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我正在尝试在 PHP 中输​​出 JSON 提要，但我一直遇到一个问题，即 JSON 提要中的第二个结果也包含第一个结果。

以下来源和输出：

来源

function fetch_tour_list($tourID) {
    include('../inc/conn.php');

    $query = mysqli_query($conn,"SELECT * FROM ticket_tour_dates WHERE TourID = $tourID");
    while($result = mysqli_fetch_array($query)) {

        $date['date'] = $result['Date'];
        $venueID = $result['VenueID'];

            $venue_query = mysqli_query($conn,"SELECT * FROM ticket_venues WHERE ID = $venueID");
            while($venue_result = mysqli_fetch_array($venue_query)) {
                $venue['id'] = $venue_result['ID'];
                $venue['name'] = $venue_result['Name'];
                $venue['location'] = $venue_result['Location'];
                $venue['latitude'] = $venue_result['Lat'];
                $venue['longitude'] = $venue_result['Long'];

                $venues[] = $venue;
            }


        $date['venue'] = $venues;
        $dates[] = $date;

    }

    echo json_encode($dates);
    mysqli_close($conn);
}

输出

[{"date":"2013-07-29","venue":[{"id":"1","name":"The Gramercy","location":"New York City","latitude":"50.00000000","longitude":"50.00000000"}]},{"date":"2013-08-02","venue":[{"id":"1","name":"The Gramercy","location":"New York City","latitude":"50.00000000","longitude":"50.00000000"},{"id":"2","name":"The Troubadour","location":"Chicago","latitude":"20.00000000","longitude":"25.00000000"}]}]
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1 回答 1

2

while在内部循环之前添加以下行：

$venues = array();
于 2013-07-21T05:03:26.477 回答