PHP - 스칼라를 배열 경고로 사용할 수 없습니다.

Question

다음 코드가 있습니다.

 $final = array();
    foreach ($words as $word) {
        $query = "SELECT Something";
        $result = $this->_db->fetchAll($query, "%".$word."%");
        foreach ($result as $row)
        {
            $id = $row['page_id'];
            if (!empty($final[$id][0]))
            {
                $final[$id][0] = $final[$id][0]+3;
            }
            else
            {
                $final[$id][0] = 3;
                $final[$id]['link'] = "/".$row['permalink'];
                $final[$id]['title'] = $row['title'];
            }
        } 
    }

.

코드가 잘 작동하는 것 같지만이 경고를 얻습니다.

Warning: Cannot use a scalar value as an array in line X, Y, Z (the line with: $final[$id][0] = 3, and the next 2).

.

누구 든지이 문제를 해결하는 방법을 말해 줄 수 있습니까?

Answer 1

요소를 추가하기 전에 배열에 $final[$id]를 배열로 설정해야합니다.로 설정하십시오

$final[$id] = array();
$final[$id][0] = 3;
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];

.

또는

$final[$id] = array(0 => 3);
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];

.

Answer 2

A bit late, but to anyone who is wondering why they are getting the "Warning: Cannot use a scalar value as an array" message;

the reason is because somewhere you have first declared your variable with a normal integer or string and then later you are trying to turn it into an array.

hope that helps

Answer 3

The Other Issue I have seen on this is when nesting arrays this tends to throw the warning, consider the following:

$data = [
"rs" => null
]

this above will work absolutely fine when used like:

$data["rs"] =  5;

But the below will throw a warning ::

$data = [
    "rs" => [
       "rs1" => null;
       ]
    ]
..

$data[rs][rs1] = 2; // this will throw the warning unless assigned to an array

Answer 4

Also make sure that you don't declare it an array and then try to assign something else to the array like a string, float, integer. I had that problem. If you do some echos of output I was seeing what I wanted the first time, but not after another pass of the same code.

Answer 5

Make sure that you don't declare it as a integer, float, string or boolean before. http://php.net/manual/en/function.is-scalar.php