PHP MySQLi从数据库中选择并推送到json

Question

$q = $db->query("SELECT * FROM user");
    while($row = mysqli_fetch_array($q)) {
        $product = array();
        $product['id'] = $row['id'];
        $product['user'] = $row['user'];
        $product['data'] = $row['data'];
    }
    $response["product"] = array();
    array_push($response["product"], $product);

I have been trying to select from a database the entire table and then loop through each result and push it to an array. The above code only seems to put to the array the last item in the table.

Answer 1

You're probably better off doing something like this:

$q = $db->query("SELECT * FROM user");
$response = array();
$response["product"] = array();
while($row = mysqli_fetch_array($q)) {
    $product = array(
        'id' => $row['id'],
        'user' => $row['user'],
        'data' => $row['data'],
    );
    array_push($response["product"], $product);
}

You were only getting the last item because you kept resetting your $response['product'] & $product array.

Answer 2

You need to push into the array for each item. At the moment you overwrite $product each time though the loop. Try this:

$q = $db->query("SELECT * FROM user");
$response["product"] = array();
while($row = mysqli_fetch_array($q)) {
    $product = array();
    $product['id'] = $row['id'];
    $product['user'] = $row['user'];
    $product['data'] = $row['data'];
    array_push($response["product"], $product);
}