使用 zip4j 的 ZipFile 类型未定义方法 extractAll(String)

Question

我正在尝试制作一个解压缩然后解析 xml.zip 中的 xml 文件的程序。添加 zip4j 的依赖项，这是我的 pom.xml

        <project xmlns="http://maven.apache.org/POM/4.0.0"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <groupId>interview</groupId>
    <artifactId>programES</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <name>xmlToES</name>
    <dependencies>
        <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.dataformat/jackson-dataformat-xml -->
        <dependency>
            <groupId>com.fasterxml.jackson.dataformat</groupId>
            <artifactId>jackson-dataformat-xml</artifactId>
            <version>2.12.1</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
        <dependency>
            <groupId>org.apache.commons</groupId>
            <artifactId>commons-lang3</artifactId>
            <version>3.11</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.12.1</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/net.lingala.zip4j/zip4j -->
        <dependency>
            <groupId>net.lingala.zip4j</groupId>
            <artifactId>zip4j</artifactId>
            <version>2.6.4</version>
        </dependency>

    </dependencies>
</project>

我的解析器代码：

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;

import model.Products;
import net.lingala.zip4j.core.ZipFile;
import org.apache.commons.lang3.StringUtils;
import java.util.zip.ZipFile;
import java.io.File;
import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;

public class Parser {
    public static void main(String[] args) throws IOException {
        ObjectMapper objectMapper = new XmlMapper();
        String source = "C:\Users\user\java_program\xml.zip";
        String destination = "C:\Users\user\java_program\xml";
        ZipFile zipFile = new ZipFile(source);
        zipFile.extractAll(destination);
        //Reads from XML and converts to POJO
        Products products = objectMapper.readValue(
                StringUtils.toEncodedString(Files.readAllBytes(Paths.get("C:\Users\user\java_program\xml\xml.xml")), StandardCharsets.UTF_8), Products.class);
        System.out.println(products);
    }

}

程序仍然写入错误：方法 extractAll(String) 未定义类型 ZipFile

非常感谢任何帮助

Answer 1

您应该导入 net.lingala.zip4j.core.ZipFile 而不是 java.util.zip.Zipfile

删除行：

import java.util.zip.ZipFile;

并删除 .core 来自您的导入 import net.lingala.zip4j.core.ZipFile;

整个class:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;
import model.Products;
import net.lingala.zip4j.ZipFile;
import net.lingala.zip4j.exception.ZipException;
import org.apache.commons.lang3.StringUtils;


public class Parser {
    public static void main(String[] args) throws IOException, ZipException {
        ObjectMapper objectMapper = new XmlMapper();
        String source = "C:\Users\user\java_program\xml.zip";
        String destination = "C:\Users\user\java_program\xml";
        ZipFile zipFile = new ZipFile(source);
        zipFile.extractAll(destination);
        //Reads from XML and converts to POJO
        Products products = objectMapper.readValue(
            StringUtils.toEncodedString(Files.readAllBytes(Paths.get("C:\Users\user\java_program\xml\xml.xml")), StandardCharsets.UTF_8), Products.class);
        System.out.println(products);
    }
}

Answer 2

在您的导入中，您正在导入 java.util.Zipfile 因此删除 -

import java.util.zip.ZipFile;

它应该可以工作。

使用 zip4j 的 ZipFile 类型未定义方法 extractAll(String)

The method extractAll(String) is undefined for the type ZipFile with zip4j

java

xml

parsing

bigdata

zip4j