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?


当前回答

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。

其他回答

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:

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();
?>

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);

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;
?>