JSON对象后到PHP脚本Android应用程序 [英] json object post to php script in android app

问题描述

我想这个Android $ C $下发送JSON对象到我的网站

I tried this android code for send json objects to my website

HttpPost httppost = new HttpPost(url);

JSONObject j = new JSONObject();
    j.put("name","name ");

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        String format = s.format(new Date()); 
    nameValuePairs.add(new BasicNameValuePair("msg",j.toString() ));

       httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                               httpclient.execute(httppost);

和这个PHP code

And this php code

<?php

$msg=$_POST["msg"]; 

$filename="androidmessages.html";

file_put_contents($filename,$msg."<br />",FILE_APPEND);

$androidmessages=file_get_contents($filename);

echo $androidmessages;
?>

这会告诉我{名：名} 但如果我用

It will show me {"name":"name "} but if i use

httppost.setHeader( "Content-Type", "application/json" );

它会显示nothing.I的遭遇有关JSON对象前，后都没有，但我认为出事wrong.I要发送一些用户信息到我的网站，并在网页上显示它可以请你告诉我，我需要改变为了克服这个问题 谢谢

it will show nothing.I have no before experince about json object post but i think something went wrong.I want to send some user information to my website and display it in web page can you please tell me what i need to change to overcome this problem Thank you

推荐答案

这取决于你想如何将数据传递给你的PHP脚本。

It depends how you want to pass the data to your PHP-script.

如果你想获得的JSON作为一个字符串在一个变量（或者其他人的补充），那么你不应该使用的内容类型应用/ JSON的。如果你只想发布JSON没有任何变量，您可以执行以下操作来代替：

If you want to get the JSON as a string in one variable (and maybe others in addition), then you should not use the content-type "application/json". If you want to only post the JSON without any variable, you can do the following instead:

HttpPost httppost = new HttpPost(url);
JSONObject j = new JSONObject();
j.put("name","name ");
httppost.setEntity(new StringEntity(j.toString());
httpclient.execute(httppost);

正如你可以从$ C $想象c您没有JSON在一个POST-VAR只，但总。

As you can imagine from the code you do not have the JSON in one POST-var only but in total.

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