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Home Backend Development PHP Tutorial php接收android发送的json数据?

php接收android发送的json数据?

Jun 23, 2016 pm 02:20 PM

android json php

$response = array();
include("conn.php");
$conn=mysql_open();
$sql =mysql_query('set names utf8'); 
mysql_query($sql);
if (isset($_POST['name'])) {……………………为甚么会提示未定义的name呢?name就是android发送过来的

回复讨论(解决方案)

你不是json么?你发送的内容是什么?你这么问别人怎么知道呢.出错的信息又是什么呢?就是个提示吧

你不是json么?你发送的内容是什么?你这么问别人怎么知道呢.出错的信息又是什么呢?就是个提示吧 谢谢斑竹提示。android端
params.add(new BasicNameValuePair("name", name));
JSONObject json = jsonParser.makeHttpRequest(url_up, "POST", params);以post方式发送
那php端$_POST['name']为什么得不到数据呢?我调试的时候是先运行android程序,然后再打开浏览器,显示没有接收到数据,数据库中也未成功插入数据

引用 1 楼 PhpNewnew 的回复:你不是json么?你发送的内容是什么?你这么问别人怎么知道呢.出错的信息又是什么呢?就是个提示吧谢谢斑竹提示。android端
params.add(new BasicNameValuePair("name", name));
JSONObject json = jsonParser.makeHttpRequest(url_u……

看是否获取到name,你可以讲值保存到文件中。因为安卓程序一打开就调用你的php文件了,你再打开文件,肯定妹纸了

先运行android程序,然后再打开浏览器

???

这样能行么?

引用 2 楼 aa77a 的回复:引用 1 楼 PhpNewnew 的回复:你不是json么?你发送的内容是什么?你这么问别人怎么知道呢.出错的信息又是什么呢?就是个提示吧谢谢斑竹提示。android端
params.add(new BasicNameValuePair("name", name));
JSONObject json = jsonParser.make……

事实证明是我的android程序写得有问题,url地址中IP写错了。但也奇怪了,我是笔记本见了一个临时局域网,然后手机连接到这个网络上了,所以我的IP就写的这个网络的IP,但是是事实证明要写电脑所连接的路由的网络的IP。

$data = file_get_contents("php://input");这样才能接收到数据,因为PHP默认只识别application/x-www.form-urlencoded标准的数据类型,因此,对型如text/xml 或者 soap 或者 application/octet-stream 之类的内容无法解析.

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