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在我的 Php 文件中，从 mysql db 中检索值时，如果 db 中存在 ny 行，则它返回正确的值，但是当数据库中没有行时，它通过 json_encode 返回错误值我的 php 代码是：

 <?php

     $con = mysql_connect("localhost","root","");
     if (!$con)
       {
         die('Could not connect: ' . mysql_error());
       }

       mysql_select_db("meetingschedulardata", $con);

        $mdate=$_GET['mdate'];
    //$mdate= '28-April-2013';  
        $date = strtotime($mdate);
    $new_date = date('Y-m-d', $date);


       $sql=mysql_query("select * from meetingdetails where mdate='$new_date'",$con);
    $row=mysql_num_rows($sql);
    if($row===0)
    {
        $output[][]=false;
        print(json_encode($output));
    }
    else
    {
        while($row=mysql_fetch_assoc($sql))
            $output[][]=$row;
        print(json_encode($output));
    }


        mysql_close();

?>

我解析 json 值的 java 代码是：

try{

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://10.0.2.2/meetingschedular/viewdetail.php?mdate="+currentDateandTime);
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        is = entity.getContent();
        }
        catch(Exception e)
        {
            Toast.makeText(getBaseContext(),e.toString() ,Toast.LENGTH_LONG).show();
        }

        //Convert response to string
        try
        {
            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"));
            sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null)
                {
                    sb.append(line + "\n");
                }
                is.close();
                result = sb.toString();                 
        }
        catch(Exception e)
        {
            Toast.makeText(getBaseContext(),e.toString() ,Toast.LENGTH_LONG).show();
        }
        //END Convert response to string

            try
            {               
                JSONArray jArray =  new JSONArray(result);

                    for(int i=0;i<jArray.length();i++)
                    {   
                        JSONArray innerJsonArray = jArray.getJSONArray(i);
                        JSONObject json_data = innerJsonArray.getJSONObject(0);                     
                        String at= json_data.getString("attendees"); 
                        String at1=json_data.getString("title");                    
                        String at2=json_data.getString("mtime");
                        String at3=json_data.getString("venue");
                        s.append(j+"Attendees:"+at);
                        s.append("\nTitle:"+at1);
                        s1.append("Time:"+at2);
                        s1.append("\nVenue:"+at3);

                        r.add(s.toString());
                        r.add(s1.toString());
                        j++;
                    }

                l.setAdapter(new ArrayAdapter<String>(this, android.R.layout.simple_expandable_list_item_1, r));

            }

            catch(Exception e1)
            {

                Toast.makeText(getBaseContext(),e1.toString(),Toast.LENGTH_LONG).show();
            }

如果 json_encode 返回 false 值并显示一些 toast 以响应 false 值，我不知道我将如何处理。请帮助我。:(

4

1 回答 1

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删除 php 代码中的 false 插入，删除 if 块，并简单地允许发送一个空的 json。您似乎已经设置了一个适当的 catch 块来处理您的 Toast 消息。

于 2013-04-29T09:38:48.940 回答