PHP解码JSON POST [英] PHP decode JSON POST

问题描述

我正在尝试以JSON形式接收POST数据.我将其卷曲为:

I"m trying to receive POST data in the form of JSON. I'm curling it as:

curl -v --header 'content-type:application/json' -X POST --data '{"content":"test content","friends":[\"38383\",\"38282\",\"38389\"],"newFriends":0,"expires":"5-20-2013","region":"35-28"}' http://testserver.com/wg/create.php?action=post

在PHP方面，我的代码是:

On the PHP side my code is:

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

    $content    = $data->{'content'};
    $friends    = $data->{'friends'};       // JSON array of FB IDs
    $newFriends = $data->{'newFriends'};
    $expires    = $data->{'expires'};
    $region     = $data->{'region'};    

但是，即使我print_r ( $data)，也什么也没有退还给我.这是处理不带表单的POST的正确方法吗?

But even when I print_r ( $data) nothing gets returned to me. Is this the right way of processing a POST without a form?

推荐答案

您提交的JSON数据无效JSON.

The JSON data you're submitting is not valid JSON.

在外壳程序中使用'时，它不会像您怀疑的那样处理\.

When you use ' in your shell, it will not handle \" as you suspect.

curl -v --header 'content-type:application/json' -X POST --data '{"content":"test content","friends": ["38383","38282","38389"],"newFriends":0,"expires":"5-20-2013","region":"35-28"}'

按预期工作.

<?php
$foo = file_get_contents("php://input");

var_dump(json_decode($foo, true));
?>

输出:

array(5) {
  ["content"]=>
  string(12) "test content"
  ["friends"]=>
  array(3) {
    [0]=>
    string(5) "38383"
    [1]=>
    string(5) "38282"
    [2]=>
    string(5) "38389"
  }
  ["newFriends"]=>
  int(0)
  ["expires"]=>
  string(9) "5-20-2013"
  ["region"]=>
  string(5) "35-28"
}

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