Question

我正在研究酒店API，它需要以JSON形式发送数据，

但我的JSON表单错误，API无效。

我的代码是： -

$json = json_encode([
    'RoomGuests' => [ json_encode(["NoOfAdults"=> 1,"NoOfChild"=> 0,"ChildAge"=> null])],
]);

我的输出是： -

{
"RoomGuests": [
"{\"NoOfAdults\":1,\"NoOfChild\":0,\"ChildAge\":null}"
 ] }

我希望结果如下： -

{
"RoomGuests": [{
    "NoOfAdults": 1,
    "NoOfChild": 0,
    "ChildAge": null
}],  }

请让我知道如何解决这个问题。

Answer 1

使用此

$json = json_encode([
    'RoomGuests' => [["NoOfAdults"=> 1,"NoOfChild"=> 0,"ChildAge"=> null]],
]);

echo $json;

这将导致

{
    "RoomGuests": [
        {
            "NoOfAdults": 1,
            "NoOfChild": 0,
            "ChildAge": null
        }
    ]
}

Answer 2

正确输出而不是：

$json = json_encode([
    'RoomGuests' => [ json_encode(["NoOfAdults"=> 1,"NoOfChild"=> 0,"ChildAge"=> null])],
]);

足以让：

$json = json_encode([
    'RoomGuests' => [ ["NoOfAdults"=> 1,"NoOfChild"=> 0,"ChildAge"=> null] ],
]);

json_encode将递归编码所有子数组。所以只召唤一次就足够了。

Answer 3

这是我的代码，以json格式打印输出。您可以按照您的编辑方式进行编辑：

<?php
include_once('connect.php');
{


    error_reporting( error_reporting() & ~E_NOTICE ); 

            $id = $_GET['id'];
            $name = $_GET['name'];
            $date = $_GET['date'];

   $select = "select * from details";
   $sel = "select id,name from details";

    $res = mysqli_query($con, $select);
   $result = mysqli_query($con,$sel);



while($row = mysqli_fetch_object($res))
{
    $output[] = $row;
}
    if(empty($output))
    {

       print(json_encode("User doesn't exist"));
    }
    else{
        while($col = mysqli_fetch_object($result))
        {
            $output[] = $col;
        }
        if(empty($output))
        {
            print(json_encode("User doesn't exits"));

        }   
        else{
            print(json_encode($output));
        }
    }   
}

?>

Answer 4

1142 : INSERT command denied to user 'readonly'@'localhost' for table 'TempPermissionsTest'

结果：

<?php

$a = (object)[
    'RoomGuests' => [
        (object)["NoOfAdults"=> 1,"NoOfChild"=> 0,"ChildAge"=> null]
    ],
];
var_export(json_encode($a));

如何将数据正确格式化为JSON

4 个答案: