Php array from sql result

Question

How can i create an array from the result? I would like to use the array in a mysql IN() query.

//$array = array();
        $get_subscategoria = mysqli_query($kapcs, "SELECT kat_id FROM termek_kategoria WHERE kat_parent = '$id'");
        if(mysqli_num_rows($get_subscategoria) > 0 )
        {
            while($sub_kat = mysqli_fetch_array($get_subscategoria))
            {
                echo $sub_kat['kat_id'];
                //$array[] = $sub_kat;
            }
        }
        //print_r( $array );

Now, this code gives back 4 row ID, that works okay. I would like an array with these ID-s, like 1,2,3,4.

use prepared statements

Rotimi
– Rotimi

2017-02-24 08:55:45 +00:00
Commented Feb 24, 2017 at 8:55 — Rotimi
– Rotimi, Commented Feb 24, 2017 at 8:55
do you need an array or a string?

RST
– RST

2017-02-24 08:59:15 +00:00
Commented Feb 24, 2017 at 8:59 — RST
– RST, Commented Feb 24, 2017 at 8:59
I want to use it in a mysqli query, with an IN() function.

Dave599
– Dave599

2017-02-24 09:04:53 +00:00
Commented Feb 24, 2017 at 9:04 — Dave599
– Dave599, Commented Feb 24, 2017 at 9:04

user2342558 · Accepted Answer · 2017-02-24 08:52:36Z

2

Instead of:

while($sub_kat = mysqli_fetch_array($get_subscategoria))
{
    echo $sub_kat['kat_id'];
    //$array[] = $sub_kat;
}

use:

$array = mysqli_fetch_assoc($get_subscategoria);

answered Feb 24, 2017 at 8:52

user2342558

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Comments

praveena · Accepted Answer · 2017-02-24 09:07:05Z

1

while($sub_kat = mysqli_fetch_array($get_subscategoria))
{
$array[] = $sub_kat['kat_id'];
}
echo implode(",",$array);

it give a result like 1,2,3,4

answered Feb 24, 2017 at 9:07

praveena

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Comments

Blank · Accepted Answer · 2017-02-24 08:59:34Z

1

For MySQL, also you can use group_concat, only gives you one record:

SELECT GROUP_CONCAT(kat_id) AS kat_id
FROM termek_kategoria 
WHERE kat_parent = '$id'
GROUP BY kat_parent

edited Feb 24, 2017 at 8:59

answered Feb 24, 2017 at 8:53

Blank

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2 Comments

Dave599 Over a year ago

It gives Undefined index: kat_id error. But the name in the table is kat_id, thats correct.

Blank Over a year ago

@Dave599 Then use kat_id as alias, check it again.

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