Convert a php resultset into an array

Question

I am trying to convert a php resultset into an array (using foreach), I'm not quite doing it right..

I have a resultset:

$result

I am looping through it as below: (horribly wrong)

while($row = mysql_fetch_array($result)) {

    foreach (mysql_fetch_array($result) as $k => $v) {
      echo "Fred's $k is ". $v['uid']. "\n";
    }

}

I want this array to be in the below format:

Array
(
    [0] => Array    //row1
        (
            [column1] => value
            [column2] => value
            [column3] => value
        .
        .

        )

    [1] => Array    //row2
        (
            [column1] => value
            [column2] => value
            [column3] => value
        .
        .
        )

    [2] => Array    //row3
        (
            [column1] => value
            [column2] => value
            [column3] => value
        .
        .
        )

)

I am a newbie, please help.

You should also stop using the mysql_ functions. They are pretty old. Better to use mysqli or PDO. — Evert
– Evert, Commented Dec 15, 2011 at 11:34

Shoogle · Accepted Answer · 2011-12-15 04:45:48Z

6

    $results = array();
    while($row = mysql_fetch_assoc($result))
    {
        $results[] = $row;
    }

    //$results has all that you need
    print_r($results);

answered Dec 15, 2011 at 4:45

Shoogle

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3 Comments

Kannan Lg Over a year ago

thats exactly what i needed.. but is there anyway, that i can store the output of print_r to a variable so i can json_encode it?

Shoogle Over a year ago

Well if your motive is to json_encode it. you can $jsonstr = json_encode($results); Check the link below link

Shoogle Over a year ago

Or check var_export($results) http://www.php.net/manual/en/function.var-export.php

Lucas · Accepted Answer · 2011-12-15 04:46:26Z

2

You don't need a foreach to do this, you can just stick with your while loop:

$result_set = array();

while($row = mysql_fetch_array($result)) {
    $result_set[] = $row;
}

Then you can use it like a normal array:

$result_set[0]['column_name'];

answered Dec 15, 2011 at 4:46

Lucas

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Comments

jogesh_pi · Accepted Answer · 2011-12-15 04:45:36Z

0

i can't understand well but hope this will help you :

$array = array();
while($row = mysql_fetch_array($result)) {
 $array[] = $row['uid'];
}

answered Dec 15, 2011 at 4:45

jogesh_pi

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Comments

KV Prajapati · Accepted Answer · 2011-12-15 04:46:09Z

0

Try this,

while($row = mysql_fetch_assoc($result)) {
   $ar[]=$row;
}

answered Dec 15, 2011 at 4:46

KV Prajapati

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Comments

Ayman Safadi · Accepted Answer · 2011-12-15 04:49:05Z

0

Try mysql_fetch_assoc instead:

$results_array = array();
while($row = mysql_fetch_assoc($result)) {
    $results_array[] = $row;
}

answered Dec 15, 2011 at 4:49

Ayman Safadi

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Comments

mickmackusa · Accepted Answer · 2021-06-01 08:05:31Z

0

As demonstrated here, you do not need to call a fetching function to access the result set values.

$resultArray = [];
foreach ($resultSetObject as $row) {
    $resultArray[] = $row;
}

Assuming you have rows in your result set, this will generate an indexed array of associative arrays.

If you want to be selective about which columns are included in your output array, then you can access them by using associative array syntax to refer to the db table's column names ($row['column_a']).

If you want EVERYTHING to go into your array, then you can using mysqli's fetch_all() to skip the manual loop.

answered Jun 1, 2021 at 8:05

mickmackusa♦

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