0

我正在尝试在 Java 中执行此操作：

if(this.ssl == true) {
    HttpsURLConnection connection = (HttpsURLConnection) new URL(address).openConnection();
}
else {
    HttpURLConnection connection = (HttpURLConnection) new URL(address).openConnection();
}
connection.setDoOutput(true);
connection.setRequestMethod("POST");

但是最后两行抛出一个错误，说找不到变量。这在Java中是不可能的吗？我知道在这种情况下使用相同类型声明一个变量（在条件之外声明它并在条件内部初始化它），但在这种情况下，类型根据条件不同。

作为参考，到目前为止，这是我的课程：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.UnsupportedEncodingException;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;
import javax.net.ssl.HttpsURLConnection;

public class Post {

    private String data;
    private boolean ssl;

    public Post(boolean ssl) {
        this.ssl = ssl;
    }

    public String sendRequest(String address) throws IOException {

        //Only send the request if there is data to be sent!
        if (!data.isEmpty()) {
            if (this.ssl == true) {
                HttpsURLConnection connection = (HttpsURLConnection) new URL(address).openConnection();
            } else {
                HttpURLConnection connection = (HttpURLConnection) new URL(address).openConnection();
            }
            connection.setDoOutput(true);
            connection.setRequestMethod("POST");
            OutputStreamWriter writer = new OutputStreamWriter(connection.getOutputStream());
            writer.write(data);
            BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
            StringBuilder response = new StringBuilder();
            reader.close();
            writer.close();
            return response.toString();
        } else {
            return null;
        }
    }

    public void setData(String[] keys, String[] values) throws UnsupportedEncodingException {
        //Take in the values and put them in the right format for a post request
        for (int i = 0; i < values.length; i++) {
            this.data += URLEncoder.encode(keys[i], "UTF-8") + "=" + URLEncoder.encode(values[i], "UTF-8");
            if (i + 1 < values.length) {
                this.data += "&";
            }
        }
    }
}
4

3 回答 3

4

这是一个范围问题。

将其更改为 

public String sendRequest(String address) throws IOException {
   HttpURLConnection connection = null;
   //Only send the request if there is data to be sent!
   if (!data.isEmpty()) {
       if (this.ssl) {
         connection = (HttpsURLConnection) new URL(address).openConnection();
       } else {
         connection = (HttpURLConnection) new URL(address).openConnection();
       }
   }
   connection.setDoOutput(true);
   connection.setRequestMethod("POST");
   ...

在语句connection中定义它时，您正试图访问。if因此，没有其他人可以访问此变量。

于 2013-04-24T18:23:40.587 回答
2

看到这个。在java中，如何根据url创建HttpsURLConnection或HttpURLConnection？

基本上，由于HttpsURLConnectionextendsHttpUrlConnection您不需要在 if 语句中声明。

于 2013-04-24T18:23:41.817 回答
0

您的connection变量在非常短的块中声明，并且在您尝试对其调用方法时它们已超出范围。

在块之前声明一个connection变量，并在每种情况下设置它，以便在调用方法时该变量仍在范围内。

HttpURLConnection connection;
if(this.ssl == true) {
    connection = (HttpsURLConnection) new URL(address).openConnection();
}
else {
    connection = (HttpURLConnection) new URL(address).openConnection();
}
connection.setDoOutput(true);
connection.setRequestMethod("POST");
于 2013-04-24T18:23:45.570 回答