How to create PHP array from MySQL table

Question

I have a MySQL table:

-------------------
|type_id|type_name| 
-------------------
|  22   | Toyota  |
|  22   | Mazda   |
|  23   | Volvo   |
|  23   |  Man    |
|  25   | Scania  |
|  25   | Iveco   |
|  25   | Fiat    |
-------------------

which is created dynamically from user input. I want to create an array from the table using PHP like this:

array(
    '22' => array('Toyota', 'Mazda'),
    '23' => array('Volvo', 'Man'),
    '25' => array('Scania', 'Iveco','Fiat'),
);

where array id will be type_id and elements will be type_name's of the type_id.I tried this:

$results = array();
    while($line = mysqli_fetch_array($result, MYSQL_ASSOC))
    {
    $results[] = $line;
    }   
    print_r($results);

But I am getting:

Array ( 
    [0] => Array ( 
            [type_id] => 22 
            [type_name] => Toyota
        ) 
    [1] => Array ( 
            [type_name] => 22 
            [type_name] => Mazda
        )
    ...so on

Please help!

did you tried anything,paste here. We can't do full code for you. Please show your effort — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 30, 2017 at 7:50
Is the order of the makes within each group important? If so, you need to include some way of defining that order. Usually this is done through the mechanism of a PRIMARY KEY — Strawberry
– Strawberry, Commented Jul 30, 2017 at 8:07
SELECT type_id, GROUP_CONCAT(type_name) FROM TABLENAME GROUP BY type_id this query will give you exactly what you are looking for. PHP processing is not needed. — Milan Chheda
– Milan Chheda, Commented Jul 30, 2017 at 8:14

Death-is-the-real-truth · Accepted Answer · 2017-07-30 08:15:12Z

1

You need to change your code a bit and it will work fine:-

$results = array(); 
while($line = mysqli_fetch_array($result, MYSQL_ASSOC)) { 
  $results[$line['type_id']][] = $line['Type_name'];  //check change here
} 
print_r($results);

Note:- check your column-names and correct them if any mistake is there.

edited Jul 30, 2017 at 8:15

answered Jul 30, 2017 at 8:05

Death-is-the-real-truth

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1 Comment

Death-is-the-real-truth Over a year ago

@BibanCS glad to help you :):)

Death-is-the-real-truth · Accepted Answer · 2017-07-30 08:19:56Z

1

Please replace your db name and table name in the code-

$conn = mysqli_connect("localhost","root","","dbname");
$query ="select * from table";
$exe = mysqli_query($conn,$query);

$input = array();
$type = array();
while ($result = mysqli_fetch_array($exe)) 
{
    if(array_key_exists($result['type_id'], $input))
    {
        $type[] = $result['type_name'];
    }
    else
    {
        $type = array($result['type_name']);
    }
    $input[$result['type_id']] = $type;
}
echo "<pre>";
print_r($input);

edited Jul 30, 2017 at 8:19

Death-is-the-real-truth

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answered Jul 30, 2017 at 8:07

Farsay

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3 Comments

Farsay Over a year ago

is there any mistake on above code ? i would like to know about it. you have several way to do this. this is one of the way. if it is wrong or it doesn't work please let me know.

Death-is-the-real-truth Over a year ago

OP don't need to change his complete code like you.Just a small change in his fetching data will lead him to desired output as he said that he is able to fetch record fine.BTW not a down-voter

Farsay Over a year ago

Thanks mate. In the starting when i see his post he didn't have the code ...later he uploaded it so i couldn't able to see his code....

aidinMC · Accepted Answer · 2017-07-30 08:01:55Z

0

select type_id and Type_name columns from table and in foreach loop do this:

$output = array();
foreach ($result as $item){
    $output[$item['type_id']][] = $item['Type_name'];
}

answered Jul 30, 2017 at 8:01

aidinMC

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