我可以使我的代码以编程方式登录到任何URL的唯一方法是使用CookieHandler.setDefault(new CookieManager);
。很好,但是我想了解如何在每个新的HttpsURLConnection
之间维护cookie。
有人可以显示如何使以下代码登录到Gmail帐户而无需使用CookieHandler.setDefault(new CookieManager);
吗?谢谢。
**笔记:
-用您自己的电子邮件和密码代替。
-CookieHandler.setDefault(new CookieManager);
在下面的代码中被注释掉。
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;
}
}
最佳答案
当您通过CookieHandler
静态设置默认java.net.CookieHandler.setDefault()
时,当前和将来的HttpURLConnection
(通过其底层的HTTP客户端类)将检查此默认cookie处理程序的存在,并找到它将用于检索以下内容的cookie:请求,并存储设置/返回的新Cookie。只要您要在所有请求中维护cookie,就会提供一个静态cookie-jar。确保设置所需的CookiePolicy
,或者默认为java.net.CookiePolicy.ACCEPT_ORIGINAL_SERVER
。
请注意,您的代码存在一些问题:
cookie.split(";", 1)[0]
此拆分不会执行任何操作,如果要从第一个分号开始截断字符串,则需要
cookie.split(";", 2)[0]
(1-> 2)。另外,您仅在GET请求后调用
setCookies
-您还需要在POST登录响应中捕获cookie。