我在Java应用程序中使用Api并触发此URL（http://checkdnd.com/api/check_dnd_no_api.php?mobiles=9999999999）。我在控制台中收到HTTP 403错误但在Web浏览器中没有发生错误并获得预期的响应。我也试过其他网址，他们工作正常没有问题或任何错误。
那么，URL中的问题是什么？我应该怎么做？
这是源代码： Main.java
import org.json.simple.*;
import org.json.simple.parser.*;
public class Main
{
public static void main(String[] args) throws Exception
{
String numb = "9999999999,8888888888";
String response = new http_client("http://checkdnd.com/api/check_dnd_no_api.php?mobiles="+numb).response;
System.out.println(response);
// encoding response
Object obj = JSONValue.parse(response);
JSONObject jObj = (JSONObject) obj;
String msg = (String) jObj.get("msg");
System.out.println("MESSAGE : "+msg);
JSONObject msg_text = (JSONObject) jObj.get("msg_text");
String[] numbers = numb.split(",");
for(String number : numbers)
{
if(number.length() != 10 || number.matches(".*[A-Za-z].*")){
System.out.println(number+" is invalid.");
}else{
if(msg_text.get(number).equals("Y"))
{
System.out.println(number+" is DND Activated.");
}else{
System.out.println(number+" is not DND Activated.");
}
}
}
}
}
现在，http_client.java
import java.net.*;
import java.io.*;
public class http_client
{
String response = "";
http_client(String URL) throws Exception
{
URL url = new URL(URL);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("GET");
BufferedReader bs = new BufferedReader(new InputStreamReader(con.getInputStream()));
String data ="";
String response = "";
while((data = bs.readLine()) != null){
response = response + data;
}
con.disconnect();
url = null;
con = null;
this.response = response;
}
}
答案 0 :(得分：0)
如果您没有向我们展示您用来访问提供的网址（http://checkdnd.com/api/check_dnd_no_api.php?mobiles=9999999999）的代码，那么很难确定您的问题究竟在哪里，但我的第一个猜测就是链接您提供的只能通过安全套接字层（SSL）访问。换句话说，该链接应以 https：// 而不是 http：//
要验证这一点，只需更改您的网址字符串：https://checkdnd.com/api/check_dnd_no_api.php?mobiles=9999999999然后重试。
您不会遇到浏览器的问题，原因很简单，因为浏览器通常会尝试使用这两种协议进行连接。网站也可以接受哪种协议，批次允许这两种协议，而另一些则不允许。
要检查网址字符串是否正在使用有效的协议，您可以使用这个小方法我快速掀起：
/**
* This method will take the supplied URL String regardless of the protocol (http or https)
* specified at the beginning of the string, and will return whether or not it is an actual
* "http" (no SSL) or "https" (is SSL) protocol. A connection to the URL is attempted first
* with the http protocol and if successful (by way of data acquisition) will then return
* that protocol. If not however, then the https protocol is attempted and if successful then
* that protocol is returned. If neither protocols were successful then Null is returned.<br><br>
*
* Returns null if the supplied URL String is invalid, a protocol does not
* exist, or a valid connection to the URL can not be established.<br><br>
*
* @param webLink (String) The full link path.<br>
*
* @return (String) Either "http" for Non SLL link, "https" for a SSL link.
* Null is returned if the supplied URL String is invalid, a protocol does
* not exist, or a valid connection to the URL can not be established.
*/
public static String isHttpOrHttps(String webLink) {
URL url;
try {
url = new URL(webLink);
} catch (MalformedURLException ex) { return null; }
String protocol = url.getProtocol();
if (protocol.equals("")) { return null; }
URLConnection yc;
try {
yc = url.openConnection();
BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream()));
in.close();
return "http";
} catch (IOException e) {
// Do nothing....check for https instead.
}
try {
yc = new URL(webLink).openConnection();
//send request for page data...
yc.setRequestProperty("User-Agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.95 Safari/537.11");
yc.connect();
BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream()));
in.close();
return "https";
} catch (IOException e) {
// Do Nothing....allow for Null to be returned.
}
return null;
}
使用此方法：
// Note that the http protocol is supplied within the url string:
String protocol = isHttpOrHttps("http://checkdnd.com/api/check_dnd_no_api.php?mobiles=9999999999");
System.out.println(protocol);
控制台的输出为：https。 isHttpOrHttps（） 方法已确定 https 协议是用于获取数据（或其他）的成功协议，即使<提供了strong> http 。
要从网页中提取页面源，您可以使用以下方法：
/**
* Returns a List ArrayList containing the page source for the supplied web
* page link.<br><br>
*
* @param link (String) The URL address of the web page to process.<br>
*
* @return (List ArrayList) A List ArrayList containing the page source for
* the supplied web page link.
*/
public static List<String> getWebPageSource(String link) {
if (link.equals("")) { return null; }
try {
URL url = new URL(link);
URLConnection yc = null;
//If url is a SSL Endpoint (using a Secure Socket Layer such as https)...
if (link.startsWith("https:")) {
yc = new URL(link).openConnection();
//send request for page data...
yc.setRequestProperty("User-Agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.95 Safari/537.11");
yc.connect();
}
//and if not a SLL Endpoint (just http)...
else { yc = url.openConnection(); }
BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream()));
String inputLine;
List<String> sourceText = new ArrayList<>();
while ((inputLine = in.readLine()) != null) {
sourceText.add(inputLine);
}
in.close();
return sourceText;
}
catch (MalformedURLException ex) {
// Do whatever you want with exception.
ex.printStackTrace();
}
catch (IOException ex) {
// Do whatever you want with exception.
ex.printStackTrace();
}
return null;
}
为了利用这里提供的两种方法，您可以尝试这样的方法：
String netLink = "http://checkdnd.com/api/check_dnd_no_api.php?mobiles=9999999999";
String protocol = isHttpOrHttps(netLink);
String netLinkProtocol = netLink.substring(0, netLink.indexOf(":"));
if (!netLinkProtocol.equals(protocol)) {
netLink = protocol + netLink.substring(netLink.indexOf(":"));
}
List<String> list = getWebPageSource(netLink);
for (int i = 0; i < list.size(); i++) {
System.out.println(list.get(i));
}
控制台输出将显示：
{"msg":"success","msg_text":{"9999999999":"N"}}