使用 PHP 读取 JSON POST [英] Reading JSON POST using PHP

问题描述

在发布这个问题之前我环顾了很多，所以如果它在另一个帖子上，我很抱歉，这只是我在这里的第二个问题，如果我没有正确格式化这个问题，我很抱歉.

I looked around a lot before posting this question so my apologies if it is on another post and this is only my second quesiton on here so apologies if I don't format this question correctly.

我创建了一个非常简单的 Web 服务，它需要获取 post 值并返回一个 JSON 编码的数组.一切都很好，直到我被告知我需要发布内容类型为 application/json 的表单数据.从那以后，我无法从 Web 服务返回任何值，这肯定与我过滤帖子值的方式有关.

I have a really simple web service that I have created that needs to take post values and return a JSON encoded array. That all worked fine until I was told I would need to post the form data with a content-type of application/json. Since then I cannot return any values from the web service and it is definitely something to do with how I am filtering their post values.

基本上在我的本地设置中，我创建了一个执行以下操作的测试页面 -

Basically in my local setup I have created a test page that does the following -

$curl = curl_init();
curl_setopt($curl, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
curl_setopt($curl, CURLOPT_POSTFIELDS, $data);                                                                  
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);                                                                      
curl_setopt($curl, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($data))                                                                       
);
curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  // Set the url path we want to call
$result = curl_exec($curl);

//see the results
$json=json_decode($result,true);
curl_close($curl);
print_r($json);

在webservice上我有这个(我去掉了一些功能)-

On the webservice I have this (I have stripped out some of the functions) -

<?php

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

/* connect to the db */
$link = mysql_connect('localhost','root','root') or die('Cannot connect to the DB');
mysql_select_db('webservice',$link) or die('Cannot select the DB');

if(isset($_POST['action']) && $_POST['action'] == 'login') {
    $statusCode = array('statusCode'=>1, 'statusDescription'=>'Login Process - Fail');
    $posts[] = array('status'=>$statusCode);
    header('Content-type: application/json');
    echo json_encode($posts);

    /* disconnect from the db */
}
@mysql_close($link);

?>

基本上我知道这是由于未设置 $_POST 值，但我找不到我需要放置的内容而不是 $_POST.我试过json_decode($_POST)、file_get_contents("php://input") 和许多其他方式，但我在黑暗中拍摄了一点.

Basically I know that it is due to the $_POST values not being set but I can't find what I need to put instead of the $_POST. I tried json_decode($_POST), file_get_contents("php://input") and a number of other ways but I was shooting in the dark a bit.

任何帮助将不胜感激.

谢谢，史蒂夫

感谢迈克尔的帮助，这是向前迈出的明确一步，我现在至少在回复帖子时得到了回复......即使它是空的

Thanks Michael for the help, that was a definite step forward I now have at least got a repsonse when I echo the post....even if it is null

更新的 CURL -

  $curl = curl_init();
  curl_setopt($curl, CURLOPT_HTTPHEADER, array('Content-Type: application/json')); 
  curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST"); 
  curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  
  curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($curl, CURLOPT_POSTFIELDS, json_encode($data));

在发布数据的页面上更新了 php -

updated php on the page that the data is posted to -

$inputJSON = file_get_contents('php://input');
$input= json_decode( $inputJSON, TRUE ); //convert JSON into array

print_r(json_encode($input));

正如我所说的，至少我现在看到了一个响应，而之前它返回的是一个空白页面

As I say at least I see a response now wheras prior it was returning a blank page

推荐答案

您的 $_POST 为空.如果您的网络服务器想要查看 json 格式的数据，您需要读取原始输入，然后使用 JSON 解码对其进行解析.

You have empty $_POST. If your web-server wants see data in json-format you need to read the raw input and then parse it with JSON decode.

你需要这样的东西:

$json = file_get_contents('php://input');
$obj = json_decode($json);

您还有用于测试 JSON 通信的错误代码...

Also you have wrong code for testing JSON-communication...

CURLOPT_POSTFIELDS 告诉 curl 将您的参数编码为 application/x-www-form-urlencoded.此处需要 JSON 字符串.

CURLOPT_POSTFIELDS tells curl to encode your parameters as application/x-www-form-urlencoded. You need JSON-string here.

更新

你的测试页面的php代码应该是这样的:

Your php code for test page should be like that:

$data_string = json_encode($data);

$ch = curl_init('http://webservice.local/');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
        'Content-Type: application/json',
        'Content-Length: ' . strlen($data_string))
);

$result = curl_exec($ch);
$result = json_decode($result);
var_dump($result);

同样在您的网络服务页面上，您应该删除 header('Content-type: application/json'); 行之一.它只能被调用一次.

Also on your web-service page you should remove one of the lines header('Content-type: application/json');. It must be called only once.

这篇关于使用 PHP 读取 JSON POST的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！