我正在尝试使用php文件从MySQL数据库中获取数据。我的java代码如下:
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");
}
当我使用浏览器访问我的网址(隐藏在变量getURL中)时,我会在屏幕上看到json字符串,就像它应该的那样。但是,当我输出阅读器的内容时(上面的代码只占第一行,但通过调整我可以的代码,当然,输出更多)它显示了显示404的网站的html代码 - 页面不存在消息。
任何人都知道出了什么问题?是的,我确实检查过拼写错误。
答案 0 :(得分:0)
好的,我不知道发生了什么,因为我没有改变任何事情。但突然间它开始起作用了吗?!?
一定是服务器端的东西,我想......
感谢您的投入和分享您的想法!