File.createTempFile in Java 中出现不兼容类型错误

Question

到目前为止，我的代码在临时目录中创建文件并处理它时工作正常。

但是现在我正在尝试提供我实际上想要在其中创建xml文件的特定目录。所以在方法createTmpXmlFile

private static Path createTmpXmlFile(final String prefix) {
try {
log.info("Creating temporary file {}{}", prefix, XML_SUFFIX);
return Files.createTempFile(Paths.get(gleifZipFile), prefix, XML_SUFFIX);
} catch (IOException e) {
throw new IllegalStateException("Could not create tmp file at " + prefix + XML_SUFFIX + ". ", e);
}
}

我从

return Files.createTempFile(prefix, XML_SUFFIX);

自

return File.createTempFile(prefix, XML_SUFFIX, "/tmp/in");

我收到以下错误：

java：

不兼容的类型：java.lang.String 无法转换为 java.io.File。

如果我在这里更改逻辑，那么它会影响调用该方法的其他方法createTmpXmlFile。

我真的不明白如何解决这个问题。下面是我的代码：

@Slf4j
public class InputCS implements Runnable {
public static final String XML_SUFFIX = ".xml";

@Value("${gleifdataimporter.file.dir}")    
private String gleifZipFile;
private void processleifZipFile() {
final AtomicBoolean isInsideLeiRecord = new AtomicBoolean();
isInsideLeiRecord.set(false);
final StringBuilder currentLeiRecordXml = new StringBuilder();
try (FileSystem zipFs = FileSystems.newFileSystem(jobRunner.getInputZipPath(), null)) {
Path tmpXMLPath = xmlFileFromLeiZipFile(zipFs);
try (Stream<String> lines = Files.lines(tmpXMLPath)) {
AtomicInteger processedLinesCounter = new AtomicInteger();
AtomicInteger currentLineNumber = new AtomicInteger();
lines.sequential().forEach(handleLineAndIncrementLineNumber(isInsideLeiRecord, currentLeiRecordXml, processedLinesCounter, currentLineNumber));
log.info("{} lines of XML file inside LEIF input ZIP file {} processed.", processedLinesCounter.get(), jobRunner.getInputZipPath());
}catch (IOException e) {
throw new IllegalStateException("Problem reading input file at " + jobRunner.getInputZipPath() + ".", e);
} finally {
Files.delete(tmpXMLPath);
}
} catch (IOException e) {
throw new IllegalStateException("Problem reading input file at " + jobRunner.getInputZipPath() + ".", e);
}
}
private Path xmlFileFromLeiZipFile(FileSystem zipFs) {       //extracts the xml file from zip file
log.info("Input file {} exists: {}", jobRunner.getInputZipPath(), Files.exists(jobRunner.getInputZipPath()));
Path tmpXmlPath = createTmpXmlFile("leif__" + System.currentTimeMillis());
for (Path rootDir : zipFs.getRootDirectories()) {
try (Stream<Path> files = treeAt(rootDir)) {
log.info("Trying to extract LEIF XML file from ZIP file into {}.", tmpXmlPath);
final Path xmlFileInsideZip = files
.filter(isNotADir())
.filter(Files::isRegularFile)
.findFirst()
.orElseThrow(() -> new IllegalStateException("No file found in LEI ZIP file."));
log.info("Path to LEIF XML file inside ZIP file: {}.", xmlFileInsideZip);
return copyReplacing(xmlFileInsideZip, tmpXmlPath);
}
}
throw new IllegalStateException("No file found in LEI ZIP file " + jobRunner.getInputZipPath() + ".");
}

private static Path createTmpXmlFile(final String prefix) {
try {
log.info("Creating temporary file {}{}", prefix, XML_SUFFIX);
return Files.createTempFile(Paths.get(gleifZipFile), prefix, XML_SUFFIX);
} catch (IOException e) {
throw new IllegalStateException("Could not create tmp file at " + prefix + XML_SUFFIX + ". ", e);
}
}

@NotNull
private static Path copyReplacing(Path from, Path to) {
requireNonNull(from, "Trying to copy from a path, which is null to path " + to + ".");   //trying to copy file where no xml file exist in root directory
requireNonNull(to, "Trying to copy from path " + from + " to a path, which is null.");
try {
return Files.copy(from, to, REPLACE_EXISTING);
} catch (IOException e) {
throw new IllegalStateException("Cannot copy from " + from + " to " + to + ". ", e);
}
}

}

Answer 1

1. 正如Slaw所建议的，使用Files#createTempFile(Path，String，String，FileAttribute...(指定创建临时文件的目录。
2. 使用 Java 7 或 8 的Paths#get(String,String...)，或 Java 11 或更高版本的Path#of(String,String...)将String转换为Path。延伸阅读： Paths.get vs Path.of

<小时 />

private static Path createTmpXmlFile(final String prefix) {
try {
// Java 11 or later
// return Files.createTempFile(Path.of("/tmp/in"), prefix, XML_SUFFIX);
// Java 8
return Files.createTempFile(Paths.get("/tmp/in"), prefix, XML_SUFFIX);
} catch (IOException e) {
throw new IllegalStateException("Could not create tmp file at " + prefix + XML_SUFFIX + ". ", e);
}
}

Answer 2

File.createTempFile期望将File对象作为第三个参数。只需将"/tmp/in"包装到文件中即可 =>return File.createTempFile(prefix, XML_SUFFIX, new File("/tmp/in"));你应该很高兴。

所以你可以做到：

File.createTempFile("prefix", "suffix", new File("/tmp/in"));

或使用蔚来汽车(推荐(

Files.createTempFile(Paths.get("/tmp/in"), "prefix", "suffix");