Question

它返回的内容如下：

[
      {
        "0": "1",
        "1": "Udon House",
        "2": "food",
        "3": "31.5076701",
        "4": "74.3544522",
        "5": "a trip to the landa.",
        "6": "images/udonHouse.jpg",
        "place_id": "1",
        "place_name": "Udon House",
        "category": "food",
        "lat": "31.5076701",
        "lon": "74.3544522",
        "description": "a trip to the landa.",
        "db_image": "images/udonHouse.jpg"
      },
      {
        "0": "5",
        "1": "Lahore Meuseum",
        "2": "attractions",
        "3": "31.5681556",
        "4": "74.3061492",
        "5": "want to discover histroy?",
        "6": "images/badshahi.jpg",
        "place_id": "5",
        "place_name": "Lahore Meuseum",
        "category": "attractions",
        "lat": "31.5681556",
        "lon": "74.3061492",
        "description": "want to discover histroy?",
        "db_image": "images/badshahi.jpg"
      }
    ]

这是我的PHP代码：

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="tourist"; // Database name 
    $con = mysqli_connect("$host", "$username", "$password","$db_name")or die("cannot connect");

if($con){

    mysqli_set_charset($con ,'utf8');

    $user_id  = $_POST['user_id'];

    $qry =

    ("SELECT p.*
FROM
  places p INNER JOIN
  favorites f ON p.place_id = f.place_id INNER JOIN
  user u ON u.id = f.user_id
WHERE
   u.id = $user_id");

    $query=mysqli_query($con ,$qry);        

    if (!$query) {
                $message  = 'Invalid query: ' . mysqli_error() . "\n";
                $message .= 'Whole query: ' . $qry;
                die($message);
    }

    $return_arr = array();


    $num_rows = mysqli_num_rows($query);
    if ($num_rows > 0) {
        while ($r = mysqli_fetch_array($query)) {


            header('Content-Type: application/json');

            array_push($return_arr,$r);
        }
        echo json_encode($return_arr);
    }
    }

?>

为什么它在一个对象中返回两次行？我只想要名字不带编号的数据我在这里做错了什么？有任何建议或替代解决方案吗？

Answer 1

默认，mysqli_fetch_array函数使用MYSQLI_BOTH作为“结果类型”。

您需要将第二个参数设置为MYSQLI_ASSOC，以使其仅返回键名。

while ($r = mysqli_fetch_array($query, MYSQLI_ASSOC)){
    array_push($return_arr,$r);
}

P.S。 header('Content-Type: application/json');只需要被称为一次，因此请将其从while循环中取出。

数组返回重复的Json数据

1 个答案: