Question

我正处于一个问题的中间，我得到一个字符串“INSERTED SUCCESSFULLY”作为我的httprequest的响应，我的目的是将它与本地字符串匹配（与上面相同）但是当我尝试匹配它们时不匹配因此不会进入我的状态。

UrlEncodedFormEntity entity = new UrlEncodedFormEntity(postParameters);
request.setEntity(entity);             
HttpResponse response= httpClient.execute(request);
bufferedReader = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
StringBuffer stringBuffer = new StringBuffer("");
String line = "";
String LineSeparator = System.getProperty("line.separator");
while ((line = bufferedReader.readLine()) != null) {
stringBuffer.append(line + LineSeparator); 
}
bufferedReader.close();
String q = "INSERTED SUCCESSFULLY";
StringBuffer got = new StringBuffer(q);
Log.d("Response", stringBuffer.toString());//MY logcat shows "INSERTED SUCCESSFULLY"
if(stringBuffer.equals(got))
   {
               // Does not enter here
       Log.d("xyz", "Getting Response" + stringBuffer.toString());
      //some TASK
   }

这与编码有关。 N.B： - 我的php代码只是回显“INSERTED SUCCESSFULLY”。

我尝试过以下代码但没有成功

if(stringBuffer.toString().equals("INSERTED SUCCESSFULLY")

Answer 1

if(stringBuffer.toString().trim().equals("INSERTED SUCCESSFULLY")

尝试这可能是有一些开始或结束空间

stringbuffer toString与相同的静态字符串不匹配

1 个答案: