Create certain JSON form

Question

The form I require is:

{
    "suggestions": [
        { "value": "United Arab Emirates", "data": "AE" },
        { "value": "United Kingdom",       "data": "UK" },
        { "value": "United States",        "data": "US" }
    ]
}

What I get with this php code:

header("Content-type: application/json");
require 'connect.inc.php';
$mysql = mysqli_connect("$db_host", "$db_username", "$db_pass", "$db_name");
$name = isset($_POST['query']) ? $_POST['query'] : "";


$sql = "SELECT MedicineName, MedicineID FROM medicinetypes WHERE MedicineName LIKE '%$name%'";

$res = mysqli_query($mysql, $sql);
$res = mysqli_fetch_all($res, MYSQLI_ASSOC);
$reply = array();
$reply['suggestions'][]=$res;
echo json_encode($reply);
mysqli_close($mysql);

is this:

{"suggestions":[[{"MedicineName":"Amoxil 500","MedicineID":"1"},{"MedicineName":"Panadol","MedicineID":"2"},{"MedicineName":"Farmako3","MedicineID":"3"},{"MedicineName":"Beer","MedicineID":"4"},{"MedicineName":"Farmako4","MedicineID":"5"},{"MedicineName":"hjkki","MedicineID":"61"},{"MedicineName":"Amoxil 1","MedicineID":"577"},{"MedicineName":"Amoxil 502","MedicineID":"580"},{"MedicineName":"Amoxil 56","MedicineID":"582"},{"MedicineName":"Amoxil 600","MedicineID":"583"},{"MedicineName":"\u03b5\u03bb\u03b1 \u03bc\u03bf\u03c5 \u03bd\u03c4\u03b5","MedicineID":"586"}]]}

That extra [ is causing a problem since it's not in the format I need. How can I do this correctly?

Here's a solution that other user posted: stackoverflow.com/questions/15678355/… — guillefix
– guillefix, Commented Aug 2, 2017 at 10:49
@guillefix — That's for hacking around the problem when the JSON is parsed (assuming it is being parsed in JS). The question is asking about fixing the data at source. — Quentin
– Quentin, Commented Aug 2, 2017 at 10:50

Quentin · Accepted Answer · 2017-08-02 10:49:55Z

6

This:

$reply['suggestions'][]=$res;

Gets $reply['suggestions']
Assigns an empty array to if it one doesn't already exist there
Puts $res (already an array) as the next item in that array

You just want to put $res as the value of $reply['suggestions'].

Remove the [].

$reply['suggestions']=$res;

answered Aug 2, 2017 at 10:49

Quentin

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Byte · Accepted Answer · 2017-08-02 10:51:08Z

2

Try changing $reply['suggestions'][]=$res; to $reply['suggestions']=$res;

answered Aug 2, 2017 at 10:51

Byte

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