array creates object with index name of variable

Question

I'm trying to create an array with two arrays inside it but when I print it or convert it to json I get an object with the index name of the variable I used to create the sub-arrays.

    $var1 = [];
    $var2 = [];

    if ($stmt = $mysqli->query('SELECT id FROM table1')) {
        while($id = $stmt->fetch_object()) {
            $var1[] = $id;
        }
        $stmt->close();
        unset($id);
    }

    if ($stmt = $mysqli->query('SELECT id FROM table2')) {
        while($id = $stmt->fetch_object()) {
            $var2[] = $id;
        }
    }

    return array($var1, $var2);

When I print it var_dump($result);:

array(2) { [0]=> array(1) { [0]=> object(stdClass)#14 (1) { ["id"]=> string(1) "1" } } [1]=> array(0) { } }

Why is there an object with the index name id?

I just need a simple array containing 2 subarrays and print them inside two select elements:

    $.get('/get?op=4', function (data) {
        var content = data,
            sale    = content[0],
            rent    = content[1];
            alert(sale);
        $('select[name="lists-sale"]').html(sale);
        $('select[name="lists-rent"]').html(rent);
    });

I echo and encode in json what is returned from the function in the /get?op=4 file.

jedrzej.kurylo · Accepted Answer · 2016-01-14 20:51:07Z

3

You're calling $stmt->fetch_object() and you get what you ask for - an object that represents a row of data returned from the database.

Replace

while($id = $stmt->fetch_object()) {
  $var1[] = $id;
}

with

while($row = $stmt->fetch_object()) {
  $var1[] = $row->id;
}

to get an array of IDs.

answered Jan 14, 2016 at 20:51

jedrzej.kurylo

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2 Comments

MajorCaiger Over a year ago

Oooh, you got there just before me :)

Chazy Chaz Over a year ago

Hmm I didn't know that, so that's why I was getting an object and the number as string instead of as an integer... Thanks! (I'll accept in 3 min)

MajorCaiger · Accepted Answer · 2016-01-14 20:55:18Z

0

$stmt->fetch_object() fetches an object containing your results.

The object represents the entire row you are selecting, not an individual column.

If you were selecting multiple columns, your object would contain multiple properties.

edited Jan 14, 2016 at 20:55

answered Jan 14, 2016 at 20:51

MajorCaiger

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4 Comments

Matt Over a year ago

Since when was mysqli depreciated?

MajorCaiger Over a year ago

My bad, got muddle up with mysql_* functions

MajorCaiger Over a year ago

PDO is still better :)

Chazy Chaz Over a year ago

mysqli is not depecreated but I'm thinking on making the switch once I have the time to learn about it.

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