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com.epam.wilma.message.search.web.support.ZipOutputStreamFactory.createZipOutputStream()方法的使用及代码示例

转载 作者:知者 更新时间:2024-03-13 12:36:21 29 4
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本文整理了Java中com.epam.wilma.message.search.web.support.ZipOutputStreamFactory.createZipOutputStream()方法的一些代码示例,展示了ZipOutputStreamFactory.createZipOutputStream()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。ZipOutputStreamFactory.createZipOutputStream()方法的具体详情如下:
包路径:com.epam.wilma.message.search.web.support.ZipOutputStreamFactory
类名称:ZipOutputStreamFactory
方法名:createZipOutputStream

ZipOutputStreamFactory.createZipOutputStream介绍

[英]This method creates a new ZipOutputStream from the given OutputStream.
[中]此方法从给定的OutputStream创建一个新的ZipOutputStream。

代码示例

代码示例来源:origin: epam/Wilma

/**
 * Compress files which is specicified in the filePaths parameter
 * and write the zipped result into the given {@link OutputStream}.
 * @param filePaths are the path of the files which will be compressed into the result
 * @param result is the compressed file which contains the files
 */
public void createZipWithFiles(final List<List<String>> filePaths, final OutputStream result) {
  ZipOutputStream zipStream = zipOutputStreamFactory.createZipOutputStream(result);
  zipFiles(filePaths, zipStream);
  closeZipStream(zipStream);
}

代码示例来源:origin: epam/Wilma

@BeforeMethod
public void setUp() throws IOException {
  MockitoAnnotations.initMocks(this);
  filePaths = new ArrayList<>();
  given(zipOutputStreamFactory.createZipOutputStream(outputStream)).willReturn(zipOutputStream);
  given(entryFactory.createZipEntry(Mockito.anyString())).willReturn(zipEntry);
}

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