-2

我让用户 通过 webservice找回他们忘记的密码。所有功能都在执行 json给出成功输出但在此调试器未进入尝试(如果其他)条件之后...

我的忘记通过响应:

{
status: "Success" 
msg: "Password has been sent your Email_id , Please check it.."
}

试着抓:

params.add(new BasicNameValuePair("email",Email));

JSONObject json = jsonParser.makeHttpRequest(url,"POST", params);
Log.d("Create Response", json.toString());
try{
    String status=json.getString(TAG_SUCCESS);
    if(status=="TAG_SUCCESS"){

        Toast.makeText(ForgetPassWord.this,"Password has been sent your Email_id , Please check it..", Toast.LENGTH_LONG).show();
        finish();
    }else{
        Toast.makeText(ForgetPassWord.this,"Please Enter Correct E-Mail", Toast.LENGTH_LONG).show();
    }
}catch(JSONException e) {
    e.printStackTrace();
}
return null;
4

2 回答 2

1

用于equals()字符串比较

喜欢

status.equals("Success")

编辑:

像下面这样改变它

  private static final String TAG_SUCCESS="status";// Because  status is the key

String status = json.getString(TAG_SUCCESS);
于 2014-05-08T10:53:06.723 回答
1

尝试这个

try {

            if (json.has((TAG_SUCCESS))) {
                String status = json.getString(TAG_SUCCESS);
                if (status.equals(TAG_SUCCESS)) {

                    Toast.makeText(
                            ForgetPassWord.this,
                            "Password has been sent your Email_id , Please check it..",
                            Toast.LENGTH_LONG).show();
                    finish();
                }

                else {
                    Toast.makeText(ForgetPassWord.this,
                            "Please Enter Correct E-Mail",
                            Toast.LENGTH_LONG).show();
                }
            }

        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
于 2014-05-08T10:59:43.717 回答