How to push values in array for JSON in php?

Question

In php, I am making an array to be turned into a JSON string.

This works, if I hardcode them:

                $data = array(
                    'firstname' => mysql_result($recordset, 0, 'first name'),
                    'lastname' => mysql_result($recordset, 0, 'last name'),
                    'email' => mysql_result($recordset, 0, 'email address'),
                    'password' => mysql_result($recordset, 0, 'password'),
                    'phone' => mysql_result($recordset, 0, 'mobile number'),
                    'website' => mysql_result($recordset, 0, 'website link'),
                    'type' => mysql_result($recordset, 0, 'type_id'),
                    'active' => mysql_result($recordset, 0, 'active'),
                    'datejoined' => mysql_result($recordset, 0, 'date joined'),
                    'dateleft' => mysql_result($recordset, 0, 'date left'),
                    'datelastactive' => mysql_result($recordset, 0, 'date last active'),
                    'status' => mysql_result($recordset, 0, 'status'),
                    'biotext' => mysql_result($recordset, 0, 'bio text'),
                    'picURL' => $picURL
                );

But if I try a loop

            $data = array();
            for ($i = 0; $i < $num_records; $i++) {
                array_push($data, "location{$i}" => mysql_result($recordset, $i, 'location'));
            }

I get the error

Parse error: syntax error, unexpected T_DOUBLE_ARROW

Does anyone know how to fix this?

Thanks

vladkras · Accepted Answer · 2013-07-24 01:31:20Z

5

try

    $data = array();
    for ($i = 0; $i < $num_records; $i++) {
       $data["location{$i}"] = mysql_result($recordset, $i, 'location');
    }

to make it associative, and then

$myJSON = json_encode($data);

to get JSON

edited Jul 24, 2013 at 1:31

answered Jul 24, 2013 at 1:24

vladkras

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Comments

Martin. · Accepted Answer · 2013-07-24 01:46:54Z

2

You need to put array as a second argument of array_push. Don't use array_push at all. Try something like this instead

$data["location{$i}"] = mysql_result($recordset, $i, 'location')));

edited Jul 24, 2013 at 1:46

answered Jul 24, 2013 at 1:21

Martin.

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2 Comments

vladkras Over a year ago

maybe because you answer is wrong? your example returns Array([0]=>Array([location{0}]=>'somevalue'),[1]=>Array([location{1}]=>'anothervalue'), ...) not Array([location{0}]=>'somevalue',[location{1}]=>'anothervalue', ...)

Martin. Over a year ago

What a begginer mistake -_- Thank you.

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