How to read zip file through getResourceAsStream with relative path?











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0
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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 at 2:07

















up vote
0
down vote

favorite












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 at 2:07















up vote
0
down vote

favorite









up vote
0
down vote

favorite











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






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asked Nov 20 at 1:51









user697911

3,079103365




3,079103365












  • 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 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 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 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 at 2:07














1 Answer
1






active

oldest

votes

















up vote
0
down vote













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 at 4:11










  • Can you post your file?
    – John Camerin
    Nov 20 at 13:22











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1 Answer
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active

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1 Answer
1






active

oldest

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oldest

votes






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oldest

votes








up vote
0
down vote













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 at 4:11










  • Can you post your file?
    – John Camerin
    Nov 20 at 13:22















up vote
0
down vote













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 at 4:11










  • Can you post your file?
    – John Camerin
    Nov 20 at 13:22













up vote
0
down vote










up vote
0
down vote









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 at 4:11









user697911

3,079103365




3,079103365










answered Nov 20 at 3:15









John Camerin

20910




20910












  • 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 at 4:11










  • Can you post your file?
    – John Camerin
    Nov 20 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 at 4:11










  • Can you post your file?
    – John Camerin
    Nov 20 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 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 at 4:11












Can you post your file?
– John Camerin
Nov 20 at 13:22




Can you post your file?
– John Camerin
Nov 20 at 13:22


















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