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我想從使用php文件的MySQL數據庫中獲取數據。我的Java代碼如下:HttpURLConnection:生成的json-code建議頁面不存在(404),即使url正確
HttpURLConnection conn = null;
URL url = null;
try {
url = new URL(getURL);
System.out.println(getURL);
conn = (HttpURLConnection)url.openConnection();
//conn.setReadTimeout(READ_TIMEOUT);
//conn.setConnectTimeout(CONNECTION_TIMEOUT);
conn.setRequestMethod("POST");
// setDoInput and setDoOutput method depict handling of both send and receive
conn.setDoInput(true);
conn.setDoOutput(true);
// Append parameters to URL
Uri.Builder builder = new Uri.Builder();
builder.appendQueryParameter("user", USER);
builder.appendQueryParameter("pass", PASS);
builder.appendQueryParameter("server", SERVER);
builder.appendQueryParameter("db", DB);
String query = builder.build().getEncodedQuery();
// Open connection for sending data
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
writer.write(query);
writer.flush();
writer.close();
os.close();
conn.connect();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e1) {
e1.printStackTrace();
}
try {
int response_code = conn.getResponseCode();
// Check if successful connection made
if (response_code == HttpURLConnection.HTTP_OK) {
// Read data sent from server
InputStream input = conn.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(input));
result = reader.readLine();
return(result);
}else{
return("unsuccessful");
}
當我去我的網址(隱藏在變量的getURL)使用的瀏覽器,我看到我的屏幕上的JSON字符串,只是因爲它應該。但是,當我輸出閱讀器的內容時(上面的代碼只是第一行,但是通過修改代碼我當然可以輸出更多),它顯示了顯示404頁面的網頁的html代碼不存在信息。
任何人有任何想法出了什麼問題?是的,我確實檢查了錯字。
也許有些人在您手動導航到該頁面時沒有注意到的重定向內容,但這會打破您的請求。 – Xatenev
哪裏ius getURL定義? –
getURL在此類擴展的類中定義。爲什麼? – MWB