java - 将 ZipEntry 数据写入字符串

Question

我已经从这样的 zip 文件中检索了一个 zip 条目。

InputStream input = params[0];
ZipInputStream zis = new ZipInputStream(input);

ZipEntry entry;
try {
    while ((entry = zis.getNextEntry())!= null) {

    }
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

这很好用，我ZipEntry没问题。

我的问题

ZipEntries由于它们是 xml 和 csv 文件，我如何将它们的内容转换为字符串。

Answer 1

你必须阅读ZipInputStream：

StringBuilder s = new StringBuilder();
byte[] buffer = new byte[1024];
int read = 0;
ZipEntry entry;
while ((entry = zis.getNextEntry())!= null) {
      while ((read = zis.read(buffer, 0, 1024)) >= 0) {
           s.append(new String(buffer, 0, read));
      }
}

当您从内部退出时while保存StringBuilder内容，并重置它。

Answer 2

使用定义的编码（例如 UTF-8）并且不创建字符串：

import java.util.zip.ZipInputStream;
import java.util.zip.ZipEntry;
import java.io.ByteArrayOutputStream;
import static java.nio.charset.StandardCharsets.UTF_8;

try (
  ZipInputStream zis = new ZipInputStream(input, UTF_8); 
  ByteArrayOutputStream baos = new ByteArrayOutputStream()
) {
  byte[] buffer = new byte[1024];
  int read = 0;
  ZipEntry entry;
  while ((entry = zis.getNextEntry()) != null)
    while ((read = zis.read(buffer, 0, buffer.length)) > 0)
      baos.write(buffer, 0, read);
  String content = baos.toString(UTF_8.name());
}

Answer 3

这是一种不会破坏 Unicode 字符的方法：

final ZipInputStream zis = new ZipInputStream(new ByteArrayInputStream(content));
final InputStreamReader isr = new InputStreamReader(zis);
final StringBuilder sb = new StringBuilder();
final char[] buffer = new char[1024];

while (isr.read(buffer, 0, buffer.length) != -1) {
    sb.append(new String(buffer));
}

System.out.println(sb.toString());

Answer 4

我会使用 apache 的 IOUtils

ZipEntry entry;
InputStream input = params[0];
ZipInputStream zis = new ZipInputStream(input);

try {
  while ((entry = zis.getNextEntry())!= null) {
    String entryAsString = IOUtils.toString(zis, StandardCharsets.UTF_8);
  }
} catch (IOException e) {
  // TODO Auto-generated catch block
  e.printStackTrace();
}
IOUtils.closeQuietly(zis);

Answer 5

Kotlin 版本，但也可以在 Java 中使用

val zipInputStream = ZipInputStream(inStream)
var zipEntry = zipInputStream.nextEntry
while(zipEntry != null) {
    println("Name of file : " + zipEntry.name)
    val fileContent = String(zipInputStream.readAllBytes(), StandardCharsets.UTF_8)
    println("File content : $fileContent")
    zipEntry = zipInputStream.nextEntry
}