java如何制作php代码来显示我发送给服务器的内容？

Question

我试图将jsonarray上载到服务器，然后获取响应文本，以查看服务器对对象做了什么。在安卓方面，我有以下代码：

            HttpPost httppost = new HttpPost("http://10.0.0.2/namedate.php");I am 
            HttpClient httpclient = new DefaultHttpClient();
            httppost.getParams().setParameter("jsonarray", json_a.toString());
            HttpResponse response = httpclient.execute(httppost);
            String responseText = EntityUtils.toString(response.getEntity());
            Log.d("ProviderTester", "The response text is "+ responseText);
            Log.i("JSONInfo","JSON object: " + json_a.toString());

jsonarray在logcat中看起来像这样：

04-10 21:29:53.293: I/JSONInfo(466): JSON object: ["[name=Mike, datetime=2012-04-10 21:29]","[name=Roger, datetime=2012-04-10 21:29]"]

目前，我只是想回显字符串，希望以后能从字符串中取出表格：

   <?php

    echo $_POST['jsonarray'];

   ?>

以下是我从logcat得到的回复：

04-10 22:22:20.033: D/ProviderTester(499): The response text is 

如何修复此问题，以便查看发送到服务器的jsonarray字符串

编辑：当我将Android代码更改为：

            ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("json_a", json_a.toString()));

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            String responseText = EntityUtils.toString(response.getEntity());

然后，我使用php脚本接受的答案在LogCat中得到以下响应：

04-10 23:05:39.833: D/ProviderTester(601): The response text is POST = array (
04-10 23:05:39.833: D/ProviderTester(601):   'json_a' => '[name=Mike, datetime=2012-04-10 21:29]\\",\\"[name=Roger, datetime=2012-03-10 21:29]\\"]\\"]',
04-10 23:05:39.833: D/ProviderTester(601): )
04-10 23:05:39.833: D/ProviderTester(601): GET = array (
04-10 23:05:39.833: D/ProviderTester(601): )
04-10 23:05:39.833: D/ProviderTester(601): request = array (
04-10 23:05:39.833: D/ProviderTester(601):   'Content-Length' => '174',
04-10 23:05:39.833: D/ProviderTester(601):   'Content-Type' => 'application/x-www-form-urlencoded',
04-10 23:05:39.833: D/ProviderTester(601):   'Host' => 'graasdfon.hostei.com',
04-10 23:05:39.833: D/ProviderTester(601):   'Connection' => 'Keep-Alive',
04-10 23:05:39.833: D/ProviderTester(601):   'User-Agent' => 'Apache-HttpClient/UNAVAILABLE (java 1.4)',
04-10 23:05:39.833: D/ProviderTester(601):   'Expect' => '100-Continue',
04-10 23:05:39.833: D/ProviderTester(601): )
04-10 23:05:39.833: D/ProviderTester(601): 
04-10 23:05:39.833: D/ProviderTester(601): <!-- www.000webhost.com Analytics Code -->

现在我只需要弄清楚如何在服务器端处理json。谢谢你的帮助

Answer 1

echo $_POST['jsonarray'];

PHP无法对非标量数据类型进行隐式toString（）表示。您需要进行显式转换

例如，如果要查看所有HTTP变量和请求头，请尝试以下操作：

<?php

$out="POST = " . var_export($_POST, true) . "\n";
$out.="GET = " . var_export($_GET, true) . "\n";
$out.="request = " . var_export(getallheaders(), true) . "\n";
print $out;

?>

（这不会拾取写入STDIN但未作为帖子正确嵌入的内容）

Answer 2

您需要在post的实体中传递json值。试试这个：

StringEntity params =new StringEntity(json_a.toString());
httppost.addHeader("content-type", "application/x-www-form-urlencoded");
httppost.setEntity(params);

而不是：

httppost.getParams().setParameter("jsonarray", json_a.toString());

Answer 3

我不知道是什么导致了您的问题（我不知道Android/Java），但最好的调试方法是在PHP中简单地执行以下操作：

<?php
    var_dump($_POST);
?>

这将导致PHP输出通过POST接收到的所有内容