我可以讓我的代碼以編程方式登錄到任何URL的唯一方法是使用CookieHandler.setDefault(new CookieManager);
。這很好,但我想了解每個新的HttpsURLConnection
之間的cookie是如何維護的。Java類CookieHandler和CookieManager如何工作?
有人可以展示如何使下面的代碼登錄到Gmail帳戶,而無需使用CookieHandler.setDefault(new CookieManager);
?謝謝。
**注意:
- 用您自己的電子郵件和密碼替換。
- CookieHandler.setDefault(new CookieManager);
已在下面的代碼中註釋掉。
public class GmailApp {
private List<String> cookies;
private HttpsURLConnection conn;
public static void main(String[] args) throws Exception {
String url = "https://accounts.google.com/ServiceLoginAuth";
String gmail = "https://mail.google.com/mail/";
GmailApp http = new GmailApp();
// CookieHandler.setDefault(new CookieManager());
String page = http.GetPageContent(url);
String postParams = http.getFormParams(page, "[email protected]", "mypassword");
http.sendPost(url, postParams);
String result = http.GetPageContent(gmail);
System.out.println(result);
}
private void sendPost(String url, String postParams) throws Exception {
URL obj = new URL(url);
conn = (HttpsURLConnection) obj.openConnection();
conn.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(conn.getOutputStream());
wr.writeBytes(postParams);
wr.flush();
wr.close();
int responseCode = conn.getResponseCode();
System.out.println("\nSending 'POST' request to URL : " + url);
System.out.println("Post parameters : " + postParams);
System.out.println("Response Code : " + responseCode);
BufferedReader in =
new BufferedReader(new InputStreamReader(conn.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
}
private String GetPageContent(String url) throws Exception {
URL obj = new URL(url);
conn = (HttpsURLConnection) obj.openConnection();
conn.setRequestMethod("GET");
conn.setUseCaches(false);
if (cookies != null) {
for (String cookie : this.cookies) {
conn.addRequestProperty("Cookie", cookie.split(";", 1)[0]);
}
}
int responseCode = conn.getResponseCode();
System.out.println("\nSending 'GET' request to URL : " + url);
System.out.println("Response Code : " + responseCode);
BufferedReader in =
new BufferedReader(new InputStreamReader(conn.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
setCookies(conn.getHeaderFields().get("Set-Cookie"));
return response.toString();
}
public String getFormParams(String html, String username, String password)
throws UnsupportedEncodingException {
System.out.println("Extracting form's data...");
Document doc = Jsoup.parse(html);
Element loginform = doc.getElementById("gaia_loginform");
Elements inputElements = loginform.getElementsByTag("input");
List<String> paramList = new ArrayList<String>();
for (Element inputElement : inputElements) {
String key = inputElement.attr("name");
String value = inputElement.attr("value");
if (key.equals("Email"))
value = username;
else if (key.equals("Passwd"))
value = password;
paramList.add(key + "=" + URLEncoder.encode(value, "UTF-8"));
}
StringBuilder result = new StringBuilder();
for (String param : paramList) {
if (result.length() == 0) {
result.append(param);
} else {
result.append("&" + param);
}
}
return result.toString();
}
public void setCookies(List<String> cookies) {
this.cookies = cookies;
}
}
我發佈的代碼只是一個示例。我想我應該簡化它。您提到的提及與我的實際問題無關。我只希望有人發佈CookieHandler和CookieManager如何管理Cookie的手動解釋。 – dbconfession
我在'CookieHandler'上添加了一些細節。 – javabrett
謝謝佈雷特。我想最終,我希望能夠跨連接登錄和維護一個cookie,而不需要使用'CookieHandler',這樣我就能從根本上理解它在做什麼。無論我做什麼,即使我將響應cookie存儲到變量並設置下一個請求,以使用'.setRequestProperty(「Cookie」,loginCookie)''使用該cookie,每個連接似乎都會失敗。您能否演示如何手動處理響應cookie並在下一個請求中有效使用它? – dbconfession