How to read zip file through getResourceAsStream with relative path?












0















I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question























  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07


















0















I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question























  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07
















0












0








0








I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question














I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?







java maven zip inputstream






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 20 '18 at 1:51









user697911user697911

3,208113472




3,208113472













  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07





















  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07



















File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

– Andrew Thompson
Nov 20 '18 at 2:07







File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

– Andrew Thompson
Nov 20 '18 at 2:07














1 Answer
1






active

oldest

votes


















0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22
















0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22














0












0








0







You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer















You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 20 '18 at 4:11









user697911

3,208113472




3,208113472










answered Nov 20 '18 at 3:15









John CamerinJohn Camerin

436111




436111













  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22



















  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22

















I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

– user697911
Nov 20 '18 at 4:11





I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

– user697911
Nov 20 '18 at 4:11













Can you post your file?

– John Camerin
Nov 20 '18 at 13:22





Can you post your file?

– John Camerin
Nov 20 '18 at 13:22




















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