如何修复关于查询中非法字符的 URIException

Posted 2023-02-26

【中文标题】如何修复关于查询中非法字符的 URIException

【英文标题】：How to fix an URIException about an illegal character in query

【发布时间】：2019-11-09 05:42:03

【问题描述】：

我正在使用 java.net 包从 Web 服务下载 xml 文件，但我收到此错误：“Illegal character in query at index 103: http://.............../current?path=//Controller/Components/Path/DataItems/DataItem[@type="PART_COUNT"]

我已经找好了职位。它应该是“=”。

import java.io.IOException;
import java.net.URI;
import java.net.URISyntaxException;
import java.net.http.HttpClient;
import java.net.http.HttpClient.Builder;
import java.net.http.HttpClient.Version;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;
import java.net.http.HttpResponse.BodyHandlers;


public class App 

    public static void main( String[] args )
    


        HttpClient client = HttpClient
                .newBuilder()
                .version(Version.HTTP_2)
                .build();

        Builder builder;
        try 
            builder = (Builder) HttpRequest.newBuilder(new URI("http://............./current?path=//Controller/Components/Path/DataItems/DataItem[@type=\"PART_COUNT\"]"));
            HttpRequest request = ((java.net.http.HttpRequest.Builder) builder).GET().build();


            HttpResponse httpResponse = client.send(request, BodyHandlers.ofString());
             String body = (String) httpResponse.body();
             System.out.println(body);

         catch (IOException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
         catch (InterruptedException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
        
        catch (Exception e) 
            System.out.println(e.getMessage());
        


    

我希望从 xml 文件中的 url 下载信息。现在我要在屏幕上打印。

<url>http://maven.apache.org</url>

  <properties>
    <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
  </properties>

  <dependencies>
    <dependency>
      <groupId>junit</groupId>
      <artifactId>junit</artifactId>
      <version>3.8.1</version>
      <scope>test</scope>
    </dependency>
  </dependencies>
  <build>
  <plugins>
    <plugin>
      <groupId>org.apache.maven.plugins</groupId>
      <artifactId>maven-compiler-plugin</artifactId>
      <version>3.7.0</version>
      <configuration>
        <source>9</source>
        <target>9</target>
        <compilerArgument>--add-modules=jdk.incubator.httpclient</compilerArgument>
      </configuration>
    </plugin>
    <plugin>
      <groupId>org.apache.maven.plugins</groupId>
      <artifactId>maven-surefire-plugin</artifactId>
      <version>2.20.1</version>
      <configuration>
      <argLine>--add-modules=jdk.incubator.httpclient</argLine>
      </configuration>
    </plugin>
  </plugins>
</build>

【问题讨论】：

Java equivalent to javascript's encodeURIComponent that produces identical output?的可能重复

【参考方案1】：

尝试对查询参数值进行urlencode

//Controller/Components/Path/DataItems/DataItem[@type=\"PART_COUNT\"]

并传递编码值

%2F%2FController%2FComponents%2FPath%2FDataItems%2FDataItem%5B%40type%3D%5C%22PART_COUNT%5C%22%5D

https://docs.oracle.com/javase/8/docs/api/index.html?java/net/URLEncoder.html

【讨论】：

这是输出“类 jdk.internal.net.http.HttpRequestBuilderImpl 不能转换为类 java.net.http.HttpClient$Builder (jdk.internal.net.http.HttpRequestBuilderImpl 和 java. net.http.HttpClient$Builder 在加载器'platform'的模块 java.net.http 中）".......上面我把我的 pom.xml 中的代码

【参考方案2】：

我解决了我的问题。我使用了错误的 java 版本，因为 Java 11 中引入了 HTTP 客户端库。我用 Java 11 创建了一个新项目。下面我将为大家展示正确的代码。希望对你有用。

public class HttpTest 
public static void main(String[] args) 
        HttpClient client = HttpClient
                .newBuilder()
                .version(Version.HTTP_2)
                .build();

        String query ="//Controller/Components/Path/DataItems/DataItem[@type=\"PART_COUNT\"]";
        String encodedQuery = encodeValue(query);
        System.out.println(encodedQuery);
        System.out.println();
        java.net.http.HttpRequest.Builder builder;



        try 
            builder = HttpRequest.newBuilder(new URI("http://.........../current?path="+encodedQuery));
            HttpRequest request =  builder.GET().build();

                HttpResponse httpResponse = client.send(request, BodyHandlers.ofString());
             String body = (String) httpResponse.body();
             System.out.println(body);

         catch (IOException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
         catch (InterruptedException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
        
        catch (Exception e) 
            System.out.println(e.getMessage());
        


    

    private static String encodeValue(String value) 
        try 
            return URLEncoder.encode(value, StandardCharsets.UTF_8.toString());
         catch (UnsupportedEncodingException ex) 
            throw new RuntimeException(ex.getCause());