PHP Add array to array with string as key

Question

I have a problem. I am using this code:

$sql = "SELECT Id, FileName FROM Templates ORDER BY DateTimeUploaded DESC";

if($result = $conn->query($sql))
{
    if($result->num_rows >= 1) 
    {           
        while($row = $result->fetch_object())
        {
            $arrTotal["Templates"] = array($row);
        }

        $result->free();

    }

    $arrTotal["Source"] = "media/templates/";
    echo json_encode($arrTotal);
}

But when I print the json, the $arrTotal["Templates"] has only one row, but it has 17 rows. What am I doing wrong?

$arrTotal["Templates"][] = array($row); Use double squares to add an entry. This is the shortcut for array_push(). — Markus Zeller
– Markus Zeller, Commented Jan 4, 2020 at 14:39
Yes, like @MarkusZeller said, you overwrite the value. Also, is not neccesary array($row), just $row when you add the object to array. — Maxi Schvindt
– Maxi Schvindt, Commented Jan 4, 2020 at 14:44
You could simplify to if($result->num_rows). Because any number not 0 will end in true. — Markus Zeller
– Markus Zeller, Commented Jan 4, 2020 at 15:03

Calos · Accepted Answer · 2020-01-04 14:40:25Z

1

You repeatedly assign values to an array member, need add [] to avoid this situation:

while($row = $result->fetch_object())
{
    $arrTotal["Templates"][] = array($row);
}

answered Jan 4, 2020 at 14:40

Calos

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2 Comments

u_mulder Over a year ago

Also it's redundant to cast object to array. fetch_assoc already fetches array.

Markus Zeller Over a year ago

@u_mulder This is no cast from object to array. It is wrapping an object in an array. Casting would look like (array) $row. But I agree, it should not be neccessary wrapping the object in an array. So $arrTotal["Templates"][] = $row; should be enough.

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