CODEWITHSUNDEEP
phpmysqlarrays
H.HISTORY
H.HISTORY

Reputation: 518

create array from mysql result?

I'm trying to create an array(); from the mysql results/query.

The Array OUTPUT should look exactly like this:

{"1":{"info":"Rooz","lat":51.503363,"lng":-0.127625},
"2":{"info":"Michelle","lat":51.503343,"lng":-0.127567}}

So I tried to do it this:

$sql = "SELECT id, name, lat, lng FROM positions ORDER BY id";
$query = mysqli_query($db_conx, $sql);
$productCount = mysqli_num_rows($query); // count the output amount
$testLocs = array();
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){

$testLocs[] = $row;

}
print_r(json_encode($testLocs));

but the OUTPUT of the code above is like this:

[{"id":"1","name":"Rooz","lat":"51.5033630","lng":"-0.1276250"},{"id":"2","name":"Michelle","lat":"51.503343","lng":"-0.127567"}]

I have no idea how to achieve the output that I want in an array.

Could someone please advise on this?

any help would be appreciated.

Upvotes: 1

Views: 92

Answers (3)

luklapp
luklapp

Reputation: 205

Addition to @Sean's answer: You could also use (array) $row instead of create the array manually with every key:

$testLocs[$row["id"]] = (array) $row;

If you change your query in the future, you don't have to change this line.

Upvotes: 1

Explosion Pills
Explosion Pills

Reputation: 191729

PHP arrays are not built in a distinct way from JSON objects:

$testLocs[$row["id"]] = $row;

If you purposely want to exclude the $id property from $row you can remove it from the $row array first.

$id = $row["id"];
unset($row["id"]);
$testLocs[$id] = $row;

Upvotes: 0

Sean
Sean

Reputation: 12433

Change

$testLocs[] = $row;

to

$testLocs[$row["id"]] = array("info"=>$row["name"], "lat"=>$row["lat"], "lng"=>$row["lng"]);

Upvotes: 3

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