我想从PHP脚本返回JSON。

我只是重复结果吗?我必须设置内容类型头吗?


用header(' content -type: application/json')设置内容类型;然后对数据进行回显。


你问题的答案在这里,

它说。

JSON文本的MIME媒体类型为 application / json。

所以如果你设置标题为那种类型,并输出你的JSON字符串,它应该工作。


是的,你需要使用echo来显示输出。Mimetype: application / json


尝试json_encode对数据进行编码,并设置带有header的content-type (' content-type: application/json');


虽然你通常没有它也没问题,但你可以也应该设置Content-Type头文件:

<?php
$data = /** whatever you're serializing **/;
header('Content-Type: application/json; charset=utf-8');
echo json_encode($data);

如果我没有使用特定的框架,我通常允许一些请求参数修改输出行为。通常对于快速故障排除,不发送报头或有时print_r数据有效负载来观察它(尽管在大多数情况下,这应该是不必要的)是很有用的。


返回JSON的完整的PHP代码如下:

$option = $_GET['option'];

if ( $option == 1 ) {
    $data = [ 'a', 'b', 'c' ];
    // will encode to JSON array: ["a","b","c"]
    // accessed as example in JavaScript like: result[1] (returns "b")
} else {
    $data = [ 'name' => 'God', 'age' => -1 ];
    // will encode to JSON object: {"name":"God","age":-1}  
    // accessed as example in JavaScript like: result.name or result['name'] (returns "God")
}

header('Content-type: application/json');
echo json_encode( $data );

您可以使用这个小型PHP库。它发送头文件并给你一个容易使用它的对象。

它看起来是这样的:

<?php
// Include the json class
include('includes/json.php');

// Then create the PHP-Json Object to suits your needs

// Set a variable ; var name = {}
$Json = new json('var', 'name'); 
// Fire a callback ; callback({});
$Json = new json('callback', 'name'); 
// Just send a raw JSON ; {}
$Json = new json();

// Build data
$object = new stdClass();
$object->test = 'OK';
$arraytest = array('1','2','3');
$jsonOnly = '{"Hello" : "darling"}';

// Add some content
$Json->add('width', '565px');
$Json->add('You are logged IN');
$Json->add('An_Object', $object);
$Json->add("An_Array",$arraytest);
$Json->add("A_Json",$jsonOnly);

// Finally, send the JSON.

$Json->send();
?>

如果你需要从发送自定义信息的php中获取json,你可以添加这个头('Content-Type: application/json');在打印任何其他东西之前,所以然后你可以打印你的自定义echo '{"monto": "'.$monto[0]->valor.'","moneda":"'.$moneda[0]->nombre.'","simbolo":"'.$moneda[0]->simbolo.'"}';


根据json_encode的手册,该方法可以返回一个非字符串(false):

成功时返回JSON编码字符串,失败时返回FALSE。

当这种情况发生时,echo json_encode($data)将输出空字符串,这是无效的JSON。

如果json_encode的参数包含一个非UTF-8字符串,则json_encode将失败(并返回false)。

这个错误情况应该在PHP中捕获,例如:

<?php
header("Content-Type: application/json");

// Collect what you need in the $data variable.

$json = json_encode($data);
if ($json === false) {
    // Avoid echo of empty string (which is invalid JSON), and
    // JSONify the error message instead:
    $json = json_encode(["jsonError" => json_last_error_msg()]);
    if ($json === false) {
        // This should not happen, but we go all the way now:
        $json = '{"jsonError":"unknown"}';
    }
    // Set HTTP response status code to: 500 - Internal Server Error
    http_response_code(500);
}
echo $json;
?>

然后,接收端当然应该知道jsonError属性的存在表明了一个错误条件,它应该相应地处理这个错误条件。

在生产模式下,最好只向客户端发送一般的错误状态,并记录更具体的错误消息,以便以后进行调查。

在PHP文档中阅读更多关于处理JSON错误的内容。


如上所述:

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

会完成这项工作。但请记住:

即使没有使用这个报头,Ajax读取json也没有问题,除非您的json包含一些HTML标记。在这种情况下,您需要将头文件设置为application/json。 确保您的文件不是用UTF8-BOM编码的。这种格式在文件顶部添加了一个字符,因此header()调用将失败。


设置访问安全性也很好——只需将*替换为您希望能够访问它的域。

<?php
header('Access-Control-Allow-Origin: *');
header('Content-type: application/json');
    $response = array();
    $response[0] = array(
        'id' => '1',
        'value1'=> 'value1',
        'value2'=> 'value2'
    );

echo json_encode($response); 
?>

这里有更多的例子:如何绕过Access-Control-Allow-Origin?


