Question

我的日期库有两列（id，name），位于localhost：88。我试图从我的Android应用程序进入它，但在解析Json时，应用程序提供了JSONException。我试过从没有Json解析的php获取数据，只是一个String。但它只获得NULL。这是我的PHP脚本：

<?php

$con = mysqli_connect('127.0.0.1:3306', 'root', '', 'test');

if(mysqli_connect_errno())
    echo 'Failed to connect to MySQL: ' . mysqli_connect_error();

$query = mysqli_query($con, "SELECT * FROM example");
$rows = array();

while($r = mysqli_fetch_assoc($query))
    $rows[] = $r;

print(json_encode($rows));
    mysqli_close($con);     
?>

这是我的方法：

public class MainActivity extends Activity
{
    TextView txt;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main_layout);

        txt = (TextView) findViewById(R.id.textView);

        txt.setText(getDataFromDB());
    }

    public String getDataFromDB()
    {
        try
        {
            HttpPost httpPost = new HttpPost("http://127.0.0.1:88/basicphp/GetData.php");

            HttpClient httpClient = new DefaultHttpClient();
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            final String response = httpClient.execute(httpPost, responseHandler);

            return response.trim();
        }
        catch(Exception e){
            return "error\n" + e.getMessage();
        }
    }
}

当我输入我的浏览器＆＃34; http://localhost:88/basicphp/GetData.php＆＃34;时，我明白了：＆＃34; [{＆＃34; id＆＃34;：＆＃34; 1＆＃34; ＆＃34;名称＆＃34;：＆＃34;约翰＆＃34;}]＆＃34 ;.所以我想在我的TextView上至少得到这个字符串，但TextView只得到null :(那么，我做错了什么？

Answer 1

更改

 HttpPost httpPost = new HttpPost("http://127.0.0.1:88/basicphp/GetData.php");

它到

 HttpPost httpPost = new HttpPost("http://10.0.2.2/basicphp/GetData.php");

访问[问题]：java.net.ConnectException: /127.0.0.1:8080 an android emulator

Android HttpPost从PHP脚本而不是JSONArray获取NULL

1 个答案: