我应该如何在下拉列表中显示多个项目？

Question

I want to display some multiple valus in dropdown with autocomplete

$(document).ready(function () {
    $('#Client-ville_id').typeahead({
        source: function (query, result) {
            $.ajax({
                url: "list_ville.php",
                data: 'query=' + query,
                dataType: "json",
                type: "POST",
                success: function (data) {
                    result($.map(data, function (item) {
                        return {
                            name: item.name,
                            cp:item.departement_code
                        }
                    }));
                }
            });
        }
    });
});

$sql = $conn->prepare("SELECT * FROM ville WHERE name LIKE ?");
$sql->bind_param("s",$search_param);
$sql->execute();
$result = $sql->get_result();
if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        $countryResult[] = $row["name"];
        $countryResult[] = $row["departement_code"];
    }
    echo json_encode($countryResult);
}
$conn->close();

But I get an error

Uncaught TypeError: Cannot read property 'name' of null at (index):531

Answer 1

The problem is that you're creating array with strings in php, you need

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        $countryResult[] = array(
          "name" => $row["name"],
          "departement_code" => $row["departement_code"]
        );
    }
    echo json_encode($countryResult);
}

or just:

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        $countryResult[] = $row;
    }
    echo json_encode($countryResult);
}

there is function that fetch all data in one function call, here is one for mysqli: mysqli_result::fetch_all .

Answer 2

In your sql file do

$sql = $conn->prepare("SELECT * FROM ville WHERE name LIKE ?");
$sql->bind_param("s",$search_param);
$sql->execute();
$result = $sql->get_result();
if ($result->num_rows > 0) {
$i = 0;
    while($row = $result->fetch_assoc()) {
        $countryResult[$i]['name'] = $row["name"];
        $countryResult[$i]['departement_code'] = $row["departement_code"];
        $i++;
    }
    echo json_encode($countryResult);
}
$conn->close();

Answer 3

The problem is that you create a non associative array, but reffer to it as associative.

while($row = $result->fetch_assoc()) {
    $countryResult[] = $row["name"];
    $countryResult[] = $row["departement_code"];
}

After this loop $countryResult array would like something like this

array(6) {
  [0] => "name1"
  [1] => "departement_code1"
  [2] => "name2"
  [3] => "departement_code2"
  [4] => "name3"
  [5] => "departement_code3"
}

You should change the while loop:

while($row = $result->fetch_assoc()) {
    $countryResult[] = $row;
}