这是一个简单的PHP脚本,返回男性女性和用户id作为json值将是任何随机值,因为你调用脚本json. PHP。

希望能有所帮助,谢谢

<?php
header("Content-type: application/json");
$myObj=new \stdClass();
$myObj->user_id = rand(0, 10);
$myObj->male = rand(0, 5);
$myObj->female = rand(0, 5);
$myJSON = json_encode($myObj);
echo $myJSON;
?>

如果你查询一个数据库,需要JSON格式的结果集,可以这样做:

<?php

$db = mysqli_connect("localhost","root","","mylogs");
//MSG
$query = "SELECT * FROM logs LIMIT 20";
$result = mysqli_query($db, $query);
//Add all records to an array
$rows = array();
while($row = $result->fetch_array()){
    $rows[] = $row;
}
//Return result to jTable
$qryResult = array();
$qryResult['logs'] = $rows;
echo json_encode($qryResult);

mysqli_close($db);

?>

有关使用jQuery解析结果的帮助,请参阅本教程。


将域对象格式化为JSON的一种简单方法是使用Marshal Serializer。 然后将数据传递给json_encode,并根据需要发送正确的Content-Type报头。 如果您正在使用Symfony这样的框架,则不需要手动设置标头。在那里您可以使用JsonResponse。

例如,处理Javascript的正确内容类型应该是application/ Javascript。

或者如果你需要支持一些相当老的浏览器,最安全的是文本/javascript。

对于所有其他用途,如移动应用程序,使用application/json作为内容类型。

这里有一个小例子:

<?php
...
$userCollection = [$user1, $user2, $user3];

$data = Marshal::serializeCollectionCallable(function (User $user) {
    return [
        'username' => $user->getUsername(),
        'email'    => $user->getEmail(),
        'birthday' => $user->getBirthday()->format('Y-m-d'),
        'followers => count($user->getFollowers()),
    ];
}, $userCollection);

header('Content-Type: application/json');
echo json_encode($data);

<?php
$data = /** whatever you're serializing **/;
header("Content-type: application/json; charset=utf-8");
echo json_encode($data);
?>

一个返回带有HTTP状态代码的JSON响应的简单函数。

function json_response($data=null, $httpStatus=200)
{
    header_remove();

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

    http_response_code($httpStatus);

    echo json_encode($data);

    exit();
}

无论何时你试图为API返回JSON响应,或者确保你有适当的标题,也确保你返回一个有效的JSON数据。

下面是示例脚本,它可以帮助您从PHP数组或返回JSON响应 来自JSON文件。

PHP脚本(代码):

<?php

// Set required headers
header('Content-Type: application/json; charset=utf-8');
header('Access-Control-Allow-Origin: *');

/**
 * Example: First
 *
 * Get JSON data from JSON file and retun as JSON response
 */

// Get JSON data from JSON file
$json = file_get_contents('response.json');

// Output, response
echo $json;

/** =. =.=. =.=. =.=. =.=. =.=. =.=. =.=. =.=. =.=. =.  */

/**
 * Example: Second
 *
 * Build JSON data from PHP array and retun as JSON response
 */

