4

我正在尝试使用 Apache POI 3.9 从字符串中读取 excel 文件，但没有成功。我对java不太熟悉。

澄清一下，在我的程序中，我已经将 excel 文件作为字符串，并且我正在使用 readFile 函数来模拟这种行为。

程序：

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.ByteBuffer;
import java.nio.charset.Charset;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;

import org.apache.poi.openxml4j.exceptions.InvalidFormatException;
import org.apache.poi.ss.usermodel.Workbook;
import org.apache.poi.ss.usermodel.WorkbookFactory;

public class Test {

    static String readFile(String path, Charset encoding) throws IOException 
    {
        byte[] encoded = Files.readAllBytes(Paths.get(path));
        return encoding.decode(ByteBuffer.wrap(encoded)).toString();
    }

    public static void main(String[] args) throws IOException, InvalidFormatException {
        String result = readFile("data.xlsx", StandardCharsets.UTF_8);

        InputStream is = new ByteArrayInputStream(result.getBytes("UTF-8"));

        Workbook book = WorkbookFactory.create(is);
    }

}

我得到的错误是：

Exception in thread "main" java.util.zip.ZipException: invalid block type
    at java.util.zip.InflaterInputStream.read(InflaterInputStream.java:164)
    at java.util.zip.ZipInputStream.read(ZipInputStream.java:193)
    at java.io.FilterInputStream.read(FilterInputStream.java:107)
    at org.apache.poi.openxml4j.util.ZipInputStreamZipEntrySource$FakeZipEntry.<init>(ZipInputStreamZipEntrySource.java:127)
    at org.apache.poi.openxml4j.util.ZipInputStreamZipEntrySource.<init>(ZipInputStreamZipEntrySource.java:55)
    at org.apache.poi.openxml4j.opc.ZipPackage.<init>(ZipPackage.java:83)
    at org.apache.poi.openxml4j.opc.OPCPackage.open(OPCPackage.java:267)
    at org.apache.poi.ss.usermodel.WorkbookFactory.create(WorkbookFactory.java:73)
    at Test.main(Test.java:28)

任何帮助，将不胜感激。

干杯

4

4 回答 4

6

所以解决我的问题是

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.file.Files;
import java.nio.file.Paths;

import org.apache.poi.openxml4j.exceptions.InvalidFormatException;
import org.apache.poi.ss.usermodel.Workbook;
import org.apache.poi.ss.usermodel.WorkbookFactory;

public class Test {

    public static void main(String[] args) throws IOException, InvalidFormatException {
        byte[] result = Files.readAllBytes(Paths.get("data.xlsx"));     
        InputStream is = new ByteArrayInputStream(result);
        Workbook book = WorkbookFactory.create(is);
    }

}
于 2013-11-11T17:04:50.720 回答
4

看起来你让这种方式太复杂了。只需遵循Apache POI 快速指南，该指南建议使用FileInputStream. 无需将字节读入字节数组并使用ByteArrayInputStream.

使用从指南中复制的以下内容之一：

// Use a file
Workbook wb = WorkbookFactory.create(new File("MyExcel.xls"));

// Use an InputStream, needs more memory
Workbook wb = WorkbookFactory.create(new FileInputStream("MyExcel.xlsx"));
于 2013-11-11T16:57:22.670 回答
0

你在做什么？您正在将二进制文件读入 abyte[]并String使用 UTF-8 将其转换为 a。稍后您将再次使用 UTF-8 将其转换回字节流。做什么的？跳过中间的所有步骤：

public static void main(String[] args) throws IOException, InvalidFormatException {
    InputStream is = new FileInputStream("data.xlsx");
    Workbook book = WorkbookFactory.create(is);
}
于 2013-11-11T16:57:07.257 回答
0

这困扰了我一段时间。没有一个建议的修复对我有用。解决这个问题的方法是在 maven-resources-plugin 中添加一个，因此

        <plugin>
            <artifactId>maven-resources-plugin</artifactId>
            <version>2.5</version>
            <configuration>
              <encoding>UTF-8</encoding>
              <nonFilteredFileExtensions>
                <nonFilteredFileExtension>docx</nonFilteredFileExtension>
                <nonFilteredFileExtension>xls</nonFilteredFileExtension>
                <nonFilteredFileExtension>xlsx</nonFilteredFileExtension>
              </nonFilteredFileExtensions>
            </configuration>
        </plugin>
于 2021-05-12T08:48:15.507 回答