","mango");
$itemsCount = count($items);
$sentence = '';
if ($itemsCount == 1) {
    $sentence = $items[0] . '.';
} else {
    $partial = array_slice($items, 0, $itemsCount-1);
    $sentence =   implode(', ', $partial).' '. $optional. ' and ' . $items[$itemsCount-1];
}
            
return $str.': '.$sentence.'.';

I want an output should be:

"Ice Cream has: vanilla flavor, chocolate flavor and mango flavor."

But I am getting output:

"Ice Cream has: vanilla, chocolate flavor and mango ."

How are you getting flavor in the output? You never use $optional anywhere. — Barmar
– Barmar, Commented Jul 8, 2022 at 21:29

Barmar · Accepted Answer · 2022-07-08 21:34:10Z

0

You need to concatenate $optional to all elements of the array, not just the result of implode(). You can use array_map() to create a new array with $optional added to each element.

$str = "Ice Cream has";
$optional= "flavor";
$items = array("vanilla","chocolate","mango");
$items1 = array_map(function($x) use ($optional) { return "$x $optional"; }, $items);
$itemsCount = count($items1);
$sentence = '';
if ($itemsCount == 1) {
    $sentence = $items[0] . '.';
} else {
    $partial = array_slice($items1, 0, $itemsCount-1);
    $sentence =   implode(', ', $partial). ' and ' . $items1[$itemsCount-1];
}
            
return $str.': '.$sentence.'.';

answered Jul 8, 2022 at 21:34

Barmar

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3 Comments

ku234 Over a year ago

Thanks. Just a quick help, what if we have $optional as an array like ["best","good","bad"] instead $optional = "flavor".

Barmar Over a year ago

How should the array be used in that case?

Barmar Over a year ago

If you want to loop over all the arrays together, you can pass multiple arrays to array_map(). The corresponding elements of each array will be arguments to the function. array_map(function($x, $option) { ... }, $items, $optional)

Mustafa · Accepted Answer · 2022-07-09 10:54:17Z

0

    $str = "Ice Cream has: ";
    $optional= "flavor";
    $items = array("vanilla","chocolate","mango");
    $count = count($items);
    $c=0;
    if ($count == 1) {
        $str .= $items[0] . ' '.$optional.'.';
    } else {
        foreach( $items as $item){
            
            $c++;
               
            if( $c == $count){
                $str = $str.' and '.$item.' '.$optional.'.';
            }else if($c == $count-1){
                $str = $str.$item.' '.$optional.'';
            }else{
                $str = $str.$item.' '.$optional.', ';
            }
        }
    }
               
    return $str;

edited Jul 9, 2022 at 10:54

answered Jul 8, 2022 at 21:47

Mustafa

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4 Comments

Barmar Over a year ago

If there's only one item in the array, you'll return and item flavor

Barmar Over a year ago

You're not using the $optional variable.

Mustafa Over a year ago

Add if statemant for if count equal 1 and change others with elseif. Change flavor strings with '.$optional.' Im on mobile cant edit now

Barmar Over a year ago

I added the if for you. I'll let you replace flavor with $optional when you get to a real computer.

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