我有一个问题，真的不知道出了什么问题:-/
我有一个简单的函数，在单击按钮后调用
public void clickFunc(View view) {
StringBuilder result = new StringBuilder();
try {
URL url = new URL("http://google.com");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(15000 /* milliseconds */);
conn.setConnectTimeout(15000 /* milliseconds */);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
conn.connect();
int responseCode=conn.getResponseCode();
if (responseCode == HttpsURLConnection.HTTP_OK) {
InputStream in = new BufferedInputStream(conn.getInputStream());
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
in.close();
httpTxt.setText(result.toString());
}
else {
httpTxt.setText("false : "+responseCode);
}
}
catch(Exception e){
httpTxt.setText("Exception: " + e.getMessage());
}
}
但是单击后我的输出是 Exception: null
在 list 中我有<uses-permission android:name="android.permission.INTERNET" />
我认为此代码与此处或互联网上提到的所有代码相同:-/
httptxt 是我的页面输出的 texview
谢谢
最佳答案
好吧，我不知道我必须在后台执行此操作...它现在正在工作，谢谢!
public void clickFunc(View view) {
new SendPostRequest().execute();
}
private class SendPostRequest extends AsyncTask<Void, Void, String> {
protected String doInBackground (Void... params) {
StringBuilder result = new StringBuilder();
HttpURLConnection conn = null;
BufferedReader reader;
try {
URL url = new URL("http://stackoverflow.com");
conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setReadTimeout(15000);
conn.setConnectTimeout(15000);
conn.setDoInput(true);
conn.setDoOutput(true);
conn.connect();
int responseCode=conn.getResponseCode();
if (responseCode == HttpsURLConnection.HTTP_OK) {
InputStream in = new BufferedInputStream(conn.getInputStream());
reader = new BufferedReader(new InputStreamReader(in));
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
in.close();
return(result.toString());
}
else {
return("false : "+responseCode);
}
}
catch(Exception e){
return("Exception: " + e.getMessage());
}
}
protected void onPostExecute(String s) {
super.onPostExecute(s);
httpTxt.setText(s);
}
}
我不是一个 Android 程序员，所以即使它是基本的我也不知道，对不起大家:)
关于我的应用程序，如果您感兴趣，为什么我想要它 - 我读取条形码，转到 php 脚本，查看数据库是否有此代码并返回 true 或 false
关于java - HttpURLConnection 不适用于 Android，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44022280/