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Java can';t将localhost mysql数据库与android应用程序连接:解析数据org.json.JSONException时出错_Java_Php_Android_Mysql_Json - Fatal编程技术网

Java can';t将localhost mysql数据库与android应用程序连接:解析数据org.json.JSONException时出错

Java can';t将localhost mysql数据库与android应用程序连接:解析数据org.json.JSONException时出错,java,php,android,mysql,json,Java,Php,Android,Mysql,Json,我正在使用JSON和php脚本从mysql数据库获取数据 try { response = CustomHttpClient.executeHttpPost( "http://192.168.1.100:61896/JsonDemoDatabase/src/jsonscript.php", postParam

我正在使用JSON和php脚本从mysql数据库获取数据

                try
                {               
                response = CustomHttpClient.executeHttpPost(
                "http://192.168.1.100:61896/JsonDemoDatabase/src/jsonscript.php",
                postParameters);

                String result = response.toString();

                try
                {
                    returnString = "";
                    JSONArray jArray = new JSONArray(result);
                    for(int i=0;i<jArray.length();i++)
                    {
                        JSONObject json_data = jArray.getJSONObject(i);
                        pass = json_data.getString("Password");
                        if(pass.equals(password.getText().toString()))
                        {

                            //Get an output to the screen
                            tv.setVisibility(1);
                            returnString += "Login Successfull";
                        }
                        else
                        {
                            tv.setVisibility(1);
                            returnString += "Login Fail";
                        }
                    }
                }
                catch(JSONException e)
                {
                    Log.e("log_tag", "Error parsing data "+e.toString());
                }
试试看
{               
响应=CustomHttpClient.executeHttpPost(
"http://192.168.1.100:61896/JsonDemoDatabase/src/jsonscript.php",
后参数);
字符串结果=response.toString();
尝试
{
returnString=“”;
JSONArray jArray=新JSONArray(结果);

对于(int i=0;iDid)您是否检查了您的响应数据?特别是,
response
上的
toString()
可能没有提供您期望的JSON。