我想从PHP脚本返回JSON。
我只是回应结果吗?我是否必须设置Content-Type
标题?
答案 0 :(得分:1369)
虽然没有它你通常很好,你可以并且应该设置Content-Type标题:
<?PHP
$data = /** whatever you're serializing **/;
header('Content-Type: application/json');
echo json_encode($data);
如果我没有使用特定的框架,我通常会允许一些请求参数来修改输出行为。它通常用于快速故障排除,不发送标题,或者有时print_r数据有效负载以引人注目(尽管在大多数情况下,它不是必需的)。
答案 1 :(得分:106)
返回JSON的一整套漂亮而清晰的PHP代码是:
$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( $data );
答案 2 :(得分:37)
尝试json_encode对数据进行编码,并使用header('Content-type: application/json');
设置内容类型。
答案 3 :(得分:36)
根据manual on json_encode
,该方法可以返回非字符串( false ):
成功时返回JSON编码的字符串或失败时返回
FALSE
。
当发生这种情况时echo json_encode($data)
将输出空字符串,即invalid JSON。
json_encode
的参数包含非UTF-8字符串,则会失败(并返回false
)。
应该在PHP中捕获此错误情况,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:
$json = json_encode(array("jsonError", json_last_error_msg()));
if ($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 属性的存在表明错误情况,它应该相应地处理。
在生产模式下,最好只向客户端发送一般错误状态,并记录更具体的错误消息以供日后调查。
阅读PHP's Documentation中有关处理JSON错误的更多信息。
答案 4 :(得分:15)
使用header('Content-type: application/json');
设置内容类型,然后回显您的数据。
答案 5 :(得分:11)
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);
?>
答案 6 :(得分:6)
这个问题有很多答案,但是没有一个问题涵盖了返回干净的JSON的整个过程,以及防止JSON响应格式错误的一切。
/*
* 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();
}
参考:
答案 7 :(得分:3)
答案 8 :(得分:3)
<?php
$data = /** whatever you're serializing **/;
header("Content-type: application/json; charset=utf-8");
echo json_encode($data);
?>
答案 9 :(得分:3)
如上所述:
header('Content-Type: application/json');
将完成这项工作。但请记住:
即使你的json包含一些HTML标签,Ajax也没有问题来读取json,即使没有使用这个标头。在这种情况下,您需要将标题设置为application / json。
确保您的文件未以UTF8-BOM编码。此格式在文件顶部添加一个字符,因此header()调用将失败。
答案 10 :(得分:1)
如果您需要从php发送自定义信息的json,您可以在打印任何其他内容之前添加此#include "provedorimagem.h"
#include <QDebug>
provedorImagem::provedorImagem() : QQuickImageProvider(QQuickImageProvider::Image)
{
}
QImage provedorImagem::requestImage(const QString &id, QSize *size, const QSize &requestedSize)
{
if(imagem.isNull())
{
qDebug() << "Erro ao prover a imagem";
}
return imagem;
}
void provedorImagem::carregaImagem(QImage imagemRecebida)
{
imagem = imagemRecebida;
}
,那么您可以打印自己的 import math
...
for i in range(bound):
temp=i
emptyFound = False
for j in range(2):
result=search_combination(data,target+(math.pow(-1, j) * temp))
if result !=[]:
emptyFound = True
break
if emptyFound == True:
break;
return [result,temp]
答案 11 :(得分:1)
这是一个简单的PHP脚本,用于返回男性女性和用户ID,因为当您调用脚本json.php时,json值将是任意随机值。
希望此帮助谢谢
<?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;
?>
答案 12 :(得分:1)
是的,您需要使用echo来显示输出。 Mimetype:application / json
答案 13 :(得分:0)
如果查询数据库并需要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,请查看this tutorial。
答案 14 :(得分:0)
将域对象格式化为JSON的简便方法是使用Marshal Serializer。
然后将数据传递给json_encode
并根据您的需要发送正确的Content-Type标头。
如果您使用的是像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Content-Type为application/javascript
。
或者,如果您需要支持一些非常古老的浏览器,最安全的是text/javascript
。
对于移动应用等所有其他用途,请使用application/json
作为内容类型。
这是一个小例子:
<?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);
答案 15 :(得分:0)
您可以使用此little PHP library。它会发送标题并为您提供一个易于使用的对象。
看起来像:
<?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.
$Json->send();
?>
答案 16 :(得分:0)
一个简单的函数,用于返回带有 HTTP状态代码的 JSON响应。
function json_response($data=null, $httpStatus=200)
{
header_remove();
header("Content-Type: application/json");
header('Status: ' . $httpStatus);
http_response_code($httpStatus);
echo json_encode($data);
}
答案 17 :(得分:0)
无论何时尝试返回API的JSON响应,或者确保您具有正确的标头,并确保您返回有效的JSON数据。
这是示例脚本,可帮助您从PHP数组或 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:
答案 18 :(得分:0)
如果您在 WordPress 中执行此操作,则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,即使您立即退出,它也会始终返回大部分头部。 parse_request
在序列中出现得更早。