如何解决在 Eclipse 中编写此代码时显示“无法解析为变量”错误使用 Spring Boot 的 maven 项目
我创建了一些 Java 项目。在这个项目中,我在 Eclipse IDE 中编写代码时发现一个错误。请帮我纠正这个错误...
CoronaVirusDataService.java(在此文件中出现错误)
包 io.covid.coronatracker.services;
import java.io.IOException;
import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;
import javax.annotation.PostConstruct;
import org.apache.commons.csv.CSVFormat;
import org.apache.commons.csv.CSVRecord;
import org.springframework.stereotype.Service;
@Service
public class CoronaVirusDataService {
private static String VIRUS_DATA_URL="https://raw.githubusercontent.com/CSSEGISandData/COVID-19/master/csse_covid_19_data/csse_covid_19_time_series/time_series_covid19_confirmed_global.csv";
@PostConstruct
public void fetchVirusData() throws IOException,InterruptedException {
HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder().uri(URI.create(VIRUS_DATA_URL)).build();
HttpResponse<String> httpResponse = client.send(request,HttpResponse.BodyHandlers.ofString());
System.out.println(httpResponse.body());
Iterable<CSVRecord> records = CSVFormat.RFC4180.withFirstRecordAsHeader().parse(**in**);
for (CSVRecord record : records) {
String id = record.get("ID");
String customerNo = record.get("CustomerNo");
String name = record.get("Name");
}
}
}
CoronaTrackerApplication.java(这是主文件) 包 io.covid.coronatracker;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
@SpringBootApplication
public class CoronaTrackerApplication {
public static void main(String[] args) {
SpringApplication.run(CoronaTrackerApplication.class,args);
}
}
解决方法
暂无找到可以解决该程序问题的有效方法,小编努力寻找整理中!
如果你已经找到好的解决方法,欢迎将解决方案带上本链接一起发送给小编。
小编邮箱:dio#foxmail.com(将#修改为@)