使用Apache POI从字符串中读取excel文件

duc*_*cky 4 java excel apache-poi

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。

为了澄清一下,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 函数来模拟该行为。

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);
    }

}
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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)
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任何帮助,将不胜感激。

干杯

duc*_*cky 6

所以我的问题的解决方法是

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);
    }

}
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  • 我这样做的原因是我在网络服务中获取此文件。 (2认同)

isn*_*bad 0

你在干什么?您正在将二进制文件读入 abyte[]String使用 UTF-8 将其转换为 a。稍后您再次使用 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);
}
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