Question

我希望将此json输出到php数组：

{
    "data": [
        {
            "id": "1",
            "name": "Roger",
            "country": "Switzerland",
            "city": "Basel"
        },
        {
            "id": "2",
            "name": "Rafael",
            "country": "Spain",
            "city": "Madrid"
        },
    ]
}

我正在尝试：

$arrResult = array();
$arrResult['data'] = array();`

while($row = mysqli_fetch_assoc($result)){`
    $id = $row['id'];
    $name = $row['name'];
    $country = $row['country'];

    $arrResult['data'][] = array(
        'id'=> $id,
        'name'=> $name,
        'country'=> $country,
    );
}

echo json_encode($arrResult, JSON_FORCE_OBJECT);

我希望从php数组获得与JSON格式相同的输出。

Answer 1

如果您使用的是PHP> = 5.2.0

，则可以使用json_decode()

以下是有关如何使用json_decode()

的示例

<?php
  // JSON string
  $someJSON = '[{"name":"Shriram","gender":"male"},{"name":"Lexa","gender":"female"},{"name":"John Doe","gender":"male"}]';

  // Convert JSON string to Array
  $someArray = json_decode($someJSON, true);
  print_r($someArray);        // Dump all data of the Array
  echo $someArray[0]["name"]; // Access Array data

  // Convert JSON string to Object
  $someObject = json_decode($someJSON);
  print_r($someObject);      // Dump all data of the Object
  echo $someObject[0]->name; // Access Object data
?>

Answer 2

您可以使用json_decode函数。例如；

<?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
echo $json;
$array = json_decode($json, true);
print_r($array);
?>

Answer 3

我找到了你的问题。删除json中的逗号（代码中已说明）。

<?php
    $json = '{
        "data": [
            {
                "id": "1",
                "name": "Roger",
                "country": "Switzerland",
                "city": "Basel"
            },
            {
                "id": "2",
                "name": "Rafael",
                "country": "Spain",
                "city": "Madrid"
            }, <--- THIS IS A PROBLEM
        ]
    }';
    /* in order for this function to work you need to delete that comma. 
    Because there isn't another list in json so comma should not be there */

    $arrResult = json_decode($json, true);
    print_r($arrResult);

Answer 4

我了解的是您想将数组编码为所示的json格式。

使用td.fc-other-month { color: white; }代替json_encode($arrResult)。

json_encode($arrResult, JSON_FORCE_OBJECT)输出一个对象而不是一个数组。

我希望将此json输出放入php数组

4 个答案: