返回信息流String api_key = "f4S2H2d6AUVsxkiOdP3YYdMFSw2urUyG7IhAXHiQ";
String pattern = "pos";
String format = "plain";
String Sentence="天下一统,吃饭了没";
String text = URLEncoder.encode(Sentence, "utf-8");
URL url = new URL("http://ltpapi.voicecloud.cn/analysis/?" + "api_key=" + api_key + "&" + "text=" + text
+ "&" + "format=" + format + "&" + "pattern=" + pattern);
//URL url = new URL(googleWeatherUrl1);
// HttpURLConnection
HttpURLConnection httpconn = (HttpURLConnection) url.openConnection();
if (httpconn.getResponseCode() == HttpURLConnection.HTTP_OK) {
Toast.makeText(getApplicationContext(), "连接API成功!",
Toast.LENGTH_SHORT).show();
// InputStreamReader
InputStreamReader isr = new InputStreamReader(httpconn.getInputStream(), "utf-8");
int i;
String content = "";
// read
while ((i = isr.read()) != -1) {
content = content + (char) i;
}
isr.close();
//设置TextView
tv.setText(content);
}
//disconnect
httpconn.disconnect();
这是一条镜像帖。来源:北邮人论坛 / java / #48830同步于 2016/3/26
该镜像源已超过 30 天没有更新,可能在源站已被删除。
Java机器人发帖
求助关于Android-WebApi连接不成功
xiongyi1995
2016/3/26镜像同步2 回复
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2 条回复
网上的代码,拿来改了改,不知道为什么,就是连不上,没改之前也连不上,是不是HttpURLConnection不能用rest 方式连接WebApi,然而HttpClient又装不上,求助啊
AndroidManifest里的《application》 前面加了一句 <uses-permission android:name="android.permission.INTERNET"></uses-permission>