php - 如何使用json检索数据到mysql android

Question

我想使用用户名从用户表中检索数据，但它无法正常工作，这是我的 php 代码:

<?php

mysql_connect("xxx","xxx","xxx");
mysql_select_db("xxx");

$username = $_POST["name"];

$sql = mysql_query("select * from user where username = '$username'");

$flag["code"] = 0;

$check = mysql_fetch_row($sql);
if ($check > 0)
{
    $flag["code"] = 1;
    echo json_encode($flag);    

    while ($row= mysql_fetch_assoc($sql)) 
    {
        $output[] = $row;
    }
}

echo json_encode($output);
mysql_close();

?>

这是我的 Java 代码:

  try
  {
     JSONObject json = new JSONObject(result);
     code = (json.getInt("code"));

if (code == 1)
{
    JSONArray array = new JSONArray(result);

    for (int i = 0; i < array.length(); i++)
    {
        JSONObject jsonObject = array.getJSONObject(i);
        firstname = jsonObject.getString("firstname");

        et.setText(firstname);
    }
}
}
catch (Exception e)
{

}

这是结果:

try
        {
            BufferedReader reader = new BufferedReader(new InputStreamReader(is, HTTP.UTF_8),8);
            StringBuilder sb = new StringBuilder();

            while ((line = reader.readLine()) != null)
            {
                sb.append(line+"\n");
            }

            is.close();

            result = sb.toString();

            Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }

没有错误，但无法检索任何数据，问题是什么？

Answer 1

请从 php 返回 json 对象而不是数组 我已将 echo 部分更改为 json_encode(array('response'=>$output,'code'=>$flag["code"]));

<?php

mysql_connect("xxx","xxx","xxx");
mysql_select_db("xxx");

$username = $_POST["name"];

$rs = mysql_query("select * from user where username = '$username'");


$flag["code"] = 0;

$check = mysql_num_rows($rs);
if ($check > 0)
{
    $flag["code"] = 1;


    while ($row= mysql_fetch_array($rs)) 
    {
        $output[] = $row;
    }
}

echo json_encode(array('response'=>$output,'code'=>$flag["code"])); //changed here
mysql_close();

?>

现在在java中，

try
  {
     JSONObject json = new JSONObject(result);
     code = (json.getInt("code"));

if (code == 1)
{
    JSONArray array = json.getJSONArray('response');

    for (int i = 0; i < array.length(); i++)
    {
        JSONObject jsonObject = array.getJSONObject(i);
        firstname = jsonObject.getString("firstname");

        et.setText(firstname);
    }
}
}
catch (Exception e)
{

}

如果失败 JSONObject json = new JSONObject(result);不是必需的，因为结果已经是 json 对象。只需删除该行并将所有 json 变量更改为结果即可。