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我在一个问题的中间,我得到一个字符串“INSERTED SUCCESSFULLY”作为来自我的httprequest的响应,我的目标是匹配它与本地字符串(与上面相同)但是当我尝试匹配它们时,它们不匹配,因此不会进入我的状态。stringbuffer toString与相同的静态字符串不匹配
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。 012B: N.B: - 我的PHP代码只是回声“插入成功”。
我尝试没有成功
if(stringBuffer.toString().equals("INSERTED SUCCESSFULLY")
问题是'LineSeparator'因为要添加新在第一个StringBuffer行,但不是在第二个,所以总是得到'FALSE'。或者在第二个字符串中添加行分隔符或者在两个字符串上调用'trim()'来比较 –