How to read zip file through getResourceAsStream with relative path?
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
add a comment |
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
File zipFile = new File(zipUrl.toURI());
AFile
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
add a comment |
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
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
java maven zip inputstream
asked Nov 20 '18 at 1:51
user697911user697911
3,208113472
3,208113472
File zipFile = new File(zipUrl.toURI());
AFile
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
add a comment |
File zipFile = new File(zipUrl.toURI());
AFile
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
add a comment |
1 Answer
1
active
oldest
votes
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?
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
add a comment |
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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
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?
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
add a comment |
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?
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
add a comment |
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?
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?
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
add a comment |
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
add a comment |
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File zipFile = new File(zipUrl.toURI());
AFile
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