我正在尝试使用Apache POI 3.9从字符串中读取excel文件,但没有任何成功。我对java不太熟悉。
只是为了澄清一下,在我的程序中我已经将excel文件作为字符串了,我正在通过使用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)
任何帮助都将不胜感激。
欢呼声
答案 0 :(得分:4)
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);
}
}
答案 1 :(得分:3)
看起来你这样做太复杂了。只需按照Apache POI Quick Guide,即建议使用FileInputStream
阅读文件。不需要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"));
答案 2 :(得分:0)
byte[]
并使用UTF-8将其转换为String
。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);
}
答案 3 :(得分:0)
这让我困扰了一段时间。没有一个建议的修复对我有用。解决该问题的方法是在 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>