// Or build JSON data from array (PHP)
$json_var = [
  'hashtag' => 'HealthMatters',
  'id' => '072b3d65-9168-49fd-a1c1-a4700fc017e0',
  'sentiment' => [
    'negative' => 44,
    'positive' => 56,
  ],
  'total' => '3400',
  'users' => [
    [
      'profile_image_url' => 'http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg',
      'screen_name' => 'rayalrumbel',
      'text' => 'Tweet (A), #HealthMatters because life is cool :) We love this life and want to spend more.',
      'timestamp' => '{{$timestamp}}',
    ],
    [
      'profile_image_url' => 'http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg',
      'screen_name' => 'mikedingdong',
      'text' => 'Tweet (B), #HealthMatters because life is cool :) We love this life and want to spend more.',
      'timestamp' => '{{$timestamp}}',
    ],
    [
      'profile_image_url' => 'http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg',
      'screen_name' => 'ScottMili',
      'text' => 'Tweet (C), #HealthMatters because life is cool :) We love this life and want to spend more.',
      'timestamp' => '{{$timestamp}}',
    ],
    [
      'profile_image_url' => 'http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg',
      'screen_name' => 'yogibawa',
      'text' => 'Tweet (D), #HealthMatters because life is cool :) We love this life and want to spend more.',
      'timestamp' => '{{$timestamp}}',
    ],
  ],
];

// Output, response
echo json_encode($json_var);

JSON文件(JSON数据):

{
    "hashtag": "HealthMatters", 
    "id": "072b3d65-9168-49fd-a1c1-a4700fc017e0", 
    "sentiment": {
        "negative": 44, 
        "positive": 56
    }, 
    "total": "3400", 
    "users": [
        {
            "profile_image_url": "http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg", 
            "screen_name": "rayalrumbel", 
            "text": "Tweet (A), #HealthMatters because life is cool :) We love this life and want to spend more.", 
            "timestamp": "{{$timestamp}}"
        }, 
        {
            "profile_image_url": "http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg", 
            "screen_name": "mikedingdong", 
            "text": "Tweet (B), #HealthMatters because life is cool :) We love this life and want to spend more.", 
            "timestamp": "{{$timestamp}}"
        }, 
        {
            "profile_image_url": "http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg", 
            "screen_name": "ScottMili", 
            "text": "Tweet (C), #HealthMatters because life is cool :) We love this life and want to spend more.", 
            "timestamp": "{{$timestamp}}"
        }, 
        {
            "profile_image_url": "http://a2.twimg.com/profile_images/1285770264/PGP_normal.jpg", 
            "screen_name": "yogibawa", 
            "text": "Tweet (D), #HealthMatters because life is cool :) We love this life and want to spend more.", 
            "timestamp": "{{$timestamp}}"
        }
    ]
}

JSON Screeshot:


这个问题有很多答案,但没有一个涵盖了返回干净JSON的整个过程,以及防止JSON响应变形所需的一切。


/*
 * returnJsonHttpResponse
 * @param $success: Boolean
 * @param $data: Object or Array
 */
function returnJsonHttpResponse($success, $data)
{
    // remove any string that could create an invalid JSON 
    // such as PHP Notice, Warning, logs...
    ob_clean();

    // this will clean up any previously added headers, to start clean
    header_remove(); 

    // Set the content type to JSON and charset 
    // (charset can be set to something else)
    header("Content-type: application/json; charset=utf-8");

    // Set your HTTP response code, 2xx = SUCCESS, 
    // anything else will be error, refer to HTTP documentation
    if ($success) {
        http_response_code(200);
    } else {
        http_response_code(500);
    }
    
    // encode your PHP Object or Array into a JSON string.
    // stdClass or array
    echo json_encode($data);

    // making sure nothing is added
    exit();
}

引用:

response_remove

ob_clean

JSON内容类型

HTTP规范

http_response_code

json_encode


如果你在WordPress中这样做,那么有一个简单的解决方案:

add_action( 'parse_request', function ($wp) {
    $data = /* Your data to serialise. */
    wp_send_json_success($data); /* Returns the data with a success flag. */
    exit(); /* Prevents more response from the server. */
})

请注意,这不在wp_head钩子中,该钩子将始终返回大部分头部,即使您立即退出。parse_request在序列中出现得更早。


如果你想要js对象,使用头content-type:

<?php
$data = /** whatever you're serializing **/;
header('Content-Type: application/json; charset=utf-8');
echo json_encode($data);

如果你只想要json: remove header content-type属性,只需要encode和echo。

<?php
$data = /** whatever you're serializing **/;
echo json_encode($data);