dongquanjie9328 2012-10-25 07:39
浏览 41
已采纳

不将我在android中的编辑文本值传递给PHP变量

I'm able to connect into my MySQL database but the problem is. My android does not passing the edit text value to PHP with this codes.

public class MySQL extends Activity {

    EditText inputName;
     private static String  url_create_product = "http://atlantis-us.com/phpFile.php?";

     JSONParser jsonParser = new JSONParser();
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_my_sql);
        inputName = (EditText) findViewById(R.id.txt1);

       ((Button) findViewById(R.id.btnsend)).setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            // TODO Auto-generated method stub
            new CreateNewProduct().execute();
        }
    });

    }

    class CreateNewProduct extends AsyncTask<String, String, String> {

        @Override
        protected String doInBackground(String... arg0) {
            // TODO Auto-generated method stub
            String name = inputName.getText().toString();
            List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("name", name));
            JSONObject json = jsonParser.makeHttpRequest(url_create_product,
                    "POST", params);

            return null;
        }

    }

}



this is my PHP codes

<?php

    $DB_HostName = "localhost";
    $DB_Name = "_atlantisdb";
    $DB_User = "atlantis_frux";
    $DB_Pass = "frux2012";
    $DB_Table = "testDB";

    if (isset ($_GET["name"]))
        $name = $_GET["name"];
    else
        $name = "Ghalia";

    $con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error()); 
    mysql_select_db($DB_Name,$con) or die(mysql_error()); 

    $sql = "insert into $DB_Table (name) values('$name');";
    $res = mysql_query($sql,$con) or die(mysql_error());

    mysql_close($con);
    if ($res) {
        echo "success";
    }else{
        echo "faild";
    }// end else

?>

can anyone trace what's worng with my codes? thanks in advance.

  • 写回答

1条回答 默认 最新

  • doushupu2521 2012-10-25 07:45
    关注

    You are sending the edit text value using POST method in your android code and getting the value using method GET in your php code.

    chage $_GET["name"] to $_POST["name"] or $_REQUEST['name']
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 素材场景中光线烘焙后灯光失效
  • ¥15 请教一下各位,为什么我这个没有实现模拟点击
  • ¥15 执行 virtuoso 命令后,界面没有,cadence 启动不起来
  • ¥50 comfyui下连接animatediff节点生成视频质量非常差的原因
  • ¥20 有关区间dp的问题求解
  • ¥15 多电路系统共用电源的串扰问题
  • ¥15 slam rangenet++配置
  • ¥15 有没有研究水声通信方面的帮我改俩matlab代码
  • ¥15 ubuntu子系统密码忘记
  • ¥15 保护模式-系统加载-段寄